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Multivariable Calculus, Seventh Edition, James Stewart, Executive Editor: Liz Covello, Assistant Editor: Liza Neustaetter, Editorial Assistant: Jennifer Staller, Media Editor : Maureen Ross, Marketing Manager: Jennifer Jones, Marketing Coordinator: Michael Ledesma, Marketing Communications Manager: Mary Anne Payumo, Content Project Manager: Cheryll Linthicum, Art Director: Vernon T. Boes, Print Buyer: Becky Cross, Rights Acquisitions Specialist: Don Schlotman, Production Service: TECH· arts, , © 2012, 2008 Brooks/Cole, Cengage Learning, ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any, form or by any means graphic, electronic, or mechanical, including, but not limited to photocopying, recording, scanning, digitizing,, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or, 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher., For product information and technology assistance, contact us at, Cengage Learning Customer & Sales Support, 1-800-354-9706., For permission to use material from this text or product,, submit all requests online at www.cengage.com/permissions., Further permissions questions can be e-mailed to, permissionrequest@cengage.com., , Library of Congress Control Number: 2010936601, , Text Designer: TECH· arts, Photo Researcher: Terri Wright, www.terriwright.com, Copy Editor: Kathi Townes, , ISBN-13: 978-0-538-49787-9, ISBN-10: 0-538-49787-4, , Cover Designer: Irene Morris, Cover Illustration: Irene Morris, Compositor: Stephanie Kuhns, TECH· arts, , Brooks/Cole, 20 Davis Drive, Belmont, CA 94002-3098, USA, Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the, United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your, local office at www.cengage.com/global., Cengage Learning products are represented in Canada by Nelson, Education, Ltd., To learn more about Brooks/Cole, visit, www.cengage.com/brookscole., , Trademarks, ExamView ® and ExamViewPro ® are registered trademarks, of FSCreations, Inc., Windows is a registered trademark of the Microsoft Corporation, and used herein under license., Macintosh and Power Macintosh are registered trademarks of, Apple Computer, Inc. Used herein under license., Derive is a registered trademark of Soft Warehouse, Inc., Maple is a registered trademark of Waterloo Maple, Inc., Mathematica is a registered trademark of Wolfram Research, Inc., Tools for Enriching is a trademark used herein under license., , K10T10, , Purchase any of our products at your local college store or at our, preferred online store www.cengagebrain.com., , Printed in the United States of America, 1 2 3 4 5 6 7 1 4 1 3 1 2 11 1 0, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Contents, Preface, , 10, , vii, , Parametric Equations and Polar Coordinates, 10.1, , Curves Defined by Parametric Equations, Laboratory Project, , 10.2, , Polar Coordinates, Laboratory Project, , Bézier Curves, , 677, , 678, N, , Families of Polar Curves, , Areas and Lengths in Polar Coordinates, , 10.5, , Conic Sections, , 10.6, , Conic Sections in Polar Coordinates, , Problems Plus, , 688, , 689, , 694, 702, , 709, 712, , Infinite Sequences and Series, 11.1, , 668, , 669, , 10.4, , Review, , 11, , N, , 660, , Running Circles around Circles, , Calculus with Parametric Curves, Laboratory Project, , 10.3, , N, , 659, , Sequences, , 713, , 714, , Laboratory Project, , N, , Logistic Sequences, , 727, , 11.2, , Series, , 727, , 11.3, , The Integral Test and Estimates of Sums, , 11.4, , The Comparison Tests, , 11.5, , Alternating Series, , 11.6, , Absolute Convergence and the Ratio and Root Tests, , 11.7, , Strategy for Testing Series, , 11.8, , Power Series, , 11.9, , Representations of Functions as Power Series, , 738, , 746, , 751, 756, , 763, , 765, 770, , iii, Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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iv, , CONTENTS, , 11.10, , Taylor and Maclaurin Series, Laboratory Project, Writing Project, , 11.11, , Review, Problems Plus, , 791, , How Newton Discovered the Binomial Series, , N, , 801, , 802, 805, , Vectors and the Geometry of Space, , 809, , 12.1, , Three-Dimensional Coordinate Systems, , 12.2, , Vectors, , 12.3, , The Dot Product, , 12.4, , The Cross Product, , 12.5, , 824, 832, The Geometry of a Tetrahedron, , N, , Equations of Lines and Planes, N, , Problems Plus, , Putting 3D in Perspective, , 851, , 861, , 863, , 13.1, , Vector Functions and Space Curves, , 13.2, , Derivatives and Integrals of Vector Functions, , 13.3, , Arc Length and Curvature, , 13.4, , Motion in Space: Velocity and Acceleration, Applied Project, , Problems Plus, , 850, , 858, , Vector Functions, , Review, , 840, , 840, , Cylinders and Quadric Surfaces, Review, , 810, , 815, , Laboratory Project, , 12.6, , 791, , 792, , Radiation from the Stars, , N, , Discovery Project, , 13, , An Elusive Limit, , Applications of Taylor Polynomials, Applied Project, , 12, , N, , 777, , N, , 864, 871, , 877, , Kepler’s Laws, , 886, , 896, , 897, 900, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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CONTENTS, , 14, , Partial Derivatives, , 901, , 14.1, , Functions of Several Variables, , 14.2, , Limits and Continuity, , 14.3, , Partial Derivatives, , 14.4, , Tangent Planes and Linear Approximations, , 14.5, , The Chain Rule, , 14.6, , Directional Derivatives and the Gradient Vector, , 14.7, , Maximum and Minimum Values, N, , Discovery Project, , 916, 924, , N, , Quadratic Approximations and Critical Points, , Rocket Science, , 988, , Applied Project, , N, , Hydro-Turbine Optimization, , 995, , 997, , Double Integrals over Rectangles, , 15.2, , Iterated Integrals, , 15.3, , Double Integrals over General Regions, , 15.4, , Double Integrals in Polar Coordinates, , 15.5, , Applications of Double Integrals, , 15.6, , Surface Area, , 15.7, , Triple Integrals, , 998, , 1006, 1012, 1021, , 1027, , 1037, 1041, , Discovery Project, , N, , Volumes of Hyperspheres, , 1051, , Triple Integrals in Cylindrical Coordinates 1051, Discovery Project, , N, , The Intersection of Three Cylinders, , Triple Integrals in Spherical Coordinates, Applied Project, , 15.10, , 990, , 991, , 15.1, , 15.9, , 980, , 981, , N, , Multiple Integrals, , 15.8, , 957, , 980, , Applied Project, , Problems Plus, , 939, , 970, , Designing a Dumpster, , Lagrange Multipliers, , Review, , 15, , 902, , 948, , Applied Project, , 14.8, , v, , N, , Roller Derby, , Problems Plus, , 1057, , 1063, , Change of Variables in Multiple Integrals, Review, , 1056, , 1064, , 1073, 1077, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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vi, , CONTENTS, , 16, , Vector Calculus, , 1079, , 16.1, , Vector Fields, , 1080, , 16.2, , Line Integrals, , 1087, , 16.3, , The Fundamental Theorem for Line Integrals, , 16.4, , Green’s Theorem, , 16.5, , Curl and Divergence, , 16.6, , Parametric Surfaces and Their Areas, , 16.7, , Surface Integrals, , 1134, , 16.8, , Stokes’ Theorem, , 1146, , Writing Project, , 1108, 1115, , The Divergence Theorem, , 16.10, , Summary, , Problems Plus, , 1123, , Three Men and Two Theorems, , 16.9, , Review, , 17, , N, , 1099, , 1152, , 1152, , 1159, 1160, 1163, , Second-Order Differential Equations, , 1165, , 17.1, , Second-Order Linear Equations, , 17.2, , Nonhomogeneous Linear Equations, , 17.3, , Applications of Second-Order Differential Equations, , 17.4, , Series Solutions, Review, , Appendixes, , 1166, , 1180, , 1188, , 1193, , A1, , F, , Proofs of Theorems, , G, , Complex Numbers, , H, , Answers to Odd-Numbered Exercises, , Index, , 1172, , A2, A5, A13, , A43, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Preface, A great discovery solves a great problem but there is a grain of discovery in the, solution of any problem. Your problem may be modest; but if it challenges your, curiosity and brings into play your inventive faculties, and if you solve it by your, own means, you may experience the tension and enjoy the triumph of discovery., GEORGE POLYA, , The art of teaching, Mark Van Doren said, is the art of assisting discovery. I have tried to, write a book that assists students in discovering calculus—both for its practical power and, its surprising beauty. In this edition, as in the first six editions, I aim to convey to the student a sense of the utility of calculus and develop technical competence, but I also strive, to give some appreciation for the intrinsic beauty of the subject. Newton undoubtedly, experienced a sense of triumph when he made his great discoveries. I want students to, share some of that excitement., The emphasis is on understanding concepts. I think that nearly everybody agrees that, this should be the primary goal of calculus instruction. In fact, the impetus for the current, calculus reform movement came from the Tulane Conference in 1986, which formulated, as their first recommendation:, Focus on conceptual understanding., I have tried to implement this goal through the Rule of Three: “Topics should be presented, geometrically, numerically, and algebraically.” Visualization, numerical and graphical experimentation, and other approaches have changed how we teach conceptual reasoning in fundamental ways. The Rule of Three has been expanded to become the Rule of Four by, emphasizing the verbal, or descriptive, point of view as well., In writing the seventh edition my premise has been that it is possible to achieve conceptual understanding and still retain the best traditions of traditional calculus. The book, contains elements of reform, but within the context of a traditional curriculum., , Alternative Versions, I have written several other calculus textbooks that might be preferable for some instructors. Most of them also come in single variable and multivariable versions., ■, , Calculus, Seventh Edition, Hybrid Version, is similar to the present textbook in, content and coverage except that all end-of-section exercises are available only in, Enhanced WebAssign. The printed text includes all end-of-chapter review material., , ■, , Calculus: Early Transcendentals, Seventh Edition, is similar to the present textbook, except that the exponential, logarithmic, and inverse trigonometric functions are covered in the first semester., vii, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_FM7eMV_FM7eMV_pi-xiv.qk_97879_FM7eMV_FM7eMV_pi-xiv 11/11/10 10:33 AM Page viii, , viii, , PREFACE, ■, , Calculus: Early Transcendentals, Seventh Edition, Hybrid Version, is similar to Calculus: Early Transcendentals, Seventh Edition, in content and coverage except that all, end-of-section exercises are available only in Enhanced WebAssign. The printed text, includes all end-of-chapter review material., , ■, , Essential Calculus is a much briefer book (800 pages), though it contains almost all, of the topics in Calculus, Seventh Edition. The relative brevity is achieved through, briefer exposition of some topics and putting some features on the website., , ■, , Essential Calculus: Early Transcendentals resembles Essential Calculus, but the, exponential, logarithmic, and inverse trigonometric functions are covered in Chapter 3., , ■, , Calculus: Concepts and Contexts, Fourth Edition, emphasizes conceptual understanding even more strongly than this book. The coverage of topics is not encyclopedic, and the material on transcendental functions and on parametric equations is woven, throughout the book instead of being treated in separate chapters., , ■, , Calculus: Early Vectors introduces vectors and vector functions in the first semester, and integrates them throughout the book. It is suitable for students taking Engineering, and Physics courses concurrently with calculus., , ■, , Brief Applied Calculus is intended for students in business, the social sciences, and, the life sciences., , What’s New in the Seventh Edition?, The changes have resulted from talking with my colleagues and students at the University, of Toronto and from reading journals, as well as suggestions from users and reviewers., Here are some of the many improvements that I’ve incorporated into this edition:, ■, , Some material has been rewritten for greater clarity or for better motivation. See, for, instance, the introduction to series on page 727 and the motivation for the cross product on page 832., , ■, , New examples have been added (see Example 4 on page 1045 for instance), and the, solutions to some of the existing examples have been amplified., , ■, , The art program has been revamped: New figures have been incorporated and a substantial percentage of the existing figures have been redrawn., , ■, , The data in examples and exercises have been updated to be more timely., , ■, , One new project has been added: Families of Polar Curves (page 688) exhibits the, fascinating shapes of polar curves and how they evolve within a family., , ■, , The section on the surface area of the graph of a function of two variables has been, restored as Section 15.6 for the convenience of instructors who like to teach it after, double integrals, though the full treatment of surface area remains in Chapter 16., , ■, , I continue to seek out examples of how calculus applies to so many aspects of the, real world. On page 933 you will see beautiful images of the earth’s magnetic field, strength and its second vertical derivative as calculated from Laplace’s equation. I, thank Roger Watson for bringing to my attention how this is used in geophysics and, mineral exploration., , ■, , More than 25% of the exercises are new. Here are some of my favorites: 11.2.49–50,, 11.10.71–72, 12.1.44, 12.4.43–44, 12.5.80, 14.6.59–60, 15.8.42, and Problems 4, 5,, and 8 on pages 861–62., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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PREFACE, , ix, , Technology Enhancements, ■, , The media and technology to support the text have been enhanced to give professors, greater control over their course, to provide extra help to deal with the varying levels, of student preparedness for the calculus course, and to improve support for conceptual, understanding. New Enhanced WebAssign features including a customizable Cengage, YouBook, Just in Time review, Show Your Work, Answer Evaluator, Personalized, Study Plan, Master Its, solution videos, lecture video clips (with associated questions),, and Visualizing Calculus (TEC animations with associated questions) have been, developed to facilitate improved student learning and flexible classroom teaching., , ■, , Tools for Enriching Calculus (TEC) has been completely redesigned and is accessible, in Enhanced WebAssign, CourseMate, and PowerLecture. Selected Visuals and, Modules are available at www.stewartcalculus.com., , Features, CONCEPTUAL EXERCISES, , The most important way to foster conceptual understanding is through the problems that, we assign. To that end I have devised various types of problems. Some exercise sets begin, with requests to explain the meanings of the basic concepts of the section. (See, for, instance, the first few exercises in Sections 11.2, 14.2, and 14.3.) Similarly, all the review, sections begin with a Concept Check and a True-False Quiz. Other exercises test conceptual understanding through graphs or tables (see Exercises 10.1.24–27, 11.10.2, 13.2.1–2,, 13.3.33–39, 14.1.1–2, 14.1.32–42, 14.3.3–10, 14.6.1–2, 14.7.3–4, 15.1.5–10, 16.1.11–18,, 16.2.17–18, and 16.3.1–2)., Another type of exercise uses verbal description to test conceptual understanding. I particularly value problems that combine and compare graphical, numerical, and algebraic, approaches., , GRADED EXERCISE SETS, , Each exercise set is carefully graded, progressing from basic conceptual exercises and skilldevelopment problems to more challenging problems involving applications and proofs., , REAL-WORLD DATA, , My assistants and I spent a great deal of time looking in libraries, contacting companies and, government agencies, and searching the Internet for interesting real-world data to introduce, motivate, and illustrate the concepts of calculus. As a result, many of the examples, and exercises deal with functions defined by such numerical data or graphs. Functions of, two variables are illustrated by a table of values of the wind-chill index as a function of air, temperature and wind speed (Example 2 in Section 14.1). Partial derivatives are introduced in Section 14.3 by examining a column in a table of values of the heat index (perceived air temperature) as a function of the actual temperature and the relative humidity., This example is pursued further in connection with linear approximations (Example 3 in, Section 14.4). Directional derivatives are introduced in Section 14.6 by using a temperature contour map to estimate the rate of change of temperature at Reno in the direction of, Las Vegas. Double integrals are used to estimate the average snowfall in Colorado on, December 20–21, 2006 (Example 4 in Section 15.1). Vector fields are introduced in Section 16.1 by depictions of actual velocity vector fields showing San Francisco Bay wind, patterns., , PROJECTS, , One way of involving students and making them active learners is to have them work (perhaps in groups) on extended projects that give a feeling of substantial accomplishment, when completed. I have included four kinds of projects: Applied Projects involve applications that are designed to appeal to the imagination of students. The project after Section, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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x, , PREFACE, , 14.8 uses Lagrange multipliers to determine the masses of the three stages of a rocket so, as to minimize the total mass while enabling the rocket to reach a desired velocity. Laboratory Projects involve technology; the one following Section 10.2 shows how to use, Bézier curves to design shapes that represent letters for a laser printer. Discovery Projects, explore aspects of geometry: tetrahedra (after Section 12.4), hyperspheres (after Section, 15.7), and intersections of three cylinders (after Section 15.8). The Writing Project after, Section 17.8 explores the historical and physical origins of Green’s Theorem and Stokes’, Theorem and the interactions of the three men involved. Many additional projects can be, found in the Instructor’s Guide., TOOLS FOR, ENRICHING™ CALCULUS, , TEC is a companion to the text and is intended to enrich and complement its contents. (It, is now accessible in Enhanced WebAssign, CourseMate, and PowerLecture. Selected, Visuals and Modules are available at www.stewartcalculus.com.) Developed by Harvey, Keynes, Dan Clegg, Hubert Hohn, and myself, TEC uses a discovery and exploratory, approach. In sections of the book where technology is particularly appropriate, marginal, icons direct students to TEC modules that provide a laboratory environment in which they, can explore the topic in different ways and at different levels. Visuals are animations of, figures in text; Modules are more elaborate activities and include exercises. Instructors can choose to become involved at several different levels, ranging from simply, encouraging students to use the Visuals and Modules for independent exploration, to, assigning specific exercises from those included with each Module, or to creating additional exercises, labs, and projects that make use of the Visuals and Modules., , HOMEWORK HINTS, , Homework Hints presented in the form of questions try to imitate an effective teaching, assistant by functioning as a silent tutor. Hints for representative exercises (usually oddnumbered) are included in every section of the text, indicated by printing the exercise, number in red. They are constructed so as not to reveal any more of the actual solution than, is minimally necessary to make further progress, and are available to students at, stewartcalculus.com and in CourseMate and Enhanced WebAssign., , ENHANCED W E B A S S I G N, , Technology is having an impact on the way homework is assigned to students, particularly, in large classes. The use of online homework is growing and its appeal depends on ease of, use, grading precision, and reliability. With the seventh edition we have been working with, the calculus community and WebAssign to develop a more robust online homework system. Up to 70% of the exercises in each section are assignable as online homework, including free response, multiple choice, and multi-part formats., The system also includes Active Examples, in which students are guided in step-by-step, tutorials through text examples, with links to the textbook and to video solutions. New, enhancements to the system include a customizable eBook, a Show Your Work feature,, Just in Time review of precalculus prerequisites, an improved Assignment Editor, and an, Answer Evaluator that accepts more mathematically equivalent answers and allows for, homework grading in much the same way that an instructor grades., , www.stewartcalculus.com, , This site includes the following., ■, , Homework Hints, , ■, , Algebra Review, , ■, , Lies My Calculator and Computer Told Me, , ■, , History of Mathematics, with links to the better historical websites, , ■, , Additional Topics (complete with exercise sets): Fourier Series, Formulas for the, Remainder Term in Taylor Series, Rotation of Axes, , ■, , Archived Problems (Drill exercises that appeared in previous editions, together with, their solutions), , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_FM7eMV_FM7eMV_pi-xiv.qk_97879_FM7eMV_FM7eMV_pi-xiv 11/9/10 4:30 PM Page xi, , PREFACE, , ■, , Challenge Problems (some from the Problems Plus sections from prior editions), Links, for particular topics, to outside web resources, , ■, , Selected Tools for Enriching Calculus (TEC) Modules and Visuals, , ■, , xi, , Content, 10 Parametric Equations, and Polar Coordinates, , This chapter introduces parametric and polar curves and applies the methods of calculus, to them. Parametric curves are well suited to laboratory projects; the three presented here, involve families of curves and Bézier curves. A brief treatment of conic sections in polar, coordinates prepares the way for Kepler’s Laws in Chapter 13., , 11 Infinite Sequences and Series, , The convergence tests have intuitive justifications (see page 738) as well as formal proofs., Numerical estimates of sums of series are based on which test was used to prove convergence. The emphasis is on Taylor series and polynomials and their applications to physics., Error estimates include those from graphing devices., , 12 Vectors and, The Geometry of Space, , The material on three-dimensional analytic geometry and vectors is divided into two chapters. Chapter 12 deals with vectors, the dot and cross products, lines, planes, and surfaces., , 13 Vector Functions, , This chapter covers vector-valued functions, their derivatives and integrals, the length and, curvature of space curves, and velocity and acceleration along space curves, culminating, in Kepler’s laws., , 14 Partial Derivatives, , Functions of two or more variables are studied from verbal, numerical, visual, and algebraic points of view. In particular, I introduce partial derivatives by looking at a specific, column in a table of values of the heat index (perceived air temperature) as a function of, the actual temperature and the relative humidity., , 15 Multiple Integrals, , Contour maps and the Midpoint Rule are used to estimate the average snowfall and average, temperature in given regions. Double and triple integrals are used to compute probabilities,, surface areas, and (in projects) volumes of hyperspheres and volumes of intersections of, three cylinders. Cylindrical and spherical coordinates are introduced in the context of evaluating triple integrals., , 16 Vector Calculus, , Vector fields are introduced through pictures of velocity fields showing San Francisco Bay, wind patterns. The similarities among the Fundamental Theorem for line integrals, Green’s, Theorem, Stokes’ Theorem, and the Divergence Theorem are emphasized., , 17 Second-Order, Differential Equations, , Since first-order differential equations are covered in Chapter 9, this final chapter deals, with second-order linear differential equations, their application to vibrating springs and, electric circuits, and series solutions., , Ancillaries, Multivariable Calculus, Seventh Edition, is supported by a complete set of ancillaries, developed under my direction. Each piece has been designed to enhance student understanding and to facilitate creative instruction. With this edition, new media and technologies have been developed that help students to visualize calculus and instructors to, customize content to better align with the way they teach their course. The tables on pages, xiii–xiv describe each of these ancillaries., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_FM7eMV_FM7eMV_pi-xiv.qk_97879_FM7eMV_FM7eMV_pi-xiv 11/9/10 4:30 PM Page xii, , xii, , PREFACE, , 0, , Acknowledgments, , The preparation of this and previous editions has involved much time spent reading the, reasoned (but sometimes contradictory) advice from a large number of astute reviewers., I greatly appreciate the time they spent to understand my motivation for the approach taken., I have learned something from each of them., SEVENTH EDITION REVIEWERS, , Amy Austin, Texas A&M University, Anthony J. Bevelacqua, University of North Dakota, Zhen-Qing Chen, University of Washington—Seattle, Jenna Carpenter, Louisiana Tech University, Le Baron O. Ferguson, University of California—Riverside, Shari Harris, John Wood Community College, Amer Iqbal, University of Washington—Seattle, Akhtar Khan, Rochester Institute of Technology, Marianne Korten, Kansas State University, Joyce Longman, Villanova University, , Richard Millspaugh, University of North Dakota, Lon H. Mitchell, Virginia Commonwealth University, Ho Kuen Ng, San Jose State University, Norma Ortiz-Robinson, Virginia Commonwealth University, Qin Sheng, Baylor University, Magdalena Toda, Texas Tech University, Ruth Trygstad, Salt Lake Community College, Klaus Volpert, Villanova University, Peiyong Wang, Wayne State University, , In addition, I would like to thank Jordan Bell, George Bergman, Leon Gerber, Mary, Pugh, and Simon Smith for their suggestions; Al Shenk and Dennis Zill for permission to, use exercises from their calculus texts; COMAP for permission to use project material;, George Bergman, David Bleecker, Dan Clegg, Victor Kaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz, Paul Sally, Lowell Smylie, and Larry Wallen for ideas for exercises;, Dan Drucker for the roller derby project; Thomas Banchoff, Tom Farmer, Fred Gass, John, Ramsay, Larry Riddle, Philip Straffin, and Klaus Volpert for ideas for projects; Dan Anderson, Dan Clegg, Jeff Cole, Dan Drucker, and Barbara Frank for solving the new exercises, and suggesting ways to improve them; Marv Riedesel and Mary Johnson for accuracy in, proofreading; and Jeff Cole and Dan Clegg for their careful preparation and proofreading, of the answer manuscript., In addition, I thank those who have contributed to past editions: Ed Barbeau, Fred, Brauer, Andy Bulman-Fleming, Bob Burton, David Cusick, Tom DiCiccio, Garret Etgen,, Chris Fisher, Stuart Goldenberg, Arnold Good, Gene Hecht, Harvey Keynes, E.L. Koh,, Zdislav Kovarik, Kevin Kreider, Emile LeBlanc, David Leep, Gerald Leibowitz, Larry, Peterson, Lothar Redlin, Carl Riehm, John Ringland, Peter Rosenthal, Doug Shaw, Dan, Silver, Norton Starr, Saleem Watson, Alan Weinstein, and Gail Wolkowicz., I also thank Kathi Townes and Stephanie Kuhns of TECHarts for their production services and the following Brooks/Cole staff: Cheryll Linthicum, content project manager;, Liza Neustaetter, assistant editor; Maureen Ross, media editor; Sam Subity, managing, media editor; Jennifer Jones, marketing manager; and Vernon Boes, art director. They have, all done an outstanding job., I have been very fortunate to have worked with some of the best mathematics editors, in the business over the past three decades: Ron Munro, Harry Campbell, Craig Barth,, Jeremy Hayhurst, Gary Ostedt, Bob Pirtle, Richard Stratton, and now Liz Covello. All of, them have contributed greatly to the success of this book., JAMES STEWART, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Ancillaries for Instructors, PowerLecture, ISBN 0-8400-5414-9, , This comprehensive DVD contains all art from the text in both, jpeg and PowerPoint formats, key equations and tables from the, text, complete pre-built PowerPoint lectures, an electronic version of the Instructor’s Guide, Solution Builder, ExamView testing software, Tools for Enriching Calculus, video instruction,, and JoinIn on TurningPoint clicker content., Instructor’s Guide, by Douglas Shaw, ISBN 0-8400-5407-6, , Each section of the text is discussed from several viewpoints., The Instructor’s Guide contains suggested time to allot, points, to stress, text discussion topics, core materials for lecture, workshop/discussion suggestions, group work exercises in a form, suitable for handout, and suggested homework assignments. An, electronic version of the Instructor’s Guide is available on the, PowerLecture DVD., Complete Solutions Manual, Multivariable, By Dan Clegg and Barbara Frank, ISBN 0-8400-4947-1, , Includes worked-out solutions to all exercises in the text., Solution Builder, www.cengage.com /solutionbuilder, This online instructor database offers complete worked out solutions to all exercises in the text. Solution Builder allows you to, create customized, secure solutions printouts (in PDF format), matched exactly to the problems you assign in class., Printed Test Bank, By William Steven Harmon, ISBN 0-8400-5408-4, , Contains text-specific multiple-choice and free response test, items., ExamView Testing, Create, deliver, and customize tests in print and online formats, with ExamView, an easy-to-use assessment and tutorial software., ExamView contains hundreds of multiple-choice and free, response test items. ExamView testing is available on the PowerLecture DVD., , ■ Electronic items, , ■ Printed items, , Ancillaries for Instructors and Students, Stewart Website, www.stewartcalculus.com, Contents: Homework Hints ■ Algebra Review ■ Additional, Topics ■ Drill exercises ■ Challenge Problems ■ Web Links ■, History of Mathematics ■ Tools for Enriching Calculus (TEC), TEC Tools for Enriching™ Calculus, By James Stewart, Harvey Keynes, Dan Clegg, and, developer Hu Hohn, Tools for Enriching Calculus (TEC) functions as both a powerful tool for instructors, as well as a tutorial environment in, which students can explore and review selected topics. The, Flash simulation modules in TEC include instructions, written and audio explanations of the concepts, and exercises., TEC is accessible in CourseMate, WebAssign, and PowerLecture. Selected Visuals and Modules are available at, www.stewartcalculus.com., Enhanced WebAssign, www.webassign.net, WebAssign’s homework delivery system lets instructors deliver,, collect, grade, and record assignments via the web. Enhanced, WebAssign for Stewart’s Calculus now includes opportunities, for students to review prerequisite skills and content both at the, start of the course and at the beginning of each section. In addition, for selected problems, students can get extra help in the, form of “enhanced feedback” (rejoinders) and video solutions., Other key features include: thousands of problems from Stewart’s Calculus, a customizable Cengage YouBook, Personal, Study Plans, Show Your Work, Just in Time Review, Answer, Evaluator, Visualizing Calculus animations and modules,, quizzes, lecture videos (with associated questions), and more!, Cengage Customizable YouBook, YouBook is a Flash-based eBook that is interactive and customizable! Containing all the content from Stewart’s Calculus,, YouBook features a text edit tool that allows instructors to modify the textbook narrative as needed. With YouBook, instructors, can quickly re-order entire sections and chapters or hide any, content they don’t teach to create an eBook that perfectly, matches their syllabus. Instructors can further customize the, text by adding instructor-created or YouTube video links., Additional media assets include: animated figures, video clips,, highlighting, notes, and more! YouBook is available in, Enhanced WebAssign., , (Table continues on page xiv.), , xiii, Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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14, , Partial Derivatives, , Graphs of functions of two variables are, surfaces that can take a variety of, shapes, including that of a saddle or, mountain pass. At this location in, southern Utah (Phipps Arch) you can, see a point that is a minimum in one, direction but a maximum in another, direction. Such surfaces are discussed, in Section 14.7., , Photo by Stan Wagon, Macalester College, , So far we have dealt with the calculus of functions of a single variable. But, in the real world, physical, quantities often depend on two or more variables, so in this chapter we turn our attention to functions of, several variables and extend the basic ideas of differential calculus to such functions., , 901, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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902, , CHAPTER 14, , PARTIAL DERIVATIVES, , Functions of Several Variables, , 14.1, , In this section we study functions of two or more variables from four points of view:, ■ verbally, (by a description in words), ■, , numerically, , (by a table of values), , ■, , algebraically, , (by an explicit formula), , ■, , visually, , (by a graph or level curves), , Functions of Two Variables, The temperature T at a point on the surface of the earth at any given time depends on the, longitude x and latitude y of the point. We can think of T as being a function of the two variables x and y, or as a function of the pair 共x, y兲. We indicate this functional dependence by, writing T 苷 f 共x, y兲., The volume V of a circular cylinder depends on its radius r and its height h. In fact, we, know that V 苷 ␲ r 2h. We say that V is a function of r and h, and we write V共r, h兲 苷 ␲ r 2h., Definition A function f of two variables is a rule that assigns to each ordered pair, , of real numbers 共x, y兲 in a set D a unique real number denoted by f 共x, y兲. The set, D is the domain of f and its range is the set of values that f takes on, that is,, 兵 f 共x, y兲 共x, y兲 僆 D其., , ⱍ, , z, , y, , f(x, y), (x, y), 0, , D, , FIGURE 1, , x, (a, b), , 0, f(a, b), , We often write z 苷 f 共x, y兲 to make explicit the value taken on by f at the general point, 共x, y兲. The variables x and y are independent variables and z is the dependent variable., [Compare this with the notation y 苷 f 共x兲 for functions of a single variable.], A function of two variables is just a function whose domain is a subset of ⺢2 and whose, range is a subset of ⺢. One way of visualizing such a function is by means of an arrow diagram (see Figure 1), where the domain D is represented as a subset of the xy-plane and the, range is a set of numbers on a real line, shown as a z-axis. For instance, if f 共x, y兲 represents, the temperature at a point 共x, y兲 in a flat metal plate with the shape of D, we can think of the, z-axis as a thermometer displaying the recorded temperatures., If a function f is given by a formula and no domain is specified, then the domain of f is, understood to be the set of all pairs 共x, y兲 for which the given expression is a well-defined, real number., EXAMPLE 1 For each of the following functions, evaluate f 共3, 2兲 and find and sketch the, , domain., (a) f 共x, y兲 苷, , sx ⫹ y ⫹ 1, x⫺1, , (b) f 共x, y兲 苷 x ln共y 2 ⫺ x兲, , SOLUTION, , (a), , f 共3, 2兲 苷, , s3 ⫹ 2 ⫹ 1, s6, 苷, 3⫺1, 2, , The expression for f makes sense if the denominator is not 0 and the quantity under the, square root sign is nonnegative. So the domain of f is, D 苷 兵共x, y兲, , ⱍ x ⫹ y ⫹ 1 艌 0,, , x 苷 1其, , The inequality x ⫹ y ⫹ 1 艌 0, or y 艌 ⫺x ⫺ 1, describes the points that lie on or above, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 903, , SECTION 14.1, , y, , x=1, , 0, , x, , Since ln共 y 2 ⫺ x兲 is defined only when y 2 ⫺ x ⬎ 0, that is, x ⬍ y 2, the domain of f is, D 苷 兵共x, y兲 x ⬍ y 2 其. This is the set of points to the left of the parabola x 苷 y 2. (See, Figure 3.), , ⱍ, , Not all functions can be represented by explicit formulas. The function in the next example is described verbally and by numerical estimates of its values., , FIGURE 2, , œ„„„„„„„, x+y+1, x-1, , EXAMPLE 2 In regions with severe winter weather, the wind-chill index is often used to, describe the apparent severity of the cold. This index W is a subjective temperature that, depends on the actual temperature T and the wind speed v. So W is a function of T and v,, and we can write W 苷 f 共T, v兲. Table 1 records values of W compiled by the National, Weather Service of the US and the Meteorological Service of Canada., , y, , x=¥, 0, , f 共3, 2兲 苷 3 ln共2 2 ⫺ 3兲 苷 3 ln 1 苷 0, , (b), , _1, , Domain of f(x, y)=, , 903, , the line y 苷 ⫺x ⫺ 1, while x 苷 1 means that the points on the line x 苷 1 must be, excluded from the domain. (See Figure 2.), , x+y+1=0, , _1, , FUNCTIONS OF SEVERAL VARIABLES, , x, , TABLE 1 Wind-chill index as a function of air temperature and wind speed, , Wind speed (km/h), T, FIGURE 3, , The New Wind-Chill Index, A new wind-chill index was introduced in, November of 2001 and is more accurate than, the old index for measuring how cold it feels, when it’s windy. The new index is based on a, model of how fast a human face loses heat. It, was developed through clinical trials in which, volunteers were exposed to a variety of temperatures and wind speeds in a refrigerated wind, tunnel., , Actual temperature (°C), , Domain of f(x, y)=x ln(¥-x), , v, , 10, , 5, , 15, , 20, , 25, , 30, , 40, , 50, , 60, , 70, , 80, , 5, , 4, , 3, , 2, , 1, , 1, , 0, , ⫺1, , ⫺1, , ⫺2, , ⫺2, , ⫺3, , 0, , ⫺2, , ⫺3, , ⫺4, , ⫺5, , ⫺6, , ⫺6, , ⫺7, , ⫺8, , ⫺9, , ⫺9, , ⫺10, , ⫺5, , ⫺7, , ⫺9, , ⫺11, , ⫺12, , ⫺12, , ⫺13, , ⫺14, , ⫺15, , ⫺16, , ⫺16, , ⫺17, , ⫺10, , ⫺13, , ⫺15, , ⫺17, , ⫺18, , ⫺19, , ⫺20, , ⫺21, , ⫺22, , ⫺23, , ⫺23, , ⫺24, , ⫺15, , ⫺19, , ⫺21, , ⫺23, , ⫺24, , ⫺25, , ⫺26, , ⫺27, , ⫺29, , ⫺30, , ⫺30, , ⫺31, , ⫺20, , ⫺24, , ⫺27, , ⫺29, , ⫺30, , ⫺32, , ⫺33, , ⫺34, , ⫺35, , ⫺36, , ⫺37, , ⫺38, , ⫺25, , ⫺30, , ⫺33, , ⫺35, , ⫺37, , ⫺38, , ⫺39, , ⫺41, , ⫺42, , ⫺43, , ⫺44, , ⫺45, , ⫺30, , ⫺36, , ⫺39, , ⫺41, , ⫺43, , ⫺44, , ⫺46, , ⫺48, , ⫺49, , ⫺50, , ⫺51, , ⫺52, , ⫺35, , ⫺41, , ⫺45, , ⫺48, , ⫺49, , ⫺51, , ⫺52, , ⫺54, , ⫺56, , ⫺57, , ⫺58, , ⫺60, , ⫺40, , ⫺47, , ⫺51, , ⫺54, , ⫺56, , ⫺57, , ⫺59, , ⫺61, , ⫺63, , ⫺64, , ⫺65, , ⫺67, , For instance, the table shows that if the temperature is ⫺5⬚C and the wind speed is, 50 km兾h, then subjectively it would feel as cold as a temperature of about ⫺15⬚C with, no wind. So, f 共⫺5, 50兲 苷 ⫺15, EXAMPLE 3 In 1928 Charles Cobb and Paul Douglas published a study in which they, modeled the growth of the American economy during the period 1899–1922. They considered a simplified view of the economy in which production output is determined by the, amount of labor involved and the amount of capital invested. While there are many other, factors affecting economic performance, their model proved to be remarkably accurate., The function they used to model production was of the form, , 1, , P共L, K 兲 苷 bL␣K 1⫺␣, , where P is the total production (the monetary value of all goods produced in a year),, L is the amount of labor (the total number of person-hours worked in a year), and K is, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 904, , 904, , PARTIAL DERIVATIVES, , CHAPTER 14, , the amount of capital invested (the monetary worth of all machinery, equipment, and, buildings). In Section 14.3 we will show how the form of Equation 1 follows from certain economic assumptions., Cobb and Douglas used economic data published by the government to obtain, Table 2. They took the year 1899 as a baseline and P, L, and K for 1899 were each, assigned the value 100. The values for other years were expressed as percentages of, the 1899 figures., Cobb and Douglas used the method of least squares to fit the data of Table 2 to the, function, , TABLE 2, ., , Year, , P, , L, , K, , 1899, 1900, 1901, 1902, 1903, 1904, 1905, 1906, 1907, 1908, 1909, 1910, 1911, 1912, 1913, 1914, 1915, 1916, 1917, 1918, 1919, 1920, 1921, 1922, , 100, 101, 112, 122, 124, 122, 143, 152, 151, 126, 155, 159, 153, 177, 184, 169, 189, 225, 227, 223, 218, 231, 179, 240, , 100, 105, 110, 117, 122, 121, 125, 134, 140, 123, 143, 147, 148, 155, 156, 152, 156, 183, 198, 201, 196, 194, 146, 161, , 100, 107, 114, 122, 131, 138, 149, 163, 176, 185, 198, 208, 216, 226, 236, 244, 266, 298, 335, 366, 387, 407, 417, 431, , P共L, K兲 苷 1.01L0.75K 0.25, , 2, , (See Exercise 79 for the details.), If we use the model given by the function in Equation 2 to compute the production in, the years 1910 and 1920, we get the values, P共147, 208兲 苷 1.01共147兲0.75共208兲0.25 ⬇ 161.9, P共194, 407兲 苷 1.01共194兲0.75共407兲0.25 ⬇ 235.8, which are quite close to the actual values, 159 and 231., The production function 1 has subsequently been used in many settings, ranging, from individual firms to global economics. It has become known as the Cobb-Douglas, production function. Its domain is 兵共L, K兲 L 艌 0, K 艌 0其 because L and K represent, labor and capital and are therefore never negative., , ⱍ, , EXAMPLE 4 Find the domain and range of t共x, y兲 苷 s9 ⫺ x 2 ⫺ y 2 ., SOLUTION The domain of t is, , y, , D 苷 兵共x, y兲, , ≈+¥=9, , ⱍ 9⫺x, , 2, , ⫺ y 2 艌 0其 苷 兵共x, y兲, , ⱍx, , 2, , ⫹ y 2 艋 9其, , which is the disk with center 共0, 0兲 and radius 3. (See Figure 4.) The range of t is, _3, , 3, , 兵 z ⱍ z 苷 s9 ⫺ x 2 ⫺ y 2 , 共x, y兲 僆 D其, , x, , Since z is a positive square root, z 艌 0. Also, because 9 ⫺ x 2 ⫺ y 2 艋 9, we have, s9 ⫺ x 2 ⫺ y 2 艋 3, So the range is, , FIGURE 4, , 兵z, , 9-≈-¥, Domain of g(x, y)=œ„„„„„„„„„, , ⱍ 0 艋 z 艋 3其 苷 关0, 3兴, , Graphs, , z, , S, , { x, y, f (x, y)}, , Another way of visualizing the behavior of a function of two variables is to consider its, graph., Definition If f is a function of two variables with domain D, then the graph of f, , is the set of all points 共x, y, z兲 in ⺢3 such that z 苷 f 共x, y兲 and 共x, y兲 is in D., , f(x, y), 0, , D, x, , FIGURE 5, , (x, y, 0), , y, , Just as the graph of a function f of one variable is a curve C with equation y 苷 f 共x兲, so, the graph of a function f of two variables is a surface S with equation z 苷 f 共x, y兲. We can, visualize the graph S of f as lying directly above or below its domain D in the xy-plane (see, Figure 5)., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 905, , SECTION 14.1, z, , FUNCTIONS OF SEVERAL VARIABLES, , 905, , EXAMPLE 5 Sketch the graph of the function f 共x, y兲 苷 6 ⫺ 3x ⫺ 2y., , (0, 0, 6), , SOLUTION The graph of f has the equation z 苷 6 ⫺ 3x ⫺ 2y, or 3x ⫹ 2y ⫹ z 苷 6,, , which represents a plane. To graph the plane we first find the intercepts. Putting, y 苷 z 苷 0 in the equation, we get x 苷 2 as the x-intercept. Similarly, the y-intercept is 3, and the z-intercept is 6. This helps us sketch the portion of the graph that lies in the first, octant in Figure 6., , (0, 3, 0), (2, 0, 0), , y, , The function in Example 5 is a special case of the function, , x, , f 共x, y兲 苷 ax ⫹ by ⫹ c, , FIGURE 6, , which is called a linear function. The graph of such a function has the equation, z 苷 ax ⫹ by ⫹ c, , or, , ax ⫹ by ⫺ z ⫹ c 苷 0, , so it is a plane. In much the same way that linear functions of one variable are important in, single-variable calculus, we will see that linear functions of two variables play a central, role in multivariable calculus., z, , 0, (3, 0, 0), , v, , (0, 0, 3), , EXAMPLE 6 Sketch the graph of t共x, y兲 苷 s9 ⫺ x 2 ⫺ y 2 ., , SOLUTION The graph has equation z 苷 s9 ⫺ x 2 ⫺ y 2 . We square both sides of this, , equation to obtain z 2 苷 9 ⫺ x 2 ⫺ y 2, or x 2 ⫹ y 2 ⫹ z 2 苷 9, which we recognize as an, equation of the sphere with center the origin and radius 3. But, since z 艌 0, the graph of, t is just the top half of this sphere (see Figure 7)., , (0, 3, 0), y, , x, , FIGURE 7, , Graph of g(x, y)=œ„„„„„„„„„, 9-≈-¥, , NOTE An entire sphere can’t be represented by a single function of x and y. As we saw, in Example 6, the upper hemisphere of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 9 is represented by the, function t共x, y兲 苷 s9 ⫺ x 2 ⫺ y 2 . The lower hemisphere is represented by the function, h共x, y兲 苷 ⫺s9 ⫺ x 2 ⫺ y 2 ., , EXAMPLE 7 Use a computer to draw the graph of the Cobb-Douglas production function, P共L, K兲 苷 1.01L0.75K 0.25., SOLUTION Figure 8 shows the graph of P for values of the labor L and capital K that lie, , between 0 and 300. The computer has drawn the surface by plotting vertical traces. We, see from these traces that the value of the production P increases as either L or K, increases, as is to be expected., , 300, 200, P, 100, 0, 300, , FIGURE 8, , v, , 200, 100, K, , 0 0, , 100, , 200, , 300, , L, , EXAMPLE 8 Find the domain and range and sketch the graph of h共x, y兲 苷 4x 2 ⫹ y 2., , SOLUTION Notice that h共x, y兲 is defined for all possible ordered pairs of real numbers, , 共x, y兲, so the domain is ⺢2, the entire xy-plane. The range of h is the set 关0, ⬁兲 of all nonnegative real numbers. [Notice that x 2 艌 0 and y 2 艌 0, so h共x, y兲 艌 0 for all x and y.], , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 906, , 906, , CHAPTER 14, , PARTIAL DERIVATIVES, , The graph of h has the equation z 苷 4x 2 ⫹ y 2, which is the elliptic paraboloid that, we sketched in Example 4 in Section 12.6. Horizontal traces are ellipses and vertical, traces are parabolas (see Figure 9)., z, , FIGURE 9, , x, , Graph of h(x, y)=4≈+¥, , y, , Computer programs are readily available for graphing functions of two variables. In most, such programs, traces in the vertical planes x 苷 k and y 苷 k are drawn for equally spaced, values of k and parts of the graph are eliminated using hidden line removal., Figure 10 shows computer-generated graphs of several functions. Notice that we get an, especially good picture of a function when rotation is used to give views from different, z, , z, , x, y, , x, , (b) f(x, y)=(≈+3¥)e _≈_¥, , (a) f(x, y)=(≈+3¥)e _≈_¥, z, , z, , x, , y, , x, , (c) f(x, y)=sin x+sin y, , y, , (d) f(x, y)=, , sin x sin y, xy, , FIGURE 10, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 907, , SECTION 14.1, , FUNCTIONS OF SEVERAL VARIABLES, , 907, , vantage points. In parts (a) and (b) the graph of f is very flat and close to the xy-plane except, near the origin; this is because e⫺x ⫺ y is very small when x or y is large., 2, , 2, , Level Curves, So far we have two methods for visualizing functions: arrow diagrams and graphs. A third, method, borrowed from mapmakers, is a contour map on which points of constant elevation, are joined to form contour lines, or level curves., , Definition The level curves of a function f of two variables are the curves with, , equations f 共x, y兲 苷 k, where k is a constant (in the range of f )., , A level curve f 共x, y兲 苷 k is the set of all points in the domain of f at which f takes on, a given value k. In other words, it shows where the graph of f has height k., You can see from Figure 11 the relation between level curves and horizontal traces. The, level curves f 共x, y兲 苷 k are just the traces of the graph of f in the horizontal plane, z 苷 k projected down to the xy-plane. So if you draw the level curves of a function and, visualize them being lifted up to the surface at the indicated height, then you can mentally, piece together a picture of the graph. The surface is steep where the level curves are close, together. It is somewhat flatter where they are farther apart., z, 40, , 45, , 00, 45, 00, 50, , 00, , LONESOME MTN., , 0, , A, 55, 00, , B, y, 50, , x, , TEC Visual 14.1A animates Figure 11 by, showing level curves being lifted up to graphs, of functions., , 0, , FIGURE 11, , 450, , f(x, y)=20, , 00, , k=45, k=40, k=35, k=30, k=25, k=20, , Lon, , eso, , me, , Cree, , k, , FIGURE 12, , One common example of level curves occurs in topographic maps of mountainous, regions, such as the map in Figure 12. The level curves are curves of constant elevation, above sea level. If you walk along one of these contour lines, you neither ascend nor descend., Another common example is the temperature function introduced in the opening paragraph, of this section. Here the level curves are called isothermals and join locations with the same, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 908, , 908, , CHAPTER 14, , PARTIAL DERIVATIVES, , temperature. Figure 13 shows a weather map of the world indicating the average January, temperatures. The isothermals are the curves that separate the colored bands., , FIGURE 13, , World mean sea-level temperatures, in January in degrees Celsius, From Atmosphere: Introduction to Meteorology, 4th Edition, 1989., © 1989 Pearson Education, Inc., , y, , EXAMPLE 9 A contour map for a function f is shown in Figure 14. Use it to estimate the, values of f 共1, 3兲 and f 共4, 5兲., , 50, , 5, , SOLUTION The point (1, 3) lies partway between the level curves with z-values 70, , 4, , and 80. We estimate that, , 3, , f 共1, 3兲 ⬇ 73, , 2, 1, 0, , 1, , 80, 70, 60, , 50, , 2, , 3, , 80, 70, 60, 4, , f 共4, 5兲 ⬇ 56, , Similarly, we estimate that, , 5, , x, , FIGURE 14, , EXAMPLE 10 Sketch the level curves of the function f 共x, y兲 苷 6 ⫺ 3x ⫺ 2y for the, values k 苷 ⫺6, 0, 6, 12., SOLUTION The level curves are, , 6 ⫺ 3x ⫺ 2y 苷 k, , 3x ⫹ 2y ⫹ 共k ⫺ 6兲 苷 0, , or, , This is a family of lines with slope ⫺ 2 . The four particular level curves with, k 苷 ⫺6, 0, 6, and 12 are 3x ⫹ 2y ⫺ 12 苷 0, 3x ⫹ 2y ⫺ 6 苷 0, 3x ⫹ 2y 苷 0, and, 3x ⫹ 2y ⫹ 6 苷 0. They are sketched in Figure 15. The level curves are equally spaced, parallel lines because the graph of f is a plane (see Figure 6)., 3, , y, , 0, , _6, k=, , 0, k=, , 6, k=, , 12, k=, , FIGURE 15, , Contour map of, f(x, y)=6-3x-2y, , x, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p901-909.qk_97817_14_ch14_p901-909 11/8/10 1:26 PM Page 909, , FUNCTIONS OF SEVERAL VARIABLES, , SECTION 14.1, , v, , 909, , EXAMPLE 11 Sketch the level curves of the function, , t共x, y兲 苷 s9 ⫺ x 2 ⫺ y 2, , k 苷 0, 1, 2, 3, , for, , SOLUTION The level curves are, , s9 ⫺ x 2 ⫺ y 2 苷 k, , x2 ⫹ y2 苷 9 ⫺ k2, , or, , This is a family of concentric circles with center 共0, 0兲 and radius s9 ⫺ k 2 . The cases, k 苷 0, 1, 2, 3 are shown in Figure 16. Try to visualize these level curves lifted up to, form a surface and compare with the graph of t (a hemisphere) in Figure 7. (See TEC, Visual 14.1A.), y, , k=3, k=2, k=1, k=0, (3, 0), , 0, , x, , FIGURE 16, , Contour map of g(x, y)=œ„„„„„„„„„, 9-≈-¥, EXAMPLE 12 Sketch some level curves of the function h共x, y兲 苷 4x 2 ⫹ y 2 ⫹ 1., SOLUTION The level curves are, , 4x 2 ⫹ y 2 ⫹ 1 苷 k, , or, , 1, 4, , x2, y2, ⫹, 苷1, 共k ⫺ 1兲, k⫺1, , which, for k ⬎ 1, describes a family of ellipses with semiaxes 12 sk ⫺ 1 and sk ⫺ 1 ., Figure 17(a) shows a contour map of h drawn by a computer. Figure 17(b) shows these, level curves lifted up to the graph of h (an elliptic paraboloid) where they become horizontal traces. We see from Figure 17 how the graph of h is put together from the level, curves., y, z, , TEC Visual 14.1B demonstrates the, connection between surfaces and their, contour maps., , x, , x, , FIGURE 17, , The graph of h(x, y)=4≈+¥+1, is formed by lifting the level curves., , y, , (a) Contour map, , (b) Horizontal traces are raised level curves, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 910, , 910, , CHAPTER 14, , PARTIAL DERIVATIVES, , K, , EXAMPLE 13 Plot level curves for the Cobb-Douglas production function of Example 3., , 300, , SOLUTION In Figure 18 we use a computer to draw a contour plot for the Cobb-, , Douglas production function, P共L, K兲 苷 1.01L 0.75K 0.25, , 200, , Level curves are labeled with the value of the production P. For instance, the level curve, labeled 140 shows all values of the labor L and capital investment K that result in a production of P 苷 140. We see that, for a fixed value of P, as L increases K decreases, and, vice versa., , 220, 180, , 100, , 140, 100, , 100, , 200, , 300 L, , FIGURE 18, , For some purposes, a contour map is more useful than a graph. That is certainly true in, Example 13. (Compare Figure 18 with Figure 8.) It is also true in estimating function values, as in Example 9., Figure 19 shows some computer-generated level curves together with the corresponding, computer-generated graphs. Notice that the level curves in part (c) crowd together near the, origin. That corresponds to the fact that the graph in part (d) is very steep near the origin., z, , y, , z, , x, x, , y, , (a) Level curves of f(x, y)=_xye_≈_¥, , (b) Two views of f(x, y)=_xye_≈_¥, , z, , y, , x, , y, x, , FIGURE 19, , (c) Level curves of f(x, y)=, , _3y, ≈+¥+1, , (d) f(x, y)=, , _3y, ≈+¥+1, , Functions of Three or More Variables, A function of three variables, f , is a rule that assigns to each ordered triple 共x, y, z兲 in a, domain D 傺 ⺢ 3 a unique real number denoted by f 共x, y, z兲. For instance, the temperature, T at a point on the surface of the earth depends on the longitude x and latitude y of the point, and on the time t, so we could write T 苷 f 共x, y, t兲., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 911, , SECTION 14.1, , FUNCTIONS OF SEVERAL VARIABLES, , 911, , EXAMPLE 14 Find the domain of f if, , f 共x, y, z兲 苷 ln共z  y兲  xy sin z, SOLUTION The expression for f 共x, y, z兲 is defined as long as z  y  0, so the domain of, , f is, D 苷 兵共x, y, z兲 僆 ⺢ 3, , ⱍ, , z  y其, , This is a half-space consisting of all points that lie above the plane z 苷 y., It’s very difficult to visualize a function f of three variables by its graph, since that would, lie in a four-dimensional space. However, we do gain some insight into f by examining its, level surfaces, which are the surfaces with equations f 共x, y, z兲 苷 k, where k is a constant., If the point 共x, y, z兲 moves along a level surface, the value of f 共x, y, z兲 remains fixed., z, , ≈+¥+z@=9, , EXAMPLE 15 Find the level surfaces of the function, , ≈+¥+z@=4, , f 共x, y, z兲 苷 x 2  y 2  z 2, SOLUTION The level surfaces are x 2  y 2  z 2 苷 k, where k  0. These form a family, , of concentric spheres with radius sk . (See Figure 20.) Thus, as 共x, y, z兲 varies over any, sphere with center O, the value of f 共x, y, z兲 remains fixed., , y, x, , ≈+¥+z@=1, FIGURE 20, , Functions of any number of variables can be considered. A function of n variables is a, rule that assigns a number z 苷 f 共x 1, x 2 , . . . , x n 兲 to an n-tuple 共x 1, x 2 , . . . , x n 兲 of real numbers. We denote by ⺢ n the set of all such n-tuples. For example, if a company uses n different, ingredients in making a food product, ci is the cost per unit of the ith ingredient, and x i units, of the ith ingredient are used, then the total cost C of the ingredients is a function of the n, variables x 1, x 2 , . . . , x n :, 3, , C 苷 f 共x 1, x 2 , . . . , x n 兲 苷 c1 x 1  c2 x 2      cn x n, , The function f is a real-valued function whose domain is a subset of ⺢ n. Sometimes we, will use vector notation to write such functions more compactly: If x 苷 具x 1, x 2 , . . . , x n 典 , we, often write f 共x兲 in place of f 共x 1, x 2 , . . . , x n 兲. With this notation we can rewrite the function, defined in Equation 3 as, f 共x兲 苷 c ⴢ x, where c 苷 具 c1, c2 , . . . , cn 典 and c ⴢ x denotes the dot product of the vectors c and x in Vn ., In view of the one-to-one correspondence between points 共x 1, x 2 , . . . , x n兲 in ⺢ n and their, position vectors x 苷 具x 1, x 2 , . . . , x n 典 in Vn , we have three ways of looking at a function f, defined on a subset of ⺢ n :, 1. As a function of n real variables x 1, x 2 , . . . , x n, 2. As a function of a single point variable 共x 1, x 2 , . . . , x n 兲, 3. As a function of a single vector variable x 苷 具x 1, x 2 , . . . , x n 典, , We will see that all three points of view are useful., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 912, , 912, , PARTIAL DERIVATIVES, , CHAPTER 14, , 14.1, , Exercises, , 1. In Example 2 we considered the function W 苷 f 共T, v兲, where, W is the wind-chill index, T is the actual temperature, and v is, , the wind speed. A numerical representation is given in Table 1., (a) What is the value of f 共15, 40兲? What is its meaning?, (b) Describe in words the meaning of the question “For what, value of v is f 共20, v兲 苷 30 ?” Then answer the question., (c) Describe in words the meaning of the question “For what, value of T is f 共T, 20兲 苷 49 ?” Then answer the question., (d) What is the meaning of the function W 苷 f 共5, v兲?, Describe the behavior of this function., (e) What is the meaning of the function W 苷 f 共T, 50兲?, Describe the behavior of this function., , discussed in Example 3 that the production will be doubled, if both the amount of labor and the amount of capital are, doubled. Determine whether this is also true for the general, production function, P共L, K 兲 苷 bLK 1, 5. A model for the surface area of a human body is given by the, , function, S 苷 f 共w, h兲 苷 0.1091w 0.425h 0.725, where w is the weight (in pounds), h is the height (in inches),, and S is measured in square feet., (a) Find f 共160, 70兲 and interpret it., (b) What is your own surface area?, , 2. The temperature-humidity index I (or humidex, for short) is the, , perceived air temperature when the actual temperature is T and, the relative humidity is h, so we can write I 苷 f 共T, h兲. The following table of values of I is an excerpt from a table compiled, by the National Oceanic & Atmospheric Administration., TABLE 3, , 6. The wind-chill index W discussed in Example 2 has been, , modeled by the following function:, W共T, v兲 苷 13.12  0.6215T  11.37v 0.16  0.3965Tv 0.16, , Apparent temperature as a function, of temperature and humidity, , Check to see how closely this model agrees with the values in, Table 1 for a few values of T and v., , Actual temperature (°F), , Relative humidity (%), h, , 20, , 30, , 40, , 50, , 60, , 70, , 80, , 77, , 78, , 79, , 81, , 82, , 83, , 85, , 82, , 84, , 86, , 88, , 90, , 93, , 90, , 87, , 90, , 93, , 96, , 100, , 106, , 95, , 93, , 96, , 101, , 107, , 114, , 124, , 100, , 99, , 104, , 110, , 120, , 132, , 144, , T, , 7. The wave heights h in the open sea depend on the speed v, , of the wind and the length of time t that the wind has been, blowing at that speed. Values of the function h 苷 f 共v, t兲 are, recorded in feet in Table 4., (a) What is the value of f 共40, 15兲? What is its meaning?, (b) What is the meaning of the function h 苷 f 共30, t兲? Describe, the behavior of this function., (c) What is the meaning of the function h 苷 f 共v, 30兲? Describe, the behavior of this function., TABLE 4, , What is the value of f 共95, 70兲? What is its meaning?, For what value of h is f 共90, h兲 苷 100?, For what value of T is f 共T, 50兲 苷 88?, What are the meanings of the functions I 苷 f 共80, h兲, and I 苷 f 共100, h兲? Compare the behavior of these two, functions of h., , 3. A manufacturer has modeled its yearly production function P, , (the monetary value of its entire production in millions of, dollars) as a Cobb-Douglas function, P共L, K兲 苷 1.47L, , 0.65, , K, , 0.35, , where L is the number of labor hours (in thousands) and K is, the invested capital (in millions of dollars). Find P共120, 20兲, and interpret it., , Duration (hours), t, , 5, , 10, , 15, , 20, , 30, , 40, , 50, , 10, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 15, , 4, , 4, , 5, , 5, , 5, , 5, , 5, , 20, , 5, , 7, , 8, , 8, , 9, , 9, , 9, , 30, , 9, , 13, , 16, , 17, , 18, , 19, , 19, , 40, , 14, , 21, , 25, , 28, , 31, , 33, , 33, , 50, , 19, , 29, , 36, , 40, , 45, , 48, , 50, , 60, , 24, , 37, , 47, , 54, , 62, , 67, , 69, , √, , Wi nd speed (knots), , (a), (b), (c), (d), , 4. Verify for the Cobb-Douglas production function, 8. A company makes three sizes of cardboard boxes: small,, , P共L, K 兲 苷 1.01L 0.75K 0.25, , ;, , Graphing calculator or computer required, , medium, and large. It costs $2.50 to make a small box, $4.00, 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 914, , 914, , PARTIAL DERIVATIVES, , CHAPTER 14, , 35. Level curves (isothermals) are shown for the water temperature, , 共in C兲 in Long Lake (Minnesota) in 1998 as a function of, depth and time of year. Estimate the temperature in the lake on, June 9 (day 160) at a depth of 10 m and on June 29 (day 180), at a depth of 5 m., , 39– 42 A contour map of a function is shown. Use it to make a, rough sketch of the graph of f ., 39., , 40., , y, , 0, 12 16, , _8, , 20, , _6, , 8, Depth (m), , y, , 14, 13, 12, 11, , x, , 5, , _4, , 20 16, 12, , 10, , x, 8, , 8, , 15, 120, , 160, , 200, , 41., 240, , 42., , y, , y, , 5, 4, , 280, 3, , Day of 1998, 2, , 36. Two contour maps are shown. One is for a function f whose, , graph is a cone. The other is for a function t whose graph is a, paraboloid. Which is which, and why?, , 2, , 1, , 3, 0, , 0, 2, , 1, , y, , I, , _3, _2, _1, 0, 1, , 3, , x, , 4 5, , 0, , x, , 43–50 Draw a contour map of the function showing several level, curves., , x, , y, , II, , 43. f 共x, y兲 苷 共 y  2x兲2, , 44. f 共x, y兲 苷 x 3  y, , 45. f 共x, y兲 苷 sx  y, , 46. f 共x, y兲 苷 ln共x 2  4y 2 兲, , 47. f 共x, y兲 苷 ye x, , 48. f 共x, y兲 苷 y sec x, , 49. f 共x, y兲 苷 sy 2  x 2, , 50. f 共x, y兲 苷 y兾共x 2  y 2 兲, , x, , 51–52 Sketch both a contour map and a graph of the function and, , compare them., 37. Locate the points A and B on the map of Lonesome Mountain, , 51. f 共x, y兲 苷 x 2  9y 2, , 52. f 共x, y兲 苷 s36  9x 2  4y 2, , (Figure 12). How would you describe the terrain near A?, Near B?, 38. Make a rough sketch of a contour map for the function whose, , graph is shown., z, , 53. A thin metal plate, located in the xy-plane, has temperature, , T共x, y兲 at the point 共x, y兲. The level curves of T are called, isothermals because at all points on such a curve the temperature is the same. Sketch some isothermals if the temperature, function is given by, T共x, y兲 苷, , 100, 1  x 2  2y 2, , 54. If V共x, y兲 is the electric potential at a point 共x, y兲 in the, y, , x, , xy-plane, then the level curves of V are called equipotential, curves because at all points on such a curve the electric, potential is the same. Sketch some equipotential curves if, V共x, y兲 苷 c兾sr 2  x 2  y 2 , where c is a positive constant., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 915, , SECTION 14.1, , ; 55–58 Use a computer to graph the function using various, domains and viewpoints. Get a printout of one that, in your opinion, gives a good view. If your software also produces level, curves, then plot some contour lines of the same function and, compare with the graph., 55. f 共x, y兲 苷 xy 2  x 3, 56. f 共x, y兲 苷 xy 3  yx 3, 共x 2y 2 兲兾3, , 57. f 共x, y兲 苷 e, , (monkey saddle), , 59–64 Match the function (a) with its graph (labeled A–F below), and (b) with its contour map (labeled I–VI). Give reasons for, your choices., 59. z 苷 sin共xy兲, , 60. z 苷 e x cos y, , 61. z 苷 sin共x  y兲, , 62. z 苷 sin x  sin y, , 共sin共x 兲  cos共 y 兲兲, 2, , 64. z 苷, , 58. f 共x, y兲 苷 cos x cos y, A, , B, , z, , 915, , 63. z 苷 共1  x 2 兲共1  y 2 兲, , (dog saddle), 2, , FUNCTIONS OF SEVERAL VARIABLES, , xy, 1  x2  y2, C, , z, , z, , y, y, , x, , y, , x, , x, z, , D, , z, , E, , z, , F, , x, , I, , II, , y, , x, , y, , x, , x, , V, , y, , x, , III, , y, , x, , IV, , y, , y, , x, , y, , VI, , y, , x, , y, , x, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 917, , LIMITS AND CONTINUITY, , SECTION 14.2, TABLE 1 Values of f 共x, y兲, , 917, , TABLE 2 Values of t共x, y兲, , y, , 1.0, , 0.5, , 0.2, , 0, , 0.2, , 0.5, , 1.0, , 1.0, , 0.455, , 0.759, , 0.829, , 0.841, , 0.829, , 0.759, , 0.455, , 1.0, , 0.000, , 0.600, , 0.923, , 0.5, , 0.759, , 0.959, , 0.986, , 0.990, , 0.986, , 0.959, , 0.759, , 0.5, , 0.600, , 0.000, , 0.2, , 0.829, , 0.986, , 0.999, , 1.000, , 0.999, , 0.986, , 0.829, , 0.2, , 0.923 0.724, , 0, , 0.841, , 0.990, , 1.000, , 1.000, , 0.990, , 0.841, , 0, , 1.000 1.000 1.000, , 0.2, , 0.829, , 0.986, , 0.999, , 1.000, , 0.999, , 0.986, , 0.829, , 0.2, , 0.923 0.724, , 0.000, , 1.000, , 0.000 0.724 0.923, , 0.5, , 0.759, , 0.959, , 0.986, , 0.990, , 0.986, , 0.959, , 0.759, , 0.5, , 0.600, , 0.000, , 0.724, , 1.000, , 0.724, , 0.000 0.600, , 1.0, , 0.455, , 0.759, , 0.829, , 0.841, , 0.829, , 0.759, , 0.455, , 1.0, , 0.000, , 0.600, , 0.923, , 1.000, , 0.923, , 0.600, , x, , y, , x, , 1.0, , 0.5, , 0.2, , 0, , 0.2, , 0.5, , 1.0, , 1.000, , 0.923, , 0.600, , 0.000, , 0.724, , 1.000, , 0.724, , 0.000 0.600, , 0.000, , 1.000, , 0.000 0.724 0.923, 1.000 1.000 1.000, , 0.000, , It appears that as 共x, y兲 approaches (0, 0), the values of f 共x, y兲 are approaching 1 whereas, the values of t共x, y兲 aren’t approaching any number. It turns out that these guesses based on, numerical evidence are correct, and we write, lim, , 共 x, y兲 l 共0, 0兲, , sin共x 2  y 2 兲, 苷1, x2  y2, , and, , lim, , 共 x, y兲 l 共0, 0兲, , x2  y2, x2  y2, , does not exist, , In general, we use the notation, lim, , 共 x, y兲 l 共 a, b兲, , f 共x, y兲 苷 L, , to indicate that the values of f 共x, y兲 approach the number L as the point 共x, y兲 approaches, the point 共a, b兲 along any path that stays within the domain of f . In other words, we can, make the values of f 共x, y兲 as close to L as we like by taking the point 共x, y兲 sufficiently, close to the point 共a, b兲, but not equal to 共a, b兲. A more precise definition follows., 1 Definition Let f be a function of two variables whose domain D includes, points arbitrarily close to 共a, b兲. Then we say that the limit of f 共x, y兲 as 共x, y兲, approaches 共a, b兲 is L and we write, , lim, , 共x, y兲 l 共a, b兲, , if for every number, if, , 共x, y兲 僆 D, , and, , f 共x, y兲 苷 L, ,  0 there is a corresponding number, s共x  a兲2  共y  b兲2, , 0, ,  0 such that, , then, , ⱍ f 共x, y兲  L ⱍ, , Other notations for the limit in Definition 1 are, lim f 共x, y兲 苷 L, , xla, ylb, , ⱍ, , and, , f 共x, y兲 l L as 共x, y兲 l 共a, b兲, , ⱍ, , Notice that f 共x, y兲  L is the distance between the numbers f 共x, y兲 and L, and, s共x  a兲 2  共y  b兲 2 is the distance between the point 共x, y兲 and the point 共a, b兲. Thus, Definition 1 says that the distance between f 共x, y兲 and L can be made arbitrarily small by, making the distance from 共x, y兲 to 共a, b兲 sufficiently small (but not 0). Figure 1 illustrates, Definition 1 by means of an arrow diagram. If any small interval 共L  , L  兲 is given, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p910-919.qk_97817_14_ch14_p910-919 11/8/10 1:27 PM Page 918, , 918, , CHAPTER 14, , PARTIAL DERIVATIVES, , around L , then we can find a disk D with center 共a, b兲 and radius  0 such that f maps, all the points in D [except possibly 共a, b兲] into the interval 共L  , L  兲., z, , z, , y, , L+∑, L, L-∑, , (x, y), , ∂, , D, , (, , x, , ), , f, , (a, b), 0, , S, , L+∑, L, L-∑, 0, , 0, , x, , FIGURE 1, , (a, b), , D∂, , y, , FIGURE 2, , y, , b, 0, , a, , x, , FIGURE 3, , Another illustration of Definition 1 is given in Figure 2 where the surface S is the graph, of f . If  0 is given, we can find  0 such that if 共x, y兲 is restricted to lie in the disk, D and 共x, y兲 苷 共a, b兲, then the corresponding part of S lies between the horizontal planes, z 苷 L  and z 苷 L  ., For functions of a single variable, when we let x approach a, there are only two possible, directions of approach, from the left or from the right. We recall from Chapter 1 that if, lim x l a f 共x兲 苷 lim x l a f 共x兲, then lim x l a f 共x兲 does not exist., For functions of two variables the situation is not as simple because we can let 共x, y兲, approach 共a, b兲 from an infinite number of directions in any manner whatsoever (see Figure 3) as long as 共x, y兲 stays within the domain of f ., Definition 1 says that the distance between f 共x, y兲 and L can be made arbitrarily small, by making the distance from 共x, y兲 to 共a, b兲 sufficiently small (but not 0). The definition, refers only to the distance between 共x, y兲 and 共a, b兲. It does not refer to the direction of, approach. Therefore, if the limit exists, then f 共x, y兲 must approach the same limit no matter how 共x, y兲 approaches 共a, b兲. Thus, if we can find two different paths of approach along, which the function f 共x, y兲 has different limits, then it follows that lim 共x, y兲 l 共a, b兲 f 共x, y兲 does, not exist., If f 共x, y兲 l L 1 as 共x, y兲 l 共a, b兲 along a path C1 and f 共x, y兲 l L 2 as, 共x, y兲 l 共a, b兲 along a path C2 , where L 1 苷 L 2 , then lim 共x, y兲 l 共a, b兲 f 共x, y兲 does, not exist., , v, , EXAMPLE 1 Show that, , lim, , 共 x, y兲 l 共0, 0兲, , x2  y2, does not exist., x2  y2, , SOLUTION Let f 共x, y兲 苷 共x 2  y 2 兲兾共x 2  y 2 兲. First let’s approach 共0, 0兲 along the, , y, , x-axis. Then y 苷 0 gives f 共x, 0兲 苷 x 2兾x 2 苷 1 for all x 苷 0, so, , f=_1, , f 共x, y兲 l 1, f=1, , x, , 共x, y兲 l 共0, 0兲 along the x-axis, , y 2, We now approach along the y-axis by putting x 苷 0. Then f 共0, y兲 苷 2 苷 1 for, y, all y 苷 0, so, f 共x, y兲 l 1, , FIGURE 4, , as, , as, , 共x, y兲 l 共0, 0兲 along the y-axis, , (See Figure 4.) Since f has two different limits along two different lines, the given limit, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p920-929.qk_97817_14_ch14_p920-929 11/8/10 1:28 PM Page 920, , 920, , CHAPTER 14, , PARTIAL DERIVATIVES, , f 共x, y兲 l 12, , so, , 共x, y兲 l 共0, 0兲 along x 苷 y 2, , as, , Since different paths lead to different limiting values, the given limit does not exist., Now let’s look at limits that do exist. Just as for functions of one variable, the calculation of limits for functions of two variables can be greatly simplified by the use of properties of limits. The Limit Laws listed in Section 1.6 can be extended to functions of two, variables: The limit of a sum is the sum of the limits, the limit of a product is the product, of the limits, and so on. In particular, the following equations are true., lim, , 2, , 共x, y兲 l 共a, b兲, , x苷a, , lim, , 共x, y兲 l 共a, b兲, , y苷b, , lim, , 共x, y兲 l 共a, b兲, , c苷c, , The Squeeze Theorem also holds., EXAMPLE 4 Find, , lim, , 共x, y兲 l 共0, 0兲, , 3x 2y, if it exists., x ⫹ y2, 2, , SOLUTION As in Example 3, we could show that the limit along any line through the, , origin is 0. This doesn’t prove that the given limit is 0, but the limits along the parabolas, y 苷 x 2 and x 苷 y 2 also turn out to be 0, so we begin to suspect that the limit does exist, and is equal to 0., Let ␧ ⬎ 0. We want to find ␦ ⬎ 0 such that, 0 ⬍ sx 2 ⫹ y 2 ⬍ ␦, , if, , that is,, , if, , then, , 冟, , 0 ⬍ sx 2 ⫹ y 2 ⬍ ␦ then, , 冟, , 3x 2 y, ⫺0 ⬍␧, x ⫹ y2, 2, , ⱍ ⱍ, , 3x 2 y, ⬍␧, x2 ⫹ y2, , But x 2 艋 x 2 ⫹ y 2 since y 2 艌 0, so x 2兾共x 2 ⫹ y 2 兲 艋 1 and therefore, , ⱍ ⱍ, , 3x 2 y, 艋 3 y 苷 3sy 2 艋 3sx 2 ⫹ y 2, x2 ⫹ y2, , 3, , ⱍ ⱍ, , Thus if we choose ␦ 苷 ␧兾3 and let 0 ⬍ sx 2 ⫹ y 2 ⬍ ␦, then, , 冟, , Another way to do Example 4 is to use the, Squeeze Theorem instead of Definition 1. From, 2 it follows that, lim, , 共 x, y兲 l 共0, 0兲, , ⱍ ⱍ, , 3 y 苷0, , 冉冊, , 冟, , 3x 2 y, ␧, ⫺ 0 艋 3sx 2 ⫹ y 2 ⬍ 3␦ 苷 3, 2, 2, x ⫹y, 3, , 苷␧, , Hence, by Definition 1,, , and so the first inequality in 3 shows that the, given limit is 0., , lim, , 共x, y兲 l 共0, 0兲, , 3x 2 y, 苷0, x ⫹ y2, 2, , Continuity, Recall that evaluating limits of continuous functions of a single variable is easy. It can be, accomplished by direct substitution because the defining property of a continuous function, is limx l a f 共x兲 苷 f 共a兲. Continuous functions of two variables are also defined by the direct, substitution property., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p920-929.qk_97817_14_ch14_p920-929 11/8/10 1:28 PM Page 925, , SECTION 14.3, , PARTIAL DERIVATIVES, , 925, , Relative humidity (%), , TABLE 1, , Heat index I as a function of, temperature and humidity, , H, , 50, , 55, , 60, , 65, , 70, , 75, , 80, , 85, , 90, , 90, , 96, , 98, , 100, , 103, , 106, , 109, , 112, , 115, , 119, , 92, , 100, , 103, , 105, , 108, , 112, , 115, , 119, , 123, , 128, , 94, , 104, , 107, , 111, , 114, , 118, , 122, , 127, , 132, , 137, , 96, , 109, , 113, , 116, , 121, , 125, , 130, , 135, , 141, , 146, , 98, , 114, , 118, , 123, , 127, , 133, , 138, , 144, , 150, , 157, , 100, , 119, , 124, , 129, , 135, , 141, , 147, , 154, , 161, , 168, , T, , Actual, temperature, (°F), , If we concentrate on the highlighted column of the table, which corresponds to a relative, humidity of H 苷 70%, we are considering the heat index as a function of the single variable T for a fixed value of H. Let’s write t共T兲 苷 f 共T, 70兲. Then t共T兲 describes how the heat, index I increases as the actual temperature T increases when the relative humidity is 70%., The derivative of t when T 苷 96⬚F is the rate of change of I with respect to T when, T 苷 96⬚F :, t⬘共96兲 苷 lim, , hl0, , t共96 ⫹ h兲 ⫺ t共96兲, f 共96 ⫹ h, 70兲 ⫺ f 共96, 70兲, 苷 lim, h, l, 0, h, h, , We can approximate t⬘共96兲 using the values in Table 1 by taking h 苷 2 and ⫺2:, t⬘共96兲 ⬇, t⬘共96兲 ⬇, , t共98兲 ⫺ t共96兲, f 共98, 70兲 ⫺ f 共96, 70兲, 133 ⫺ 125, 苷, 苷, 苷4, 2, 2, 2, t共94兲 ⫺ t共96兲, f 共94, 70兲 ⫺ f 共96, 70兲, 118 ⫺ 125, 苷, 苷, 苷 3.5, ⫺2, ⫺2, ⫺2, , Averaging these values, we can say that the derivative t⬘共96兲 is approximately 3.75. This, means that, when the actual temperature is 96⬚F and the relative humidity is 70%, the, apparent temperature (heat index) rises by about 3.75⬚F for every degree that the actual, temperature rises!, Now let’s look at the highlighted row in Table 1, which corresponds to a fixed temperature of T 苷 96⬚F. The numbers in this row are values of the function G共H兲 苷 f 共96, H兲,, which describes how the heat index increases as the relative humidity H increases when the, actual temperature is T 苷 96⬚F. The derivative of this function when H 苷 70% is the rate, of change of I with respect to H when H 苷 70%:, G⬘共70兲 苷 lim, , hl0, , G共70 ⫹ h兲 ⫺ G共70兲, f 共96, 70 ⫹ h兲 ⫺ f 共96, 70兲, 苷 lim, hl0, h, h, , By taking h 苷 5 and ⫺5, we approximate G⬘共70兲 using the tabular values:, G⬘共70兲 ⬇, G⬘共70兲 ⬇, , G共75兲 ⫺ G共70兲, f 共96, 75兲 ⫺ f 共96, 70兲, 130 ⫺ 125, 苷, 苷, 苷1, 5, 5, 5, G共65兲 ⫺ G共70兲, f 共96, 65兲 ⫺ f 共96, 70兲, 121 ⫺ 125, 苷, 苷, 苷 0.8, ⫺5, ⫺5, ⫺5, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p920-929.qk_97817_14_ch14_p920-929 11/8/10 1:28 PM Page 928, , 928, , CHAPTER 14, , PARTIAL DERIVATIVES, , z, , z=4-≈-2¥, , As we have seen in the case of the heat index function, partial derivatives can also be, interpreted as rates of change. If z 苷 f 共x, y兲, then ⭸z兾⭸x represents the rate of change of z, with respect to x when y is fixed. Similarly, ⭸z兾⭸y represents the rate of change of z with, respect to y when x is fixed., , C¡, , EXAMPLE 2 If f 共x, y兲 苷 4 ⫺ x 2 ⫺ 2y 2, find fx 共1, 1兲 and fy 共1, 1兲 and interpret these num-, , y=1, , bers as slopes., , (1, 1, 1), , x, , SOLUTION We have, y, , (1, 1), , 2, , fx 共x, y兲 苷 ⫺2x, , fy 共x, y兲 苷 ⫺4y, , fx 共1, 1兲 苷 ⫺2, , fy 共1, 1兲 苷 ⫺4, , FIGURE 2, z, , z=4-≈-2¥, , C™, , The graph of f is the paraboloid z 苷 4 ⫺ x 2 ⫺ 2y 2 and the vertical plane y 苷 1 intersects it in the parabola z 苷 2 ⫺ x 2, y 苷 1. (As in the preceding discussion, we label, it C1 in Figure 2.) The slope of the tangent line to this parabola at the point 共1, 1, 1兲 is, fx 共1, 1兲 苷 ⫺2. Similarly, the curve C2 in which the plane x 苷 1 intersects the paraboloid is the parabola z 苷 3 ⫺ 2y 2, x 苷 1, and the slope of the tangent line at 共1, 1, 1兲 is, fy 共1, 1兲 苷 ⫺4. (See Figure 3.), , x=1, (1, 1, 1), y, x, , 2, , Figure 4 is a computer-drawn counterpart to Figure 2. Part (a) shows the plane y 苷 1, intersecting the surface to form the curve C1 and part (b) shows C1 and T1 . [We have used, the vector equations r共t兲 苷 具 t, 1, 2 ⫺ t 2 典 for C1 and r共t兲 苷 具1 ⫹ t, 1, 1 ⫺ 2t 典 for T1 .], Similarly, Figure 5 corresponds to Figure 3., , (1, 1), , FIGURE 3, 4, , 4, , 3, , 3, , z 2, , z 2, , 1, , 1, , 0, , 0, 0, , 1, y, , 1, , FIGURE 4, , x, , 0, 0, , 1, y, , 1, , (a), , 4, , 3, , 3, , z 2, , z 2, , 1, , 1, 0, 0, , 1, y, , 1, , 2, , x, , (b), , 4, , 0, , FIGURE 5, , 2, , 0, , 2, , x, , 0, , 0, 0, , 1, y, , 1, , 2, , x, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p920-929.qk_97817_14_ch14_p920-929 11/8/10 1:28 PM Page 929, , SECTION 14.3, , v, , EXAMPLE 3 If f 共x, y兲 苷 sin, , PARTIAL DERIVATIVES, , 929, , 冉 冊, , x, ⭸f, ⭸f, , calculate, and, ., 1⫹y, ⭸x, ⭸y, , SOLUTION Using the Chain Rule for functions of one variable, we have, , 冉 冊 冉 冊 冉 冊, 冉 冊 冉 冊 冉 冊, , Some computer algebra systems can plot, surfaces defined by implicit equations in three, variables. Figure 6 shows such a plot of the, surface defined by the equation in Example 4., , v, , ⭸f, x, 苷 cos, ⭸x, 1⫹y, , ⴢ, , ⭸, ⭸x, , x, 1⫹y, , 苷 cos, , ⭸f, x, 苷 cos, ⭸y, 1⫹y, , ⴢ, , ⭸, ⭸y, , x, 1⫹y, , 苷 ⫺cos, , x, 1⫹y, , ⴢ, , x, 1⫹y, , 1, 1⫹y, ⴢ, , x, 共1 ⫹ y兲2, , EXAMPLE 4 Find ⭸z兾⭸x and ⭸z兾⭸y if z is defined implicitly as a function of x and y by, , the equation, x 3 ⫹ y 3 ⫹ z 3 ⫹ 6xyz 苷 1, SOLUTION To find ⭸z兾⭸x, we differentiate implicitly with respect to x, being careful to, , treat y as a constant:, 3x 2 ⫹ 3z 2, , ⭸z, ⭸z, ⫹ 6yz ⫹ 6xy, 苷0, ⭸x, ⭸x, , Solving this equation for ⭸z兾⭸x, we obtain, , FIGURE 6, , ⭸z, x 2 ⫹ 2yz, 苷⫺ 2, ⭸x, z ⫹ 2xy, Similarly, implicit differentiation with respect to y gives, ⭸z, y 2 ⫹ 2xz, 苷⫺ 2, ⭸y, z ⫹ 2xy, , Functions of More Than Two Variables, Partial derivatives can also be defined for functions of three or more variables. For example,, if f is a function of three variables x, y, and z, then its partial derivative with respect to x is, defined as, fx 共x, y, z兲 苷 lim, , hl0, , f 共x ⫹ h, y, z兲 ⫺ f 共x, y, z兲, h, , and it is found by regarding y and z as constants and differentiating f 共x, y, z兲 with respect, to x. If w 苷 f 共x, y, z兲, then fx 苷 ⭸w兾⭸x can be interpreted as the rate of change of w with, respect to x when y and z are held fixed. But we can’t interpret it geometrically because the, graph of f lies in four-dimensional space., In general, if u is a function of n variables, u 苷 f 共x 1, x 2 , . . . , x n 兲, its partial derivative, with respect to the ith variable x i is, ⭸u, f 共x1 , . . . , xi⫺1 , xi ⫹ h, xi⫹1 , . . . , xn 兲 ⫺ f 共x1 , . . . , xi , . . . , xn兲, 苷 lim, hl0, ⭸xi, h, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p930-939.qk_97817_14_ch14_p930-939 11/8/10 1:28 PM Page 931, , PARTIAL DERIVATIVES, , SECTION 14.3, , 931, , 20, z 0, _20, Figure 7 shows the graph of the function f, in Example 6 and the graphs of its first- and, second-order partial derivatives for ⫺2 艋 x 艋 2,, ⫺2 艋 y 艋 2. Notice that these graphs are consistent with our interpretations of fx and fy as, slopes of tangent lines to traces of the graph of f., For instance, the graph of f decreases if we start, at 共0, ⫺2兲 and move in the positive x-direction., This is reflected in the negative values of fx. You, should compare the graphs of fy x and fyy with the, graph of fy to see the relationships., , _40, _2, , _1, y, , 0, , _2, _1, 1 0 x, 2 2, , 1, , f, , 40, 40, z, , 20, , z 20, , 0, _20, _2, , _1, y, , 0, , 1, , _2, _1, 1 0 x, 2 2, , 0, _2, , _1, y, , 0, , fx, , 20, z 0, _20, _2, , _1, y, , 0, , 1, , _2, _1, 1 0 x, 2 2, , fxx, , 40, , 20, z 0, , 20, z 0, , _20, , _20, , _40, _1, y, , 0, , 1, , _2, _1, 1 0 x, 2 2, , fy, , 40, , _2, , 1, , _2, _1, 1 0 x, 2 2, , 1, , _2, _1, 1 0 x, 2 2, , _40, _2, , _1, y, , fxy ⫽ fyx, , 0, , fyy, , FIGURE 7, , Notice that fx y 苷 fyx in Example 6. This is not just a coincidence. It turns out that the, mixed partial derivatives fx y and fyx are equal for most functions that one meets in practice., The following theorem, which was discovered by the French mathematician Alexis Clairaut, (1713–1765), gives conditions under which we can assert that fx y 苷 fyx . The proof is given in, Appendix F., Clairaut, , Clairaut’s Theorem Suppose f is defined on a disk D that contains the point 共a, b兲., , Alexis Clairaut was a child prodigy in mathematics: he read l’Hospital’s textbook on, calculus when he was ten and presented a, paper on geometry to the French Academy of, Sciences when he was 13. At the age of 18,, Clairaut published Recherches sur les courbes à, double courbure, which was the first systematic, treatise on three-dimensional analytic geometry, and included the calculus of space curves., , If the functions fx y and fyx are both continuous on D, then, fx y 共a, b兲 苷 fyx 共a, b兲, , Partial derivatives of order 3 or higher can also be defined. For instance,, fx yy 苷 共 fx y 兲y 苷, , ⭸, ⭸y, , 冉 冊, ⭸2 f, ⭸y ⭸x, , 苷, , ⭸3 f, ⭸y 2 ⭸x, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p930-939.qk_97817_14_ch14_p930-939 11/11/10 9:31 AM Page 933, , SECTION 14.3, , PARTIAL DERIVATIVES, , 933, , Partial differential equations involving functions of three variables are also very important in science and engineering. The three-dimensional Laplace equation is, 5, , ⭸2u, ⭸2u, ⭸2u, ⫹, ⫹, 苷0, ⭸x 2, ⭸y 2, ⭸z 2, , and one place it occurs is in geophysics. If u共x, y, z兲 represents magnetic field strength at, position 共x, y, z兲, then it satisfies Equation 5. The strength of the magnetic field indicates, the distribution of iron-rich minerals and reflects different rock types and the location of, faults. Figure 9 shows a contour map of the earth’s magnetic field as recorded from an aircraft carrying a magnetometer and flying 200 m above the surface of the ground. The contour map is enhanced by color-coding of the regions between the level curves., , 0.103, , 0.040, , FPO, New Art to, come, , 0.002, , -0.019, , FIGURE 9, , Magnetic field strength of the earth, , Courtesy Roger Watson, , -0.037, , -0.051, , -0.066, , -0.109, , Nano Teslas, per meter, , Figure 10 shows a contour map for the second-order partial derivative of u in the vertical direction, that is, u zz. It turns out that the values of the partial derivatives uxx and u yy, are relatively easily measured from a map of the magnetic field. Then values of u zz can be, calculated from Laplace’s equation 5 ., , 0.000117, , 0.000037, , 0.000002, , -0.000017, , -0.000036, , FIGURE 10, , Second vertical derivative, of the magnetic field, , Courtesy Roger Watson, , -0.000064, , -0.000119, , -0.000290, , Nano Teslas, per m/m, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p930-939.qk_97817_14_ch14_p930-939 11/8/10 1:28 PM Page 934, , 934, , CHAPTER 14, , PARTIAL DERIVATIVES, , The Cobb-Douglas Production Function, In Example 3 in Section 14.1 we described the work of Cobb and Douglas in modeling the, total production P of an economic system as a function of the amount of labor L and the, capital investment K. Here we use partial derivatives to show how the particular form of, their model follows from certain assumptions they made about the economy., If the production function is denoted by P 苷 P共L, K兲, then the partial derivative ⭸P兾⭸L, is the rate at which production changes with respect to the amount of labor. Economists, call it the marginal production with respect to labor or the marginal productivity of labor., Likewise, the partial derivative ⭸P兾⭸K is the rate of change of production with respect to, capital and is called the marginal productivity of capital. In these terms, the assumptions, made by Cobb and Douglas can be stated as follows., (i) If either labor or capital vanishes, then so will production., (ii) The marginal productivity of labor is proportional to the amount of production, per unit of labor., (iii) The marginal productivity of capital is proportional to the amount of production, per unit of capital., Because the production per unit of labor is P兾L, assumption (ii) says that, ⭸P, P, 苷␣, ⭸L, L, for some constant ␣. If we keep K constant 共K 苷 K0 兲, then this partial differential equation, becomes an ordinary differential equation:, dP, P, 苷␣, dL, L, , 6, , If we solve this separable differential equation by the methods of Section 9.3 (see also Exercise 85), we get, 7, , P共L, K0 兲 苷 C1共K0 兲L␣, , Notice that we have written the constant C1 as a function of K0 because it could depend on, the value of K0 ., Similarly, assumption (iii) says that, ⭸P, P, 苷␤, ⭸K, K, and we can solve this differential equation to get, 8, , P共L 0 , K兲 苷 C2共L 0 兲K ␤, , Comparing Equations 7 and 8, we have, 9, , P共L, K兲 苷 bL␣K ␤, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p930-939.qk_97817_14_ch14_p930-939 11/8/10 1:28 PM Page 935, , PARTIAL DERIVATIVES, , SECTION 14.3, , 935, , where b is a constant that is independent of both L and K. Assumption (i) shows that ␣ ⬎ 0, and ␤ ⬎ 0., Notice from Equation 9 that if labor and capital are both increased by a factor m, then, P共mL, mK兲 苷 b共mL兲␣共mK 兲␤ 苷 m␣⫹␤bL␣K ␤ 苷 m␣⫹␤P共L, K 兲, If ␣ ⫹ ␤ 苷 1, then P共mL, mK兲 苷 mP共L, K兲, which means that production is also increased, by a factor of m. That is why Cobb and Douglas assumed that ␣ ⫹ ␤ 苷 1 and therefore, P共L, K兲 苷 bL␣K 1⫺␣, This is the Cobb-Douglas production function that we discussed in Section 14.1., , 14.3, , Exercises, (b) In general, what can you say about the signs of ⭸W兾⭸T, and ⭸W兾⭸v ?, (c) What appears to be the value of the following limit?, , 1. The temperature T (in ⬚C兲 at a location in the Northern Hemi-, , sphere depends on the longitude x, latitude y, and time t, so we, can write T 苷 f 共x, y, t兲. Let’s measure time in hours from the, beginning of January., (a) What are the meanings of the partial derivatives ⭸T兾⭸x,, ⭸T兾⭸y, and ⭸T兾⭸t ?, (b) Honolulu has longitude 158⬚ W and latitude 21⬚ N. Suppose that at 9:00 AM on January 1 the wind is blowing hot, air to the northeast, so the air to the west and south is warm, and the air to the north and east is cooler. Would you expect, fx 共158, 21, 9兲, fy 共158, 21, 9兲, and ft 共158, 21, 9兲 to be positive or negative? Explain., , lim, , vl⬁, , 4. The wave heights h in the open sea depend on the speed v, , of the wind and the length of time t that the wind has been, blowing at that speed. Values of the function h 苷 f 共v, t兲 are, recorded in feet in the following table., Duration (hours), , 2. At the beginning of this section we discussed the function, , 3. The wind-chill index W is the perceived temperature when the, actual temperature is T and the wind speed is v, so we can, write W 苷 f 共T, v兲. The following table of values is an excerpt, , from Table 1 in Section 14.1., , Actual temperature (°C), , Wind speed (km/h), v, , 20, , 30, , 40, , 50, , 60, , 70, , ⫺10, , ⫺18, , ⫺20, , ⫺21, , ⫺22, , ⫺23, , ⫺23, , ⫺15, , ⫺24, , ⫺26, , ⫺27, , ⫺29, , ⫺30, , ⫺30, , ⫺20, , ⫺30, , ⫺33, , ⫺34, , ⫺35, , ⫺36, , ⫺37, , ⫺25, , ⫺37, , ⫺39, , ⫺41, , ⫺42, , ⫺43, , ⫺44, , T, , t, , 5, , 10, , 15, , 20, , 30, , 40, , 50, , 10, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 15, , 4, , 4, , 5, , 5, , 5, , 5, , 5, , 20, , 5, , 7, , 8, , 8, , 9, , 9, , 9, , 30, , 9, , 13, , 16, , 17, , 18, , 19, , 19, , 40, , 14, , 21, , 25, , 28, , 31, , 33, , 33, , 50, , 19, , 29, , 36, , 40, , 45, , 48, , 50, , 60, , 24, , 37, , 47, , 54, , 62, , 67, , 69, , v, , Wind speed (knots), , I 苷 f 共T, H 兲, where I is the heat index, T is the temperature,, and H is the relative humidity. Use Table 1 to estimate, fT 共92, 60兲 and fH 共92, 60兲. What are the practical interpretations, of these values?, , (a) What are the meanings of the partial derivatives ⭸h兾⭸v, and ⭸h兾⭸t ?, (b) Estimate the values of fv 共40, 15兲 and ft 共40, 15兲. What are, the practical interpretations of these values?, (c) What appears to be the value of the following limit?, , (a) Estimate the values of f T 共⫺15, 30兲 and fv 共⫺15, 30兲. What, are the practical interpretations of these values?, , ;, , Graphing calculator or computer required, , ⭸W, ⭸v, , CAS Computer algebra system required, , lim, , tl⬁, , ⭸h, ⭸t, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p930-939.qk_97817_14_ch14_p930-939 11/8/10 1:29 PM Page 938, , 938, , CHAPTER 14, , PARTIAL DERIVATIVES, , 80. If u 苷 e a1 x1⫹a2 x2⫹⭈⭈⭈⫹an x n, where a 12 ⫹ a 22 ⫹ ⭈ ⭈ ⭈ ⫹ a n2 苷 1,, , show that, ⭸2u, ⭸2u, ⭸2u, ⫹, ⫹ ⭈⭈⭈ ⫹, 苷u, 2, 2, ⭸x1, ⭸x 2, ⭸x n2, 81. Verify that the function z 苷 ln共e ⫹ e 兲 is a solution of the, x, , y, , differential equations, ⭸z, ⭸z, ⫹, 苷1, ⭸x, ⭸y, and, ⭸ 2z ⭸ 2z, ⫺, ⭸x 2 ⭸y 2, , 2, , 苷0, , 82. The temperature at a point 共x, y兲 on a flat metal plate is given, , by T共x, y兲 苷 60兾共1 ⫹ x 2 ⫹ y 2 兲, where T is measured in ⬚C, and x, y in meters. Find the rate of change of temperature with, respect to distance at the point 共2, 1兲 in (a) the x-direction and, (b) the y-direction., , 83. The total resistance R produced by three conductors with resis-, , tances R1 , R2 , R3 connected in a parallel electrical circuit is, given by the formula, 1, 1, 1, 1, 苷, ⫹, ⫹, R, R1, R2, R3, Find ⭸R兾⭸R1., 84. Show that the Cobb-Douglas production function P 苷 bL␣K ␤, , satisfies the equation, L, , ⭸P, ⭸P, ⫹K, 苷 共␣ ⫹ ␤兲P, ⭸L, ⭸K, , 85. Show that the Cobb-Douglas production function satisfies, , P共L, K0 兲 苷 C1共K0 兲L␣ by solving the differential equation, dP, P, 苷␣, dL, L, (See Equation 6.), 86. Cobb and Douglas used the equation P共L, K兲 苷 1.01L 0.75 K 0.25, , to model the American economy from 1899 to 1922, where L, is the amount of labor and K is the amount of capital. (See, Example 3 in Section 14.1.), (a) Calculate PL and PK., (b) Find the marginal productivity of labor and the marginal, productivity of capital in the year 1920, when L 苷 194 and, K 苷 407 (compared with the assigned values L 苷 100 and, K 苷 100 in 1899). Interpret the results., (c) In the year 1920 which would have benefited production, more, an increase in capital investment or an increase in, spending on labor?, 87. The van der Waals equation for n moles of a gas is, , 冉, , P⫹, , 88. The gas law for a fixed mass m of an ideal gas at absolute tem-, , perature T, pressure P, and volume V is PV 苷 mRT, where R is, the gas constant. Show that, ⭸P ⭸V ⭸T, 苷 ⫺1, ⭸V ⭸T ⭸P, , 冉 冊, ⭸ 2z, ⭸x ⭸y, , ture of the gas. The constant R is the universal gas constant, and a and b are positive constants that are characteristic of a, particular gas. Calculate ⭸T兾⭸P and ⭸P兾⭸V ., , 冊, , n 2a, 共V ⫺ nb兲 苷 nRT, V2, , where P is the pressure, V is the volume, and T is the tempera-, , 89. For the ideal gas of Exercise 88, show that, , T, , ⭸P ⭸V, 苷 mR, ⭸T ⭸T, , 90. The wind-chill index is modeled by the function, , W 苷 13.12 ⫹ 0.6215T ⫺ 11.37v 0.16 ⫹ 0.3965T v 0.16, where T is the temperature 共⬚C兲 and v is the wind speed, 共km兾h兲. When T 苷 ⫺15⬚C and v 苷 30 km兾h, by how much, would you expect the apparent temperature W to drop if the, actual temperature decreases by 1⬚C ? What if the wind speed, increases by 1 km兾h ?, 91. The kinetic energy of a body with mass m and velocity v is, K 苷 12 mv 2. Show that, , ⭸K ⭸2K, 苷K, ⭸m ⭸v 2, 92. If a, b, c are the sides of a triangle and A, B, C are the opposite, , angles, find ⭸A兾⭸a, ⭸A兾⭸b, ⭸A兾⭸c by implicit differentiation of, the Law of Cosines., 93. You are told that there is a function f whose partial deriva-, , tives are fx 共x, y兲 苷 x ⫹ 4y and fy 共x, y兲 苷 3x ⫺ y. Should you, believe it?, 2, 2, ; 94. The paraboloid z 苷 6 ⫺ x ⫺ x ⫺ 2y intersects the plane, , x 苷 1 in a parabola. Find parametric equations for the tangent, line to this parabola at the point 共1, 2, ⫺4兲. Use a computer to, graph the paraboloid, the parabola, and the tangent line on the, same screen., , 95. The ellipsoid 4x 2 ⫹ 2y 2 ⫹ z 2 苷 16 intersects the plane y 苷 2, , in an ellipse. Find parametric equations for the tangent line to, this ellipse at the point 共1, 2, 2兲., 96. In a study of frost penetration it was found that the temperature, , T at time t (measured in days) at a depth x (measured in feet), can be modeled by the function, T共x, t兲 苷 T0 ⫹ T1 e⫺␭ x sin共␻ t ⫺ ␭ x兲, where ␻ 苷 2␲兾365 and ␭ is a positive constant., (a) Find ⭸T兾⭸x. What is its physical significance?, (b) Find ⭸T兾⭸t. What is its physical significance?, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 14.4, , (c) Show that T satisfies the heat equation Tt 苷 kTxx for a, certain constant k., (d) If ␭ 苷 0.2, T0 苷 0, and T1 苷 10, use a computer to, graph T共x, t兲., (e) What is the physical significance of the term ⫺␭ x in the, expression sin共␻ t ⫺ ␭ x兲?, , ;, , 939, , 99. If f 共x, y兲 苷 x共x 2 ⫹ y 2 兲⫺3兾2e sin共x y兲, find fx 共1, 0兲., 2, , [Hint: Instead of finding fx 共x, y兲 first, note that it’s easier, to use Equation 1 or Equation 2.], , 3, 100. If f 共x, y兲 苷 s, x 3 ⫹ y 3 , find fx 共0, 0兲., , 101. Let, , 97. Use Clairaut’s Theorem to show that if the third-order partial, , fx yy 苷 fyx y 苷 fyyx, 98. (a) How many nth-order partial derivatives does a function, , ;, , of two variables have?, (b) If these partial derivatives are all continuous, how many, of them can be distinct?, (c) Answer the question in part (a) for a function of three, variables., , 再, , x 3y ⫺ xy 3, x2 ⫹ y2, f 共x, y兲 苷, 0, , derivatives of f are continuous, then, , 14.4, , TANGENT PLANES AND LINEAR APPROXIMATIONS, , CAS, , (a), (b), (c), (d), (e), , if 共x, y兲 苷 共0, 0兲, if 共x, y兲 苷 共0, 0兲, , Use a computer to graph f ., Find fx 共x, y兲 and fy 共x, y兲 when 共x, y兲 苷 共0, 0兲., Find fx 共0, 0兲 and fy 共0, 0兲 using Equations 2 and 3., Show that fxy 共0, 0兲 苷 ⫺1 and fyx 共0, 0兲 苷 1., Does the result of part (d) contradict Clairaut’s Theorem?, Use graphs of fxy and fyx to illustrate your answer., , Tangent Planes and Linear Approximations, One of the most important ideas in single-variable calculus is that as we zoom in toward, a point on the graph of a differentiable function, the graph becomes indistinguishable, from its tangent line and we can approximate the function by a linear function. (See Section 2.9.) Here we develop similar ideas in three dimensions. As we zoom in toward a point, on a surface that is the graph of a differentiable function of two variables, the surface looks, more and more like a plane (its tangent plane) and we can approximate the function by a, linear function of two variables. We also extend the idea of a differential to functions of, two or more variables., , Tangent Planes, z, , T¡, C¡, P, T™, , C™, , 0, y, , x, , FIGURE 1, , Suppose a surface S has equation z 苷 f 共x, y兲, where f has continuous first partial derivatives, and let P共x 0 , y0 , z0 兲 be a point on S. As in the preceding section, let C1 and C2 be the, curves obtained by intersecting the vertical planes y 苷 y0 and x 苷 x 0 with the surface S., Then the point P lies on both C1 and C2. Let T1 and T2 be the tangent lines to the curves C1, and C2 at the point P. Then the tangent plane to the surface S at the point P is defined to be, the plane that contains both tangent lines T1 and T2 . (See Figure 1.), We will see in Section 14.6 that if C is any other curve that lies on the surface S and, passes through P, then its tangent line at P also lies in the tangent plane. Therefore you can, think of the tangent plane to S at P as consisting of all possible tangent lines at P to curves, that lie on S and pass through P. The tangent plane at P is the plane that most closely approximates the surface S near the point P., We know from Equation 12.5.7 that any plane passing through the point P共x 0 , y0 , z0 兲 has, an equation of the form, , The tangent plane contains the, tangent lines T¡, T and T™, T., , A共x ⫺ x 0 兲 ⫹ B共 y ⫺ y0 兲 ⫹ C共z ⫺ z0 兲 苷 0, By dividing this equation by C and letting a 苷 ⫺A兾C and b 苷 ⫺B兾C, we can write it in, the form, 1, , z ⫺ z0 苷 a共x ⫺ x 0兲 ⫹ b共y ⫺ y0 兲, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p940-949.qk_97817_14_ch14_p940-949 11/8/10 1:29 PM Page 943, , SECTION 14.4, , TANGENT PLANES AND LINEAR APPROXIMATIONS, , 943, , The corresponding linear approximation is, xe xy ⬇ x ⫹ y, f 共1.1, ⫺0.1兲 ⬇ 1.1 ⫺ 0.1 苷 1, , so, , Compare this with the actual value of f 共1.1, ⫺0.1兲 苷 1.1e ⫺0.11 ⬇ 0.98542., EXAMPLE 3 At the beginning of Section 14.3 we discussed the heat index (perceived, temperature) I as a function of the actual temperature T and the relative humidity H and, gave the following table of values from the National Weather Service., , Relative humidity (%), , Actual, temperature, (°F), , H, , 50, , 55, , 60, , 65, , 70, , 75, , 80, , 85, , 90, , 90, , 96, , 98, , 100, , 103, , 106, , 109, , 112, , 115, , 119, , 92, , 100, , 103, , 105, , 108, , 112, , 115, , 119, , 123, , 128, , 94, , 104, , 107, , 111, , 114, , 118, , 122, , 127, , 132, , 137, , 96, , 109, , 113, , 116, , 121, , 125, , 130, , 135, , 141, , 146, , 98, , 114, , 118, , 123, , 127, , 133, , 138, , 144, , 150, , 157, , 100, , 119, , 124, , 129, , 135, , 141, , 147, , 154, , 161, , 168, , T, , Find a linear approximation for the heat index I 苷 f 共T, H兲 when T is near 96⬚F and H is, near 70%. Use it to estimate the heat index when the temperature is 97⬚F and the relative, humidity is 72%., SOLUTION We read from the table that f 共96, 70兲 苷 125. In Section 14.3 we used the tabular values to estimate that fT 共96, 70兲 ⬇ 3.75 and fH 共96, 70兲 ⬇ 0.9. (See pages 925–26.), So the linear approximation is, , f 共T, H兲 ⬇ f 共96, 70兲 ⫹ fT 共96, 70兲共T ⫺ 96兲 ⫹ fH 共96, 70兲共H ⫺ 70兲, ⬇ 125 ⫹ 3.75共T ⫺ 96兲 ⫹ 0.9共H ⫺ 70兲, In particular,, f 共97, 72兲 ⬇ 125 ⫹ 3.75共1兲 ⫹ 0.9共2兲 苷 130.55, Therefore, when T 苷 97⬚F and H 苷 72%, the heat index is, I ⬇ 131⬚F, , Differentials, , y, , For a differentiable function of one variable, y 苷 f 共x兲, we define the differential dx to be an, independent variable; that is, dx can be given the value of any real number. The differential, of y is then defined as, , y=ƒ, , Îy, dx=Îx, 0, , a, , a+Îx, , tangent line, y=f(a)+fª(a)(x-a), FIGURE 6, , 9, , dy, , x, , dy 苷 f ⬘共x兲 dx, , (See Section 2.9.) Figure 6 shows the relationship between the increment ⌬y and the differential dy : ⌬y represents the change in height of the curve y 苷 f 共x兲 and dy represents the, change in height of the tangent line when x changes by an amount dx 苷 ⌬x., For a differentiable function of two variables, z 苷 f 共x, y兲, we define the differentials, dx and dy to be independent variables; that is, they can be given any values. Then the, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p940-949.qk_97817_14_ch14_p940-949 11/8/10 1:29 PM Page 945, , SECTION 14.4, , TANGENT PLANES AND LINEAR APPROXIMATIONS, , 945, , The increment of z is, ⌬z 苷 f 共2.05, 2.96兲 ⫺ f 共2, 3兲, 苷 关共2.05兲2 ⫹ 3共2.05兲共2.96兲 ⫺ 共2.96兲2 兴 ⫺ 关2 2 ⫹ 3共2兲共3兲 ⫺ 3 2 兴, 苷 0.6449, Notice that ⌬z ⬇ dz but dz is easier to compute., EXAMPLE 5 The base radius and height of a right circular cone are measured as 10 cm, and 25 cm, respectively, with a possible error in measurement of as much as 0.1 cm in, each. Use differentials to estimate the maximum error in the calculated volume of the, cone., SOLUTION The volume V of a cone with base radius r and height h is V 苷, , ␲ r 2h兾3. So, , the differential of V is, dV 苷, , ⭸V, ⭸V, 2␲rh, ␲r 2, dr ⫹, dh 苷, dr ⫹, dh, ⭸r, ⭸h, 3, 3, , ⱍ ⱍ, , ⱍ ⱍ, , Since each error is at most 0.1 cm, we have ⌬r 艋 0.1, ⌬h 艋 0.1. To estimate the, largest error in the volume we take the largest error in the measurement of r and of h., Therefore we take dr 苷 0.1 and dh 苷 0.1 along with r 苷 10, h 苷 25. This gives, dV 苷, , 500␲, 100␲, 共0.1兲 ⫹, 共0.1兲 苷 20␲, 3, 3, , Thus the maximum error in the calculated volume is about 20␲ cm3 ⬇ 63 cm.3, , Functions of Three or More Variables, Linear approximations, differentiability, and differentials can be defined in a similar manner for functions of more than two variables. A differentiable function is defined by an, expression similar to the one in Definition 7. For such functions the linear approximation, is, f 共x, y, z兲 ⬇ f 共a, b, c兲 ⫹ fx 共a, b, c兲共x ⫺ a兲 ⫹ fy 共a, b, c兲共y ⫺ b兲 ⫹ fz共a, b, c兲共z ⫺ c兲, and the linearization L共x, y, z兲 is the right side of this expression., If w 苷 f 共x, y, z兲, then the increment of w is, ⌬w 苷 f 共x ⫹ ⌬x, y ⫹ ⌬y, z ⫹ ⌬z兲 ⫺ f 共x, y, z兲, The differential dw is defined in terms of the differentials dx, dy, and dz of the independent variables by, ⭸w, ⭸w, ⭸w, dw 苷, dx ⫹, dy ⫹, dz, ⭸x, ⭸y, ⭸z, EXAMPLE 6 The dimensions of a rectangular box are measured to be 75 cm, 60 cm,, and 40 cm, and each measurement is correct to within 0.2 cm. Use differentials to estimate the largest possible error when the volume of the box is calculated from these, measurements., SOLUTION If the dimensions of the box are x, y, and z, its volume is V 苷 xyz and so, , dV 苷, , ⭸V, ⭸V, ⭸V, dx ⫹, dy ⫹, dz 苷 yz dx ⫹ xz dy ⫹ xy dz, ⭸x, ⭸y, ⭸z, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p940-949.qk_97817_14_ch14_p940-949 11/8/10 1:29 PM Page 947, , SECTION 14.4, , 23. Use the table in Example 3 to find a linear approximation to, , the heat index function when the temperature is near 94⬚F, and the relative humidity is near 80%. Then estimate the heat, index when the temperature is 95⬚F and the relative humidity, is 78%., 24. The wind-chill index W is the perceived temperature when the, actual temperature is T and the wind speed is v, so we can, write W 苷 f 共T, v兲. The following table of values is an excerpt, , from Table 1 in Section 14.1. Use the table to find a linear, approximation to the wind-chill index function when T is near, ⫺15⬚C and v is near 50 km兾h. Then estimate the wind-chill, index when the temperature is ⫺17⬚C and the wind speed, is 55 km兾h., , TANGENT PLANES AND LINEAR APPROXIMATIONS, , 947, , possible error of ⫾2 km兾h, and the temperature is measured, as ⫺11⬚C, with a possible error of ⫾1⬚C. Use differentials to, estimate the maximum error in the calculated value of W due, to the measurement errors in T and v., 37. The tension T in the string of the yo-yo in the figure is, , T苷, , mtR, 2r 2 ⫹ R 2, , where m is the mass of the yo-yo and t is acceleration due to, gravity. Use differentials to estimate the change in the tension, if R is increased from 3 cm to 3.1 cm and r is increased from, 0.7 cm to 0.8 cm. Does the tension increase or decrease?, T, , Actual temperature (°C), , Wind speed (km/h), v, , 20, , 30, , 40, , 50, , 60, , 70, , ⫺10, , ⫺18, , ⫺20, , ⫺21, , ⫺22, , ⫺23, , ⫺23, , ⫺15, , ⫺24, , ⫺26, , ⫺27, , ⫺29, , ⫺30, , ⫺30, , ⫺20, , ⫺30, , ⫺33, , ⫺34, , ⫺35, , ⫺36, , ⫺37, , ⫺25, , ⫺37, , ⫺39, , ⫺41, , ⫺42, , ⫺43, , ⫺44, , T, , 25–30 Find the differential of the function., 25. z 苷 e ⫺2x cos 2␲ t, , 26. u 苷 sx 2 ⫹ 3y 2, , 27. m 苷 p 5q 3, , 28. T 苷, , 29. R 苷 ␣␤ 2 cos ␥, , 30. L 苷 xze⫺y ⫺z, , 1 ⫹ u vw, 2, , 31. If z 苷 5x ⫹ y and 共x, y兲 changes from 共1, 2兲 to 共1.05, 2.1兲,, 2, , r, , 38. The pressure, volume, and temperature of a mole of an ideal, , gas are related by the equation PV 苷 8.31T, where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approximate change in the pressure if the, volume increases from 12 L to 12.3 L and the temperature, decreases from 310 K to 305 K., 39. If R is the total resistance of three resistors, connected in par-, , v, , 2, , R, , 2, , compare the values of ⌬z and dz., , 32. If z 苷 x 2 ⫺ xy ⫹ 3y 2 and 共x, y兲 changes from 共3, ⫺1兲 to, , 共2.96, ⫺0.95兲, compare the values of ⌬z and dz., , 33. The length and width of a rectangle are measured as 30 cm and, , 24 cm, respectively, with an error in measurement of at most, 0.1 cm in each. Use differentials to estimate the maximum, error in the calculated area of the rectangle., 34. Use differentials to estimate the amount of metal in a closed, , cylindrical can that is 10 cm high and 4 cm in diameter if the, metal in the top and bottom is 0.1 cm thick and the metal in the, sides is 0.05 cm thick., , allel, with resistances R1 , R2 , R3 , then, 1, 1, 1, 1, 苷, ⫹, ⫹, R, R1, R2, R3, If the resistances are measured in ohms as R1 苷 25 ⍀,, R2 苷 40 ⍀, and R3 苷 50 ⍀, with a possible error of 0.5% in, each case, estimate the maximum error in the calculated value, of R., 40. Four positive numbers, each less than 50, are rounded to the, , first decimal place and then multiplied together. Use differentials to estimate the maximum possible error in the computed, product that might result from the rounding., 41. A model for the surface area of a human body is given by, S 苷 0.1091w 0.425 h 0.725, where w is the weight (in pounds), h is, , the height (in inches), and S is measured in square feet. If the, errors in measurement of w and h are at most 2%, use differentials to estimate the maximum percentage error in the calculated surface area., , 35. Use differentials to estimate the amount of tin in a closed tin, , can with diameter 8 cm and height 12 cm if the tin is 0.04 cm, thick., 36. The wind-chill index is modeled by the function, , W 苷 13.12 ⫹ 0.6215T ⫺ 11.37v 0.16 ⫹ 0.3965T v 0.16, where T is the temperature 共in ⬚C兲 and v is the wind speed, 共in km兾h兲. The wind speed is measured as 26 km兾h, with a, , 42. Suppose you need to know an equation of the tangent plane to, , a surface S at the point P共2, 1, 3兲. You don’t have an equation, for S but you know that the curves, r1共t兲 苷 具2 ⫹ 3t, 1 ⫺ t 2, 3 ⫺ 4t ⫹ t 2 典, r2共u兲 苷 具1 ⫹ u 2, 2u 3 ⫺ 1, 2u ⫹ 1典, both lie on S. Find an equation of the tangent plane at P., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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948, , CHAPTER 14, , PARTIAL DERIVATIVES, , 43– 44 Show that the function is differentiable by finding values, of ␧1 and ␧2 that satisfy Definition 7., 43. f 共x, y兲 苷 x 2 ⫹ y 2, , 46. (a) The function, , 44. f 共x, y兲 苷 xy ⫺ 5y 2, , f 共x, y兲 苷, , 45. Prove that if f is a function of two variables that is differen-, , lim, , 14.5, , if 共x, y兲 苷 共0, 0兲, if 共x, y兲 苷 共0, 0兲, , was graphed in Figure 4. Show that fx 共0, 0兲 and fy 共0, 0兲, both exist but f is not differentiable at 共0, 0兲. [Hint: Use, the result of Exercise 45.], (b) Explain why fx and fy are not continuous at 共0, 0兲., , tiable at 共a, b兲, then f is continuous at 共a, b兲., Hint: Show that, 共⌬x, ⌬y兲 l 共0, 0兲, , 再, , xy, x2 ⫹ y2, 0, , f 共a ⫹ ⌬x, b ⫹ ⌬y兲 苷 f 共a, b兲, , The Chain Rule, Recall that the Chain Rule for functions of a single variable gives the rule for differentiating, a composite function: If y 苷 f 共x兲 and x 苷 t共t兲, where f and t are differentiable functions,, then y is indirectly a differentiable function of t and, dy, dy dx, 苷, dt, dx dt, , 1, , For functions of more than one variable, the Chain Rule has several versions, each of, them giving a rule for differentiating a composite function. The first version (Theorem 2), deals with the case where z 苷 f 共x, y兲 and each of the variables x and y is, in turn, a function of a variable t. This means that z is indirectly a function of t, z 苷 f 共 t共t兲, h共t兲兲, and the, Chain Rule gives a formula for differentiating z as a function of t. We assume that f is differentiable (Definition 14.4.7). Recall that this is the case when fx and fy are continuous, (Theorem 14.4.8)., 2 The Chain Rule (Case 1) Suppose that z 苷 f 共x, y兲 is a differentiable function of, x and y, where x 苷 t共t兲 and y 苷 h共t兲 are both differentiable functions of t. Then z, is a differentiable function of t and, , dz, ⭸f dx, ⭸f dy, 苷, ⫹, dt, ⭸x dt, ⭸y dt, , PROOF A change of ⌬t in t produces changes of ⌬x in x and ⌬y in y. These, in turn, produce a change of ⌬z in z, and from Definition 14.4.7 we have, , ⌬z 苷, , ⭸f, ⭸f, ⌬x ⫹, ⌬y ⫹ ␧1 ⌬x ⫹ ␧2 ⌬y, ⭸x, ⭸y, , where ␧1 l 0 and ␧2 l 0 as 共⌬x, ⌬y兲 l 共0, 0兲. [If the functions ␧1 and ␧2 are not, defined at 共0, 0兲, we can define them to be 0 there.] Dividing both sides of this equation, by ⌬t, we have, ⌬z, ⭸f ⌬x, ⭸f ⌬y, ⌬x, ⌬y, 苷, ⫹, ⫹ ␧1, ⫹ ␧2, ⌬t, ⭸x ⌬t, ⭸y ⌬t, ⌬t, ⌬t, If we now let ⌬t l 0, then ⌬x 苷 t共t ⫹ ⌬t兲 ⫺ t共t兲 l 0 because t is differentiable and, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p950-959.qk_97817_14_ch14_p950-959 11/8/10 1:30 PM Page 950, , 950, , PARTIAL DERIVATIVES, , CHAPTER 14, , the Chain Rule gives, dP, P dT, P dV, 8.31 dT, 8.31T dV, 苷, , 苷, , dt, T dt, V dt, V dt, V 2 dt, 苷, , 8.31, 8.31共300兲, 共0.1兲 , 共0.2兲 苷 0.04155, 100, 100 2, , The pressure is decreasing at a rate of about 0.042 kPa兾s., We now consider the situation where z 苷 f 共x, y兲 but each of x and y is a function of two, variables s and t : x 苷 t共s, t兲, y 苷 h共s, t兲. Then z is indirectly a function of s and t and we, wish to find z兾s and z兾t. Recall that in computing z兾t we hold s fixed and compute, the ordinary derivative of z with respect to t. Therefore we can apply Theorem 2 to obtain, z, z x, z y, 苷, , t, x t, y t, A similar argument holds for z兾s and so we have proved the following version of the, Chain Rule., 3 The Chain Rule (Case 2) Suppose that z 苷 f 共x, y兲 is a differentiable function of, x and y, where x 苷 t共s, t兲 and y 苷 h共s, t兲 are differentiable functions of s and t., Then, , z, z x, z y, 苷, , s, x s, y s, , z, z x, z y, 苷, , t, x t, y t, , EXAMPLE 3 If z 苷 e x sin y, where x 苷 st 2 and y 苷 s 2t, find z兾s and z兾t., SOLUTION Applying Case 2 of the Chain Rule, we get, , z, z x, z y, 苷, , 苷 共e x sin y兲共t 2 兲  共e x cos y兲共2st兲, s, x s, y s, 2, , 2, , 苷 t 2e st sin共s 2t兲  2ste st cos共s 2t兲, z, z x, z y, 苷, , 苷 共e x sin y兲共2st兲  共e x cos y兲共s 2 兲, t, x t, y t, 2, , 2, , 苷 2ste st sin共s 2t兲  s 2e st cos共s 2t兲, z, ⳵z, ⳵x, ⳵x, ⳵s, , s, , x, , ⳵z, ⳵y, ⳵x, ⳵t, , t, , FIGURE 2, , ⳵y, ⳵s, , s, , y, , ⳵y, ⳵t, , t, , Case 2 of the Chain Rule contains three types of variables: s and t are independent variables, x and y are called intermediate variables, and z is the dependent variable. Notice that, Theorem 3 has one term for each intermediate variable and each of these terms resembles, the one-dimensional Chain Rule in Equation 1., To remember the Chain Rule, it’s helpful to draw the tree diagram in Figure 2. We draw, branches from the dependent variable z to the intermediate variables x and y to indicate that, z is a function of x and y. Then we draw branches from x and y to the independent variables, s and t. On each branch we write the corresponding partial derivative. To find z兾s, we, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p950-959.qk_97817_14_ch14_p950-959 11/8/10 1:30 PM Page 951, , SECTION 14.5, , THE CHAIN RULE, , 951, , find the product of the partial derivatives along each path from z to s and then add these, products:, z, z x, z y, 苷, , s, x s, y s, Similarly, we find z兾t by using the paths from z to t., Now we consider the general situation in which a dependent variable u is a function of, n intermediate variables x 1 , . . . , x n , each of which is, in turn, a function of m independent, variables t1 , . . . , tm . Notice that there are n terms, one for each intermediate variable. The, proof is similar to that of Case 1., , 4 The Chain Rule (General Version) Suppose that u is a differentiable function of, the n variables x 1 , x 2 , . . . , x n and each x j is a differentiable function of the m variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and, , u, u x 1, u x 2, u x n, 苷, ,   , ti, x 1 ti, x 2 ti, x n ti, for each i 苷 1, 2, . . . , m., , v EXAMPLE 4 Write out the Chain Rule for the case where w 苷 f 共x, y, z, t兲 and, x 苷 x共u, v兲, y 苷 y共u, v兲, z 苷 z共u, v兲, and t 苷 t共u, v兲., SOLUTION We apply Theorem 4 with n 苷 4 and m 苷 2. Figure 3 shows the tree diagram., , w, x, , y, v, , u, , v, , u, , v, , u, , Although we haven’t written the derivatives on the branches, it’s understood that if a, branch leads from y to u, then the partial derivative for that branch is y兾u. With the aid, of the tree diagram, we can now write the required expressions:, , t, , z, u, , v, , w, w x, w y, w z, w t, 苷, , , , u, x u, y u, z u, t u, , FIGURE 3, , w, w x, w y, w z, w t, 苷, , , , v, x v, y v, z v, t v, , v, , EXAMPLE 5 If u 苷 x 4 y  y 2 z 3, where x 苷 rse t, y 苷 rs 2e t, and z 苷 r 2s sin t, find the, , value of u兾s when r 苷 2, s 苷 1, t 苷 0., , SOLUTION With the help of the tree diagram in Figure 4, we have, , u, x, r, , s, , y, t, , FIGURE 4, , r, , s, , u, u x, u y, u z, 苷, , , s, x s, y s, z s, , z, t, , r, , s, , t, , 苷 共4x 3 y兲共re t 兲  共x 4  2yz 3 兲共2rset 兲  共3y 2z 2 兲共r 2 sin t兲, When r 苷 2, s 苷 1, and t 苷 0, we have x 苷 2, y 苷 2, and z 苷 0, so, u, 苷 共64兲共2兲  共16兲共4兲  共0兲共0兲 苷 192, s, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p950-959.qk_97817_14_ch14_p950-959 11/8/10 1:30 PM Page 953, , SECTION 14.5, , THE CHAIN RULE, , 953, , y 苷 f 共x兲, where F共x, f 共x兲兲 苷 0 for all x in the domain of f . If F is differentiable, we can, apply Case 1 of the Chain Rule to differentiate both sides of the equation F共x, y兲 苷 0 with, respect to x. Since both x and y are functions of x, we obtain, F dx, F dy, , 苷0, x dx, y dx, But dx兾dx 苷 1, so if F兾y 苷 0 we solve for dy兾dx and obtain, F, dy, x, Fx, 苷, 苷, dx, F, Fy, y, , 6, , To derive this equation we assumed that F共x, y兲 苷 0 defines y implicitly as a function of, x. The Implicit Function Theorem, proved in advanced calculus, gives conditions under, which this assumption is valid: It states that if F is defined on a disk containing 共a, b兲, where, F共a, b兲 苷 0, Fy 共a, b兲 苷 0, and Fx and Fy are continuous on the disk, then the equation, F共x, y兲 苷 0 defines y as a function of x near the point 共a, b兲 and the derivative of this function is given by Equation 6., EXAMPLE 8 Find y if x 3  y 3 苷 6xy., SOLUTION The given equation can be written as, , F共x, y兲 苷 x 3  y 3  6xy 苷 0, so Equation 6 gives, dy, Fx, 3x 2  6y, x 2  2y, 苷, 苷 2, 苷 2, dx, Fy, 3y  6x, y  2x, , The solution to Example 8 should be, compared to the one in Example 2 in, Section 2.6., , Now we suppose that z is given implicitly as a function z 苷 f 共x, y兲 by an equation of the, form F共x, y, z兲 苷 0. This means that F共x, y, f 共x, y兲兲 苷 0 for all 共x, y兲 in the domain, of f . If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F共x, y, z兲 苷 0 as follows:, F x, F y, F z, , , 苷0, x x, y x, z x, , But, , , 共x兲 苷 1, x, , and, , , 共y兲 苷 0, x, , so this equation becomes, F, F z, , 苷0, x, z x, If F兾z 苷 0, we solve for z兾x and obtain the first formula in Equations 7 on page 954., The formula for z兾y is obtained in a similar manner., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p950-959.qk_97817_14_ch14_p950-959 11/8/10 1:30 PM Page 955, , SECTION 14.5, , 17–20 Use a tree diagram to write out the Chain Rule for the given, , case. Assume all functions are differentiable., 17. u 苷 f 共x, y兲,, , where x 苷 x共r, s, t兲, y 苷 y共r, s, t兲, , 18. R 苷 f 共x, y, z, t兲, where x 苷 x共u, v, w兲, y 苷 y共u, v, w兲,, z 苷 z共u, v, w兲, t 苷 t共u, v, w兲, 19. w 苷 f 共r, s, t兲,, , where r 苷 r共x, y兲, s 苷 s共x, y兲, t 苷 t共x, y兲, , 20. t 苷 f 共u, v, w兲, where u 苷 u共 p, q, r, s兲, v 苷 v 共 p, q, r, s兲,, w 苷 w 共 p, q, r, s兲, , THE CHAIN RULE, , 955, , and rainfall is decreasing at a rate of 0.1 cm兾year. They also, estimate that, at current production levels, W兾T 苷 2, and W兾R 苷 8., (a) What is the significance of the signs of these partial, derivatives?, (b) Estimate the current rate of change of wheat production,, dW兾dt., 37. The speed of sound traveling through ocean water with salinity, , 35 parts per thousand has been modeled by the equation, C 苷 1449.2  4.6T  0.055T 2  0.00029T 3  0.016D, , 21–26 Use the Chain Rule to find the indicated partial derivatives., 21. z 苷 x 4  x 2 y,, , z z z, ,, ,, s t u, , x 苷 s  2t  u,, , y 苷 stu 2;, , when s 苷 4, t 苷 2, u 苷 1, , v, , 22. T 苷, , , u 苷 pqsr , v 苷 psq r ;, 2u  v, T T T, ,, ,, when p 苷 2, q 苷 1, r 苷 4, p q r, x 苷 r cos , y 苷 r sin ,, , 23. w 苷 xy  yz  zx,, w w, , r, , ,, , , , when r 苷 2,  苷 兾2, , 24. P 苷 su 2  v 2  w 2 ,, , P P, ,, x y, , z 苷 r ;, , u 苷 xe y,, , v 苷 ye x,, , w 苷 e xy ;, , when x 苷 0, y 苷 2, , pq, , p 苷 u  vw, q 苷 v  u w, r 苷 w  u v ;, pr, N N N, ,, ,, when u 苷 2, v 苷 3, w 苷 4, u v w, , 26. u 苷 xe ty,, , x苷 2 , y苷 2 , t苷, u u u, ,, ,, when 苷 1, 苷 2,, , , , , 2, , ;, 苷1, , 27–30 Use Equation 6 to find dy兾dx., , 1, , 29. tan 共x y兲 苷 x  xy, 2, , 28. cos共xy兲 苷 1  sin y, 2, , 30. e y sin x 苷 x  xy, , 31–34 Use Equations 7 to find z兾x and z兾y ., 31. x 2  2y 2  3z 2 苷 1, , 32. x 2  y 2  z 2  2z 苷 4, , 33. e 苷 xyz, , 34. yz  x ln y 苷 z 2, , z, , D, , T, 16, , 20, , 14, , 15, , 12, , 10, , 10, , 5, , 8, 10, , 25. N 苷, , 27. y cos x 苷 x 2  y 2, , where C is the speed of sound (in meters per second), T is the, temperature (in degrees Celsius), and D is the depth below the, ocean surface (in meters). A scuba diver began a leisurely dive, into the ocean water; the diver’s depth and the surrounding, water temperature over time are recorded in the following, graphs. Estimate the rate of change (with respect to time) of, the speed of sound through the ocean water experienced by the, diver 20 minutes into the dive. What are the units?, , 20, , 30, , 40 t, (min), , 10, , 20, , 30, , 40 t, (min), , 38. The radius of a right circular cone is increasing at a rate of, , 1.8 in兾s while its height is decreasing at a rate of 2.5 in兾s. At, what rate is the volume of the cone changing when the radius, is 120 in. and the height is 140 in.?, 39. The length 艎, width w, and height h of a box change with, , time. At a certain instant the dimensions are 艎 苷 1 m and, w 苷 h 苷 2 m, and 艎 and w are increasing at a rate of 2 m兾s, while h is decreasing at a rate of 3 m兾s. At that instant find the, rates at which the following quantities are changing., (a) The volume, (b) The surface area, (c) The length of a diagonal, , 40. The voltage V in a simple electrical circuit is slowly decreasing, , as the battery wears out. The resistance R is slowly increasing, as the resistor heats up. Use Ohm’s Law, V 苷 IR, to find how, the current I is changing at the moment when R 苷 400 ,, I 苷 0.08 A, dV兾dt 苷 0.01 V兾s, and dR兾dt 苷 0.03 兾s., 41. The pressure of 1 mole of an ideal gas is increasing at a rate, , 35. The temperature at a point 共x, y兲 is T共x, y兲, measured in degrees, , Celsius. A bug crawls so that its position after t seconds is, given by x 苷 s1  t , y 苷 2  13 t, where x and y are measured, in centimeters. The temperature function satisfies Tx 共2, 3兲 苷 4, and Ty 共2, 3兲 苷 3. How fast is the temperature rising on the, bug’s path after 3 seconds?, 36. Wheat production W in a given year depends on the average, , temperature T and the annual rainfall R. Scientists estimate, that the average temperature is rising at a rate of 0.15°C兾year, , of 0.05 kPa兾s and the temperature is increasing at a rate of, 0.15 K兾s. Use the equation in Example 2 to find the rate of, change of the volume when the pressure is 20 kPa and the, temperature is 320 K., 42. A manufacturer has modeled its yearly production function P, , (the value of its entire production in millions of dollars) as a, Cobb-Douglas function, P共L, K兲 苷 1.47L 0.65 K 0.35, where L is the number of labor hours (in thousands) and K is, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p950-959.qk_97817_14_ch14_p950-959 11/8/10 1:30 PM Page 956, , 956, , PARTIAL DERIVATIVES, , CHAPTER 14, , the invested capital (in millions of dollars). Suppose that when, L 苷 30 and K 苷 8, the labor force is decreasing at a rate of, 2000 labor hours per year and capital is increasing at a rate of, $500,000 per year. Find the rate of change of production., 43. One side of a triangle is increasing at a rate of 3 cm兾s and a, , second side is decreasing at a rate of 2 cm兾s. If the area of the, triangle remains constant, at what rate does the angle between, the sides change when the first side is 20 cm long, the second, side is 30 cm, and the angle is 兾6 ?, 44. If a sound with frequency fs is produced by a source traveling, along a line with speed vs and an observer is traveling with, speed vo along the same line from the opposite direction toward, , the source, then the frequency of the sound heard by the, observer is, c  vo, fo 苷, fs, c  vs, , 冉 冊, , where c is the speed of sound, about 332 m兾s. (This is the, Doppler effect.) Suppose that, at a particular moment, you, are in a train traveling at 34 m兾s and accelerating at 1.2 m兾s 2., A train is approaching you from the opposite direction on the, other track at 40 m兾s, accelerating at 1.4 m兾s 2, and sounds its, whistle, which has a frequency of 460 Hz. At that instant, what, is the perceived frequency that you hear and how fast is it, changing?, 45– 48 Assume that all the given functions are differentiable., 45. If z 苷 f 共x, y兲, where x 苷 r cos  and y 苷 r sin , (a) find z兾r, , and z兾 and (b) show that, , 冉 冊 冉 冊 冉冊 冉 冊, z, x, , z, y, , 2, , , , 2, , 苷, , z, r, , 2, , 1, r2, , , , z, , , u, x, , 2, , , , u, y, , 2, , 苷 e2s, , 47. If z 苷 f 共x  y兲, show that, , 2, , , , u, t, , 2, , z, z, , 苷 0., x, y, , 48. If z 苷 f 共x, y兲, where x 苷 s  t and y 苷 s  t, show that, , 冉 冊 冉 冊, z, x, , 2, , , , z, y, , 2, , 苷, , z z, s t, , 49–54 Assume that all the given functions have continuous, second-order partial derivatives., 49. Show that any function of the form, , z 苷 f 共x  at兲  t共x  at兲, is a solution of the wave equation, 2z, 2z, 2, 苷, a, t 2, x 2, [Hint: Let u 苷 x  at, v 苷 x  at.], , 册, , 51. If z 苷 f 共x, y兲, where x 苷 r 2  s 2 and y 苷 2rs, find 2z兾r s., , (Compare with Example 7.), 52. If z 苷 f 共x, y兲, where x 苷 r cos  and y 苷 r sin , find, , (a) z兾r, (b) z兾, and (c) 2z兾r ., , 53. If z 苷 f 共x, y兲, where x 苷 r cos  and y 苷 r sin , show that, , 2z, 2z, 2z, 1 2z, 1 z, , 苷, , , x 2, y 2, r 2, r 2  2, r r, 54. Suppose z 苷 f 共x, y兲, where x 苷 t共s, t兲 and y 苷 h共s, t兲., , (a) Show that, 2z, 2z, 2 苷, t, x 2, , 冉 冊, , x 2, 2z x y, 2z, 2, , t, x y t t, y 2, z 2x, z 2 y, , 2 , x t, y t 2, , 冉 冊, y, t, , 2, , (b) Find a similar formula for 2z兾s t., 55. A function f is called homogeneous of degree n if it satisfies, , the equation f 共t x, t y兲 苷 t n f 共x, y兲 for all t, where n is a positive, integer and f has continuous second-order partial derivatives., (a) Verify that f 共x, y兲 苷 x 2 y  2x y 2  5y 3 is homogeneous, of degree 3., (b) Show that if f is homogeneous of degree n, then, x, , 冋冉 冊 冉 冊 册, u, s, , 冋, , 2u, 2u, 2u, 2u, , 苷 e2s,  2, 2, 2, 2, x, y, s, t, , 2, , 46. If u 苷 f 共x, y兲, where x 苷 e s cos t and y 苷 e s sin t, show that, , 冉 冊 冉 冊, , 50. If u 苷 f 共x, y兲, where x 苷 e s cos t and y 苷 e s sin t, show that, , f, f, y, 苷 n f 共x, y兲, x, y, , [Hint: Use the Chain Rule to differentiate f 共tx, t y兲 with, respect to t.], 56. If f is homogeneous of degree n, show that, , x2, , 2f, 2f, 2f,  2xy,  y 2 2 苷 n共n  1兲 f 共x, y兲, 2, x, x y, y, , 57. If f is homogeneous of degree n, show that, , fx 共t x, t y兲 苷 t n1fx 共x, y兲, 58. Suppose that the equation F共x, y, z兲 苷 0 implicitly defines each, , of the three variables x, y, and z as functions of the other two:, z 苷 f 共x, y兲, y 苷 t共x, z兲, x 苷 h共 y, z兲. If F is differentiable and, Fx , Fy , and Fz are all nonzero, show that, z x y, 苷 1, x y z, 59. Equation 6 is a formula for the derivative dy兾dx of a function, , defined implicitly by an equation F 共x, y兲 苷 0, provided that F, is differentiable and Fy 苷 0. Prove that if F has continuous second derivatives, then a formula for the second derivative of y is, d 2y, Fxx Fy2  2Fxy Fx Fy  Fyy Fx2, 2 苷 , dx, Fy3, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 14.6, , DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR, , 957, , Directional Derivatives and the Gradient Vector, , 14.6, , 60, , 50, Reno, , San, Francisco, , 60, Las, Vegas, , 70, 70, , 80, , The weather map in Figure 1 shows a contour map of the temperature function T共x, y兲 for, the states of California and Nevada at 3:00 PM on a day in October. The level curves, or, isothermals, join locations with the same temperature. The partial derivative Tx at a location, such as Reno is the rate of change of temperature with respect to distance if we travel east, from Reno; Ty is the rate of change of temperature if we travel north. But what if we want, to know the rate of change of temperature when we travel southeast (toward Las Vegas), or, in some other direction? In this section we introduce a type of derivative, called a directional derivative, that enables us to find the rate of change of a function of two or more, variables in any direction., , Los Angeles, 0, , Directional Derivatives, , 50 100 150 200, (Distance in miles), , Recall that if z 苷 f 共x, y兲, then the partial derivatives fx and fy are defined as, , FIGURE 1, , fx 共x0 , y0 兲 苷 lim, , f 共x0  h, y0 兲  f 共x0 , y0 兲, h, , fy 共x0 , y0 兲 苷 lim, , f 共x0 , y0  h兲  f 共x0 , y0 兲, h, , hl0, , 1, hl0, , y, , u, , sin ¨, , ¨, (x¸, y¸), , cos ¨, , 0, , x, , and represent the rates of change of z in the x- and y-directions, that is, in the directions of, the unit vectors i and j., Suppose that we now wish to find the rate of change of z at 共x 0 , y0 兲 in the direction of an, arbitrary unit vector u 苷 具a, b 典 . (See Figure 2.) To do this we consider the surface S with, the equation z 苷 f 共x, y兲 (the graph of f ) and we let z0 苷 f 共x 0 , y0 兲. Then the point, P共x 0 , y0 , z0 兲 lies on S. The vertical plane that passes through P in the direction of u intersects S in a curve C. (See Figure 3.) The slope of the tangent line T to C at the point P is the, rate of change of z in the direction of u., , FIGURE 2, z, , A unit vector u=ka, bl=kcos ¨, sin ¨l, , T, , P(x¸, y¸, z¸), , Q(x, y, z), , TEC Visual 14.6A animates Figure 3 by, , S, , rotating u and therefore T., , C, , Pª(x ¸, y¸, 0), , ha, , u, , y, , h, hb, Qª (x, y, 0), , FIGURE 3, , x, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p950-959.qk_97817_14_ch14_p950-959 11/8/10 1:30 PM Page 958, , 958, , CHAPTER 14, , PARTIAL DERIVATIVES, , If Q共x, y, z兲 is another point on C and P, Q are the projections of P, Q onto the xy-plane,, B is parallel to u and so, then the vector PQ, B 苷 hu 苷 具ha, hb典, PQ, for some scalar h. Therefore x  x 0 苷 ha, y  y0 苷 hb, so x 苷 x 0  ha, y 苷 y0  hb,, and, z, z  z0, f 共x 0  ha, y0  hb兲  f 共x 0 , y0 兲, 苷, 苷, h, h, h, If we take the limit as h l 0, we obtain the rate of change of z (with respect to distance) in, the direction of u, which is called the directional derivative of f in the direction of u., 2 Definition The directional derivative of f at 共x 0 , y0 兲 in the direction of a unit, vector u 苷 具a, b典 is, , Du f 共x 0 , y0 兲 苷 lim, , hl0, , f 共x 0  ha, y0  hb兲  f 共x 0 , y0 兲, h, , if this limit exists., By comparing Definition 2 with Equations 1 , we see that if u 苷 i 苷 具 1, 0典 , then, Di f 苷 fx and if u 苷 j 苷 具0, 1典 , then Dj f 苷 fy . In other words, the partial derivatives of f, with respect to x and y are just special cases of the directional derivative., EXAMPLE 1 Use the weather map in Figure 1 to estimate the value of the directional, derivative of the temperature function at Reno in the southeasterly direction., SOLUTION The unit vector directed toward the southeast is u 苷 共i  j兲兾s2 , but we, , won’t need to use this expression. We start by drawing a line through Reno toward the, southeast (see Figure 4)., , 60, , 50, Reno, , San, Francisco, , 60, Las, Vegas, , 70, 70, 0, , FIGURE 4, , 50 100 150 200, (Distance in miles), , 80, , Los Angeles, , We approximate the directional derivative Du T by the average rate of change of the, temperature between the points where this line intersects the isothermals T 苷 50 and, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p960-969.qk_97817_14_ch14_p960-969 11/8/10 1:31 PM Page 966, , 966, , CHAPTER 14, , PARTIAL DERIVATIVES, , Significance of the Gradient Vector, We now summarize the ways in which the gradient vector is significant. We first consider, a function f of three variables and a point P共x 0 , y0 , z0 兲 in its domain. On the one hand, we, know from Theorem 15 that the gradient vector ⵜf 共x0 , y0, z0 兲 gives the direction of fastest, increase of f . On the other hand, we know that ⵜf 共x0 , y0 , z0 兲 is orthogonal to the level surface S of f through P. (Refer to Figure 9.) These two properties are quite compatible intuitively because as we move away from P on the level surface S, the value of f does not, change at all. So it seems reasonable that if we move in the perpendicular direction, we get, the maximum increase., In like manner we consider a function f of two variables and a point P共x 0 , y0 兲 in its, domain. Again the gradient vector ⵜf 共x0 , y0 兲 gives the direction of fastest increase of f ., Also, by considerations similar to our discussion of tangent planes, it can be shown that, ⵜf 共x0 , y0 兲 is perpendicular to the level curve f 共x, y兲 苷 k that passes through P. Again this, is intuitively plausible because the values of f remain constant as we move along the curve., (See Figure 11.), y, , ±f(x¸, y¸), , P (x¸, y¸), , level curve, f(x, y)=k, 0, , 300, 200, , curve of, steepest, ascent, , x, , FIGURE 11, , 100, , FIGURE 12, , If we consider a topographical map of a hill and let f 共x, y兲 represent the height above sea, level at a point with coordinates 共x, y兲, then a curve of steepest ascent can be drawn as in, Figure 12 by making it perpendicular to all of the contour lines. This phenomenon can also, be noticed in Figure 12 in Section 14.1, where Lonesome Creek follows a curve of steepest descent., Computer algebra systems have commands that plot sample gradient vectors. Each gradient vector ⵜf 共a, b兲 is plotted starting at the point 共a, b兲. Figure 13 shows such a plot, (called a gradient vector field ) for the function f 共x, y兲 苷 x 2 ⫺ y 2 superimposed on a contour map of f. As expected, the gradient vectors point “uphill” and are perpendicular to the, level curves., y, _9, _6, _3, 0, , 3 6 9, x, , FIGURE 13, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p960-969.qk_97817_14_ch14_p960-969 11/8/10 1:31 PM Page 968, , 968, , CHAPTER 14, , PARTIAL DERIVATIVES, , 27. (a) Show that a differentiable function f decreases most, , rapidly at x in the direction opposite to the gradient vector,, that is, in the direction of ⫺ⵜ f 共x兲., (b) Use the result of part (a) to find the direction in which the, function f 共x, y兲 苷 x 4 y ⫺ x 2 y 3 decreases fastest at the, point 共2, ⫺3兲., 28. Find the directions in which the directional derivative of, , f 共x, y兲 苷 ye⫺xy at the point 共0, 2兲 has the value 1., , 35. Let f be a function of two variables that has continuous, , partial derivatives and consider the points A共1, 3兲, B共3, 3兲,, derivative of f at A in, C共1, 7兲, and D共6, 15兲. The directional, l, the direction of the vector AB is 3 and the directional derival, tive at A in the direction of AC is 26. Find the directional, l, derivative of f at A in the direction of the vector AD., 36. Shown is a topographic map of Blue River Pine Provincial, , Park in British Columbia. Draw curves of steepest descent, from point A (descending to Mud Lake) and from point B., , 29. Find all points at which the direction of fastest change of the, , function f 共x, y兲 苷 x 2 ⫹ y 2 ⫺ 2 x ⫺ 4y is i ⫹ j., Blue River, , Blue River, , 30. Near a buoy, the depth of a lake at the point with coordinates, , Mud Lake, , Blue River Pine Provincial Park, , 共x, y兲 is z 苷 200 ⫹ 0.02x 2 ⫺ 0.001y 3, where x, y, and z are, measured in meters. A fisherman in a small boat starts at the, point 共80, 60兲 and moves toward the buoy, which is located at, 共0, 0兲. Is the water under the boat getting deeper or shallower, when he departs? Explain., , Mud Creek, Smoke Creek, , A, 2200 m, , 31. The temperature T in a metal ball is inversely proportional to, , the distance from the center of the ball, which we take to be the, origin. The temperature at the point 共1, 2, 2兲 is 120⬚., (a) Find the rate of change of T at 共1, 2, 2兲 in the direction, toward the point 共2, 1, 3兲., (b) Show that at any point in the ball the direction of greatest, increase in temperature is given by a vector that points, toward the origin., 32. The temperature at a point 共x, y, z兲 is given by, , T共x, y, z兲 苷 200e⫺x, , 2, , ⫺3y 2⫺9z 2, , where T is measured in ⬚C and x, y, z in meters., (a) Find the rate of change of temperature at the point, P共2, ⫺1, 2兲 in the direction toward the point 共3, ⫺3, 3兲., (b) In which direction does the temperature increase fastest, at P ?, (c) Find the maximum rate of increase at P., , 2000 m, , 34. Suppose you are climbing a hill whose shape is given by the, , equation z 苷 1000 ⫺ 0.005x 2 ⫺ 0.01y 2, where x, y, and z are, measured in meters, and you are standing at a point with coordinates 共60, 40, 966兲. The positive x-axis points east and the, positive y-axis points north., (a) If you walk due south, will you start to ascend or descend?, At what rate?, (b) If you walk northwest, will you start to ascend or descend?, At what rate?, (c) In which direction is the slope largest? What is the rate of, ascent in that direction? At what angle above the horizontal, does the path in that direction begin?, , 2200 m, , North Thompson River, Reproduced with the permission of Natural Resources Canada 2009,, courtesy of the Centre of Topographic Information., , 37. Show that the operation of taking the gradient of a function has, the given property. Assume that u and v are differentiable func-, , tions of x and y and that a, b are constants., (a) ⵜ共au ⫹ b v兲 苷 a ⵜu ⫹ b ⵜv, , 冉冊, , (c) ⵜ, , u, v, , 苷, , (b) ⵜ共u v兲 苷 u ⵜv ⫹ v ⵜu, , v ⵜu ⫺ u ⵜv, , (d) ⵜu n 苷 nu n⫺1 ⵜu, , v2, , 38. Sketch the gradient vector ⵜ f 共4, 6兲 for the function f whose, , level curves are shown. Explain how you chose the direction, and length of this vector., y, , 33. Suppose that over a certain region of space the electrical poten-, , tial V is given by V共x, y, z兲 苷 5x 2 ⫺ 3xy ⫹ xyz., (a) Find the rate of change of the potential at P共3, 4, 5兲 in the, direction of the vector v 苷 i ⫹ j ⫺ k., (b) In which direction does V change most rapidly at P ?, (c) What is the maximum rate of change at P ?, , 2200 m, , B, 1000 m, , _5, 6, , (4, 6), , _3, _1, , 4, , 0, 1, , 3, , 5, , 2, , 0, , 2, , 4, , 6, , x, , 39. The second directional derivative of f 共x, y兲 is, , Du2 f 共x, y兲 苷 Du 关Du f 共x, y兲兴, If f 共x, y兲 苷 x 3 ⫹ 5x 2 y ⫹ y 3 and u 苷, Du2 f 共2, 1兲., , 具 35 , 45 典 , calculate, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p960-969.qk_97817_14_ch14_p960-969 11/8/10 1:31 PM Page 969, , SECTION 14.6, , 40. (a) If u 苷 具a, b典 is a unit vector and f has continuous, , x 2 ⫹ y 2 ⫹ z 2 ⫺ 8x ⫺ 6y ⫺ 8z ⫹ 24 苷 0 are tangent to each, other at the point 共1, 1, 2兲. (This means that they have a common tangent plane at the point.), , Du2 f 苷 fxx a 2 ⫹ 2 fxy ab ⫹ fyy b 2, (b) Find the second directional derivative of f 共x, y兲 苷 xe 2y in, the direction of v 苷 具 4, 6典 ., , 57. Show that every plane that is tangent to the cone, , x 2 ⫹ y 2 苷 z 2 passes through the origin., 58. Show that every normal line to the sphere x 2 ⫹ y 2 ⫹ z 2 苷 r 2, , 41– 46 Find equations of (a) the tangent plane and (b) the normal, , passes through the center of the sphere., , line to the given surface at the specified point., , 42. y 苷 x 2 ⫺ z 2,, 43. xyz 2 苷 6,, , 59. Where does the normal line to the paraboloid z 苷 x 2 ⫹ y 2 at, , 共3, 3, 5兲, , the point 共1, 1, 2兲 intersect the paraboloid a second time?, , 共4, 7, 3兲, , 60. At what points does the normal line through the point, , 共3, 2, 1兲, , 44. xy ⫹ yz ⫹ zx 苷 5,, 45. x ⫹ y ⫹ z 苷 e ,, , 共1, 2, 1兲 on the ellipsoid 4x 2 ⫹ y 2 ⫹ 4z 2 苷 12 intersect the, sphere x 2 ⫹ y 2 ⫹ z 2 苷 102?, , 共1, 2, 1兲, , 61. Show that the sum of the x-, y-, and z-intercepts of any, , 共0, 0, 1兲, , xyz, , 46. x 4 ⫹ y 4 ⫹ z 4 苷 3x 2 y 2z 2,, , 969, , 56. Show that the ellipsoid 3x 2 ⫹ 2y 2 ⫹ z 2 苷 9 and the sphere, , second partial derivatives, show that, , 41. 2共x ⫺ 2兲 2 ⫹ 共 y ⫺ 1兲 2 ⫹ 共z ⫺ 3兲 2 苷 10,, , DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR, , tangent plane to the surface sx ⫹ sy ⫹ sz 苷 sc is a, constant., , 共1, 1, 1兲, , 62. Show that the pyramids cut off from the first octant by any, , tangent planes to the surface xyz 苷 1 at points in the first, octant must all have the same volume., , ; 47– 48 Use a computer to graph the surface, the tangent plane, and, the normal line on the same screen. Choose the domain carefully, so that you avoid extraneous vertical planes. Choose the, viewpoint so that you get a good view of all three objects., 47. x y ⫹ yz ⫹ zx 苷 3,, , 共1, 1, 1兲, , 48. x yz 苷 6,, , 63. Find parametric equations for the tangent line to the curve of, , intersection of the paraboloid z 苷 x 2 ⫹ y 2 and the ellipsoid, 4x 2 ⫹ y 2 ⫹ z 2 苷 9 at the point 共⫺1, 1, 2兲., , 共1, 2, 3兲, , 64. (a) The plane y ⫹ z 苷 3 intersects the cylinder x 2 ⫹ y 2 苷 5, , 49. If f 共x, y兲 苷 xy, find the gradient vector ⵜ f 共3, 2兲 and use it, , to find the tangent line to the level curve f 共x, y兲 苷 6 at the, point 共3, 2兲. Sketch the level curve, the tangent line, and the, gradient vector., , ;, , in an ellipse. Find parametric equations for the tangent, line to this ellipse at the point 共1, 2, 1兲., (b) Graph the cylinder, the plane, and the tangent line on the, same screen., 65. (a) Two surfaces are called orthogonal at a point of inter-, , 50. If t共x, y兲 苷 x 2 ⫹ y 2 ⫺ 4x, find the gradient vector ⵜt共1, 2兲, and use it to find the tangent line to the level curve, t共x, y兲 苷 1 at the point 共1, 2兲. Sketch the level curve, the tangent line, and the gradient vector., , section if their normal lines are perpendicular at that, point. Show that surfaces with equations F共x, y, z兲 苷 0, and G共x, y, z兲 苷 0 are orthogonal at a point P where, ⵜF 苷 0 and ⵜG 苷 0 if and only if, Fx Gx ⫹ Fy Gy ⫹ Fz Gz 苷 0, , 51. Show that the equation of the tangent plane to the ellipsoid, , x 兾a ⫹ y 兾b ⫹ z 兾c 苷 1 at the point 共x 0 , y0 , z0 兲 can be, written as, xx 0, yy0, zz0, ⫹ 2 ⫹ 2 苷1, a2, b, c, 2, , 2, , 2, , 2, , 2, , 2, , (b) Use part (a) to show that the surfaces z 2 苷 x 2 ⫹ y 2 and, x 2 ⫹ y 2 ⫹ z 2 苷 r 2 are orthogonal at every point of, intersection. Can you see why this is true without using, calculus?, 3, x y is continuous and, 66. (a) Show that the function f 共x, y兲 苷 s, , 52. Find the equation of the tangent plane to the hyperboloid, , x 2兾a 2 ⫹ y 2兾b 2 ⫺ z 2兾c 2 苷 1 at 共x 0 , y0 , z0 兲 and express it in a, form similar to the one in Exercise 51., , ;, 53. Show that the equation of the tangent plane to the elliptic, , paraboloid z兾c 苷 x 2兾a 2 ⫹ y 2兾b 2 at the point 共x 0 , y0 , z0 兲 can, be written as, 2xx 0, 2yy0, z ⫹ z0, ⫹, 苷, a2, b2, c, 54. At what point on the paraboloid y 苷 x 2 ⫹ z 2 is the tangent, , plane parallel to the plane x ⫹ 2y ⫹ 3z 苷 1?, , 55. Are there any points on the hyperboloid x 2 ⫺ y 2 ⫺ z 2 苷 1, , where the tangent plane is parallel to the plane z 苷 x ⫹ y?, , at P, , the partial derivatives fx and fy exist at the origin but the, directional derivatives in all other directions do not exist., (b) Graph f near the origin and comment on how the graph, confirms part (a)., 67. Suppose that the directional derivatives of f 共x, y兲 are known, , at a given point in two nonparallel directions given by unit, vectors u and v. Is it possible to find ⵜ f at this point? If so,, how would you do it?, 68. Show that if z 苷 f 共x, y兲 is differentiable at x 0 苷 具x 0 , y0 典, then, , lim, , x l x0, , f 共x兲 ⫺ f 共x 0 兲 ⫺ ⵜ f 共x 0 兲 ⴢ 共x ⫺ x 0 兲, 苷0, x ⫺ x0, , ⱍ, , ⱍ, , [Hint: Use Definition 14.4.7 directly.], , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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970, , CHAPTER 14, , PARTIAL DERIVATIVES, , Maximum and Minimum Values, , 14.7, , z, , absolute, maximum, , local, maximum, , y, , x, , absolute, minimum, , local, minimum, , FIGURE 1, , As we saw in Chapter 3, one of the main uses of ordinary derivatives is in finding maximum and minimum values (extreme values). In this section we see how to use partial, derivatives to locate maxima and minima of functions of two variables. In particular, in, Example 6 we will see how to maximize the volume of a box without a lid if we have a fixed, amount of cardboard to work with., Look at the hills and valleys in the graph of f shown in Figure 1. There are two points, 共a, b兲 where f has a local maximum, that is, where f 共a, b兲 is larger than nearby values of, f 共x, y兲. The larger of these two values is the absolute maximum. Likewise, f has two local, minima, where f 共a, b兲 is smaller than nearby values. The smaller of these two values is the, absolute minimum., 1 Definition A function of two variables has a local maximum at 共a, b兲 if, f 共x, y兲 艋 f 共a, b兲 when 共x, y兲 is near 共a, b兲. [This means that f 共x, y兲 艋 f 共a, b兲 for, all points 共x, y兲 in some disk with center 共a, b兲.] The number f 共a, b兲 is called a, local maximum value. If f 共x, y兲 艌 f 共a, b兲 when 共x, y兲 is near 共a, b兲, then f has a, local minimum at 共a, b兲 and f 共a, b兲 is a local minimum value., , If the inequalities in Definition 1 hold for all points 共x, y兲 in the domain of f , then f has, an absolute maximum (or absolute minimum) at 共a, b兲., Notice that the conclusion of Theorem 2 can, be stated in the notation of gradient vectors, as ⵜf 共a, b兲 苷 0., , 2, , Theorem If f has a local maximum or minimum at 共a, b兲 and the first-order, partial derivatives of f exist there, then fx 共a, b兲 苷 0 and fy 共a, b兲 苷 0., , PROOF Let t共x兲 苷 f 共x, b兲. If f has a local maximum (or minimum) at 共a, b兲, then t has a, , local maximum (or minimum) at a, so t⬘共a兲 苷 0 by Fermat’s Theorem (see Theorem 3.1.4)., But t⬘共a兲 苷 fx 共a, b兲 (see Equation 14.3.1) and so fx 共a, b兲 苷 0. Similarly, by applying Fermat’s Theorem to the function G共 y兲 苷 f 共a, y兲, we obtain fy 共a, b兲 苷 0., If we put fx 共a, b兲 苷 0 and fy 共a, b兲 苷 0 in the equation of a tangent plane (Equation, 14.4.2), we get z 苷 z0 . Thus the geometric interpretation of Theorem 2 is that if the graph, of f has a tangent plane at a local maximum or minimum, then the tangent plane must be, horizontal., A point 共a, b兲 is called a critical point (or stationary point) of f if fx 共a, b兲 苷 0 and, fy共a, b兲 苷 0, or if one of these partial derivatives does not exist. Theorem 2 says that if f, has a local maximum or minimum at 共a, b兲, then 共a, b兲 is a critical point of f . However, as, in single-variable calculus, not all critical points give rise to maxima or minima. At a critical point, a function could have a local maximum or a local minimum or neither., , z, , EXAMPLE 1 Let f 共x, y兲 苷 x 2 ⫹ y 2 ⫺ 2x ⫺ 6y ⫹ 14. Then, , fx 共x, y兲 苷 2x ⫺ 2, (1, 3, 4), , These partial derivatives are equal to 0 when x 苷 1 and y 苷 3, so the only critical point, is 共1, 3兲. By completing the square, we find that, , 0, x, , FIGURE 2, , z=≈+¥-2x-6y+14, , fy 共x, y兲 苷 2y ⫺ 6, , y, , f 共x, y兲 苷 4 ⫹ 共x ⫺ 1兲2 ⫹ 共 y ⫺ 3兲2, Since 共x ⫺ 1兲2 艌 0 and 共y ⫺ 3兲2 艌 0, we have f 共x, y兲 艌 4 for all values of x and y., Therefore f 共1, 3兲 苷 4 is a local minimum, and in fact it is the absolute minimum of f ., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p970-979.qk_97817_14_ch14_p970-979 11/8/10 1:32 PM Page 971, , SECTION 14.7, , MAXIMUM AND MINIMUM VALUES, , 971, , This can be confirmed geometrically from the graph of f, which is the elliptic paraboloid, with vertex 共1, 3, 4兲 shown in Figure 2., EXAMPLE 2 Find the extreme values of f 共x, y兲 苷 y 2 ⫺ x 2., SOLUTION Since fx 苷 ⫺2x and fy 苷 2y, the only critical point is 共0, 0兲. Notice that, , z, , x, , FIGURE 3, , Example 2 illustrates the fact that a function need not have a maximum or minimum, value at a critical point. Figure 3 shows how this is possible. The graph of f is the hyperbolic paraboloid z 苷 y 2 ⫺ x 2, which has a horizontal tangent plane (z 苷 0) at the origin., You can see that f 共0, 0兲 苷 0 is a maximum in the direction of the x-axis but a minimum in, the direction of the y-axis. Near the origin the graph has the shape of a saddle and so 共0, 0兲, is called a saddle point of f ., A mountain pass also has the shape of a saddle. As the photograph of the geological formation illustrates, for people hiking in one direction the saddle point is the lowest point on, their route, while for those traveling in a different direction the saddle point is the highest, point., We need to be able to determine whether or not a function has an extreme value at a critical point. The following test, which is proved at the end of this section, is analogous to the, Second Derivative Test for functions of one variable., 3 Second Derivatives Test Suppose the second partial derivatives of f are continuous on a disk with center 共a, b兲, and suppose that fx 共a, b兲 苷 0 and fy 共a, b兲 苷 0, [that is, 共a, b兲 is a critical point of f ]. Let, , Photo by Stan Wagon, Macalester College, , z=¥-≈, , y, , for points on the x-axis we have y 苷 0, so f 共x, y兲 苷 ⫺x 2 ⬍ 0 (if x 苷 0). However, for, points on the y-axis we have x 苷 0, so f 共x, y兲 苷 y 2 ⬎ 0 (if y 苷 0). Thus every disk, with center 共0, 0兲 contains points where f takes positive values as well as points where, f takes negative values. Therefore f 共0, 0兲 苷 0 can’t be an extreme value for f , so f has, no extreme value., , D 苷 D共a, b兲 苷 fxx 共a, b兲 fyy 共a, b兲 ⫺ 关 fx y 共a, b兲兴 2, (a) If D ⬎ 0 and fxx 共a, b兲 ⬎ 0, then f 共a, b兲 is a local minimum., (b) If D ⬎ 0 and fxx 共a, b兲 ⬍ 0, then f 共a, b兲 is a local maximum., (c) If D ⬍ 0, then f 共a, b兲 is not a local maximum or minimum., NOTE 1 In case (c) the point 共a, b兲 is called a saddle point of f and the graph of f, crosses its tangent plane at 共a, b兲., NOTE 2 If D 苷 0, the test gives no information: f could have a local maximum or local, minimum at 共a, b兲, or 共a, b兲 could be a saddle point of f ., NOTE 3 To remember the formula for D, it’s helpful to write it as a determinant:, , D苷, , 冟, , 冟, , fxx fx y, 苷 fxx fyy ⫺ 共 fx y 兲2, fyx fyy, , v, , EXAMPLE 3 Find the local maximum and minimum values and saddle points of, f 共x, y兲 苷 x 4 ⫹ y 4 ⫺ 4xy ⫹ 1., , SOLUTION We first locate the critical points:, , fx 苷 4x 3 ⫺ 4y, , fy 苷 4y 3 ⫺ 4x, , Setting these partial derivatives equal to 0, we obtain the equations, x3 ⫺ y 苷 0, , and, , y3 ⫺ x 苷 0, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p970-979.qk_97817_14_ch14_p970-979 11/8/10 1:32 PM Page 973, , SECTION 14.7, , MAXIMUM AND MINIMUM VALUES, , 973, , In the first case (x 苷 0), Equation 5 becomes ⫺4y共1 ⫹ y 2 兲 苷 0, so y 苷 0 and we, have the critical point 共0, 0兲., In the second case 共10y ⫺ 5 ⫺ 2x 2 苷 0兲, we get, x 2 苷 5y ⫺ 2.5, , 6, , and, putting this in Equation 5, we have 25y ⫺ 12.5 ⫺ 4y ⫺ 4y 3 苷 0. So we have to, solve the cubic equation, 4y 3 ⫺ 21y ⫹ 12.5 苷 0, , 7, , Using a graphing calculator or computer to graph the function, t共y兲 苷 4y 3 ⫺ 21y ⫹ 12.5, _3, , 2.7, , as in Figure 6, we see that Equation 7 has three real roots. By zooming in, we can find, the roots to four decimal places:, y ⬇ ⫺2.5452, , FIGURE 6, , y ⬇ 0.6468, , y ⬇ 1.8984, , (Alternatively, we could have used Newton’s method or a rootfinder to locate these, roots.) From Equation 6, the corresponding x-values are given by, x 苷 ⫾s5y ⫺ 2.5, If y ⬇ ⫺2.5452, then x has no corresponding real values. If y ⬇ 0.6468, then, x ⬇ ⫾0.8567. If y ⬇ 1.8984, then x ⬇ ⫾2.6442. So we have a total of five critical, points, which are analyzed in the following chart. All quantities are rounded to two, decimal places., Critical point, , Value of f, , fxx, , D, , Conclusion, , 共0, 0兲, , 0.00, , ⫺10.00, , 80.00, , local maximum, , 共⫾2.64, 1.90兲, , 8.50, , ⫺55.93, , 2488.72, , local maximum, , 共⫾0.86, 0.65兲, , ⫺1.48, , ⫺5.87, , ⫺187.64, , saddle point, , Figures 7 and 8 give two views of the graph of f and we see that the surface opens, downward. [This can also be seen from the expression for f 共x, y兲: The dominant terms, are ⫺x 4 ⫺ 2y 4 when x and y are large.] Comparing the values of f at its local maximum points, we see that the absolute maximum value of f is f 共⫾2.64, 1.90兲 ⬇ 8.50. In, other words, the highest points on the graph of f are 共⫾2.64, 1.90, 8.50兲., , ⱍ ⱍ, , ⱍ ⱍ, , z, , z, , TEC Visual 14.7 shows several families, of surfaces. The surface in Figures 7 and 8, is a member of one of these families., , x, x, , FIGURE 7, , y, y, , FIGURE 8, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p970-979.qk_97817_14_ch14_p970-979 11/8/10 1:32 PM Page 975, , SECTION 14.7, , MAXIMUM AND MINIMUM VALUES, , 975, , Solving this equation for z, we get z 苷 共12 ⫺ xy兲兾关2共x ⫹ y兲兴, so the expression for V, becomes, 12 ⫺ xy, 12xy ⫺ x 2 y 2, V 苷 xy, 苷, 2共x ⫹ y兲, 2共x ⫹ y兲, We compute the partial derivatives:, ⭸V, y 2共12 ⫺ 2xy ⫺ x 2 兲, 苷, ⭸x, 2共x ⫹ y兲2, , ⭸V, x 2共12 ⫺ 2xy ⫺ y 2 兲, 苷, ⭸y, 2共x ⫹ y兲2, , If V is a maximum, then ⭸V兾⭸x 苷 ⭸V兾⭸y 苷 0, but x 苷 0 or y 苷 0 gives V 苷 0, so we, must solve the equations, 12 ⫺ 2xy ⫺ x 2 苷 0, , 12 ⫺ 2xy ⫺ y 2 苷 0, , These imply that x 2 苷 y 2 and so x 苷 y. (Note that x and y must both be positive in this, problem.) If we put x 苷 y in either equation we get 12 ⫺ 3x 2 苷 0, which gives x 苷 2,, y 苷 2, and z 苷 共12 ⫺ 2 ⴢ 2兲兾关2共2 ⫹ 2兲兴 苷 1., We could use the Second Derivatives Test to show that this gives a local maximum, of V, or we could simply argue from the physical nature of this problem that there must, be an absolute maximum volume, which has to occur at a critical point of V, so it must, occur when x 苷 2, y 苷 2, z 苷 1. Then V 苷 2 ⴢ 2 ⴢ 1 苷 4, so the maximum volume of, the box is 4 m3., , Absolute Maximum and Minimum Values, , (a) Closed sets, , For a function f of one variable, the Extreme Value Theorem says that if f is continuous on, a closed interval 关a, b兴, then f has an absolute minimum value and an absolute maximum, value. According to the Closed Interval Method in Section 3.1, we found these by evaluating f not only at the critical numbers but also at the endpoints a and b., There is a similar situation for functions of two variables. Just as a closed interval contains its endpoints, a closed set in ⺢ 2 is one that contains all its boundary points. [A boundary point of D is a point 共a, b兲 such that every disk with center 共a, b兲 contains points in D, and also points not in D.] For instance, the disk, , ⱍ, , D 苷 兵共x, y兲 x 2 ⫹ y 2 艋 1其, , (b) Sets that are not closed, FIGURE 11, , which consists of all points on and inside the circle x 2 ⫹ y 2 苷 1, is a closed set because it, contains all of its boundary points (which are the points on the circle x 2 ⫹ y 2 苷 1). But if, even one point on the boundary curve were omitted, the set would not be closed. (See, Figure 11.), A bounded set in ⺢ 2 is one that is contained within some disk. In other words, it is finite, in extent. Then, in terms of closed and bounded sets, we can state the following counterpart, of the Extreme Value Theorem in two dimensions., , 8 Extreme Value Theorem for Functions of Two Variables If f is continuous on a, closed, bounded set D in ⺢ 2, then f attains an absolute maximum value f 共x 1, y1兲, and an absolute minimum value f 共x 2 , y2 兲 at some points 共x 1, y1兲 and 共x 2 , y2兲 in D., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p970-979.qk_97817_14_ch14_p970-979 11/8/10 1:32 PM Page 976, , 976, , CHAPTER 14, , PARTIAL DERIVATIVES, , To find the extreme values guaranteed by Theorem 8, we note that, by Theorem 2, if f, has an extreme value at 共x 1, y1兲, then 共x 1, y1兲 is either a critical point of f or a boundary, point of D. Thus we have the following extension of the Closed Interval Method., 9 To find the absolute maximum and minimum values of a continuous function, f on a closed, bounded set D :, 1. Find the values of f at the critical points of f in D., 2. Find the extreme values of f on the boundary of D., 3. The largest of the values from steps 1 and 2 is the absolute maximum value;, , the smallest of these values is the absolute minimum value., , EXAMPLE 7 Find the absolute maximum and minimum values of the function, f 共x, y兲 苷 x 2 ⫺ 2xy ⫹ 2y on the rectangle D 苷 兵共x, y兲 0 艋 x 艋 3, 0 艋 y 艋 2其., , ⱍ, , SOLUTION Since f is a polynomial, it is continuous on the closed, bounded rectangle D,, , so Theorem 8 tells us there is both an absolute maximum and an absolute minimum., According to step 1 in 9 , we first find the critical points. These occur when, fx 苷 2x ⫺ 2y 苷 0, , so the only critical point is 共1, 1兲, and the value of f there is f 共1, 1兲 苷 1., In step 2 we look at the values of f on the boundary of D, which consists of the four, line segments L 1 , L 2 , L 3 , L 4 shown in Figure 12. On L 1 we have y 苷 0 and, , y, (0, 2), , L£, , (2, 2), , L¢, , (3, 2), , f 共x, 0兲 苷 x 2, , L™, , (0, 0), , L¡, , (3, 0), , fy 苷 ⫺2x ⫹ 2 苷 0, , x, , FIGURE 12, , 0艋x艋3, , This is an increasing function of x, so its minimum value is f 共0, 0兲 苷 0 and its maximum value is f 共3, 0兲 苷 9. On L 2 we have x 苷 3 and, f 共3, y兲 苷 9 ⫺ 4y, , 0艋y艋2, , This is a decreasing function of y, so its maximum value is f 共3, 0兲 苷 9 and its minimum, value is f 共3, 2兲 苷 1. On L 3 we have y 苷 2 and, f 共x, 2兲 苷 x 2 ⫺ 4x ⫹ 4, 9, , 0艋x艋3, , By the methods of Chapter 3, or simply by observing that f 共x, 2兲 苷 共x ⫺ 2兲2, we see, that the minimum value of this function is f 共2, 2兲 苷 0 and the maximum value is, f 共0, 2兲 苷 4. Finally, on L 4 we have x 苷 0 and, f 共0, y兲 苷 2y, , 0, , D, L¡, 30, , FIGURE 13, f(x, y)=≈-2xy+2y, , 2, , L™, , 0艋y艋2, , with maximum value f 共0, 2兲 苷 4 and minimum value f 共0, 0兲 苷 0. Thus, on the boundary, the minimum value of f is 0 and the maximum is 9., In step 3 we compare these values with the value f 共1, 1兲 苷 1 at the critical point and, conclude that the absolute maximum value of f on D is f 共3, 0兲 苷 9 and the absolute, minimum value is f 共0, 0兲 苷 f 共2, 2兲 苷 0. Figure 13 shows the graph of f ., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p970-979.qk_97817_14_ch14_p970-979 11/8/10 1:32 PM Page 979, , MAXIMUM AND MINIMUM VALUES, , SECTION 14.7, , an absolute maximum. But this is not true for functions of two, variables. Show that the function, f 共x, y兲 苷 3xe y ⫺ x 3 ⫺ e 3y, has exactly one critical point, and that f has a local maximum, there that is not an absolute maximum. Then use a computer to, produce a graph with a carefully chosen domain and viewpoint, to see how this is possible., 39. Find the shortest distance from the point 共2, 0, ⫺3兲 to the plane, , x ⫹ y ⫹ z 苷 1., , 40. Find the point on the plane x ⫺ 2y ⫹ 3z 苷 6 that is closest to, , the point 共0, 1, 1兲., , (b) Find the dimensions that minimize heat loss. (Check both, the critical points and the points on the boundary of the, domain.), (c) Could you design a building with even less heat loss, if the restrictions on the lengths of the walls were removed?, 53. If the length of the diagonal of a rectangular box must be L ,, , what is the largest possible volume?, 54. Three alleles (alternative versions of a gene) A, B, and O, , determine the four blood types A (AA or AO), B (BB or BO),, O (OO), and AB. The Hardy-Weinberg Law states that the proportion of individuals in a population who carry two different, alleles is, P 苷 2pq ⫹ 2pr ⫹ 2rq, , 41. Find the points on the cone z 2 苷 x 2 ⫹ y 2 that are closest to the, , point 共4, 2, 0兲., , 42. Find the points on the surface y 苷 9 ⫹ xz that are closest to, 2, , the origin., 43. Find three positive numbers whose sum is 100 and whose, , product is a maximum., 44. Find three positive numbers whose sum is 12 and the sum of, , whose squares is as small as possible., 45. Find the maximum volume of a rectangular box that is, , inscribed in a sphere of radius r., 46. Find the dimensions of the box with volume 1000 cm3 that has, , minimal surface area., , 979, , where p, q, and r are the proportions of A, B, and O in the, population. Use the fact that p ⫹ q ⫹ r 苷 1 to show that P is, at most 23., 55. Suppose that a scientist has reason to believe that two quanti-, , ties x and y are related linearly, that is, y 苷 mx ⫹ b, at least, approximately, for some values of m and b. The scientist, performs an experiment and collects data in the form of points, 共x 1, y1兲, 共x 2 , y2 兲, . . . , 共x n , yn 兲, and then plots these points. The, points don’t lie exactly on a straight line, so the scientist wants, to find constants m and b so that the line y 苷 mx ⫹ b “fits” the, points as well as possible (see the figure)., y, (x i, yi ), , 47. Find the volume of the largest rectangular box in the first, , di, , octant with three faces in the coordinate planes and one, vertex in the plane x ⫹ 2y ⫹ 3z 苷 6., , (⁄, ›), , mx i+b, , 48. Find the dimensions of the rectangular box with largest, 2, , volume if the total surface area is given as 64 cm ., 49. Find the dimensions of a rectangular box of maximum volume, , 0, , x, , such that the sum of the lengths of its 12 edges is a constant c., 50. The base of an aquarium with given volume V is made of slate, , and the sides are made of glass. If slate costs five times as, much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials., , Let di 苷 yi ⫺ 共mx i ⫹ b兲 be the vertical deviation of the point, 共x i , yi兲 from the line. The method of least squares determines, m and b so as to minimize 冘ni苷1 di2 , the sum of the squares of, these deviations. Show that, according to this method, the line, of best fit is obtained when, , 51. A cardboard box without a lid is to have a volume of, , n, , 32,000 cm3. Find the dimensions that minimize the amount, of cardboard used., 52. A rectangular building is being designed to minimize, , heat loss. The east and west walls lose heat at a rate of, 10 units兾m2 per day, the north and south walls at a rate of, 8 units兾m2 per day, the floor at a rate of 1 unit兾m2 per day, and, the roof at a rate of 5 units兾m2 per day. Each wall must be at, least 30 m long, the height must be at least 4 m, and the, volume must be exactly 4000 m3., (a) Find and sketch the domain of the heat loss as a function of, the lengths of the sides., , m, , 兺x, , n, , i, , ⫹ bn 苷, , i苷1, n, , m, , 兺x, , i苷1, , ⫹b, , i, , i苷1, n, , 2, i, , 兺y, , 兺x, , i苷1, , n, , i, , 苷, , 兺xy, , i i, , i苷1, , Thus the line is found by solving these two equations in the, two unknowns m and b. (See Section 1.2 for a further discussion and applications of the method of least squares.), 56. Find an equation of the plane that passes through the point, , 共1, 2, 3兲 and cuts off the smallest volume in the first octant., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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980, , CHAPTER 14, , PARTIAL DERIVATIVES, , APPLIED PROJECT, , DESIGNING A DUMPSTER, For this project we locate a rectangular trash Dumpster in order to study its shape and construction. We then attempt to determine the dimensions of a container of similar design that minimize, construction cost., 1. First locate a trash Dumpster in your area. Carefully study and describe all details of its, , construction, and determine its volume. Include a sketch of the container., 2. While maintaining the general shape and method of construction, determine the dimensions, , such a container of the same volume should have in order to minimize the cost of construction. Use the following assumptions in your analysis:, ■, , The sides, back, and front are to be made from 12-gauge (0.1046 inch thick) steel sheets,, which cost $0.70 per square foot (including any required cuts or bends)., , ■, , The base is to be made from a 10-gauge (0.1345 inch thick) steel sheet, which costs $0.90, per square foot., , ■, , Lids cost approximately $50.00 each, regardless of dimensions., , ■, , Welding costs approximately $0.18 per foot for material and labor combined., , Give justification of any further assumptions or simplifications made of the details of, construction., 3. Describe how any of your assumptions or simplifications may affect the final result., 4. If you were hired as a consultant on this investigation, what would your conclusions be?, , Would you recommend altering the design of the Dumpster? If so, describe the savings that, would result., , DISCOVERY PROJECT, , QUADRATIC APPROXIMATIONS AND CRITICAL POINTS, The Taylor polynomial approximation to functions of one variable that we discussed in Chapter 11, can be extended to functions of two or more variables. Here we investigate quadratic approximations to functions of two variables and use them to give insight into the Second Derivatives Test, for classifying critical points., In Section 14.4 we discussed the linearization of a function f of two variables at a point 共a, b兲:, L共x, y兲 苷 f 共a, b兲 ⫹ fx 共a, b兲共x ⫺ a兲 ⫹ fy 共a, b兲共 y ⫺ b兲, Recall that the graph of L is the tangent plane to the surface z 苷 f 共x, y兲 at 共a, b, f 共a, b兲兲 and the, corresponding linear approximation is f 共x, y兲 ⬇ L共x, y兲. The linearization L is also called the, first-degree Taylor polynomial of f at 共a, b兲., 1. If f has continuous second-order partial derivatives at 共a, b兲, then the second-degree, , Taylor polynomial of f at 共a, b兲 is, , Q共x, y兲 苷 f 共a, b兲 ⫹ fx 共a, b兲共x ⫺ a兲 ⫹ fy 共a, b兲共 y ⫺ b兲, ⫹ 12 fxx 共a, b兲共x ⫺ a兲2 ⫹ fx y 共a, b兲共x ⫺ a兲共y ⫺ b兲 ⫹ 12 fyy 共a, b兲共y ⫺ b兲2, and the approximation f 共x, y兲 ⬇ Q共x, y兲 is called the quadratic approximation to f at, 共a, b兲. Verify that Q has the same first- and second-order partial derivatives as f at 共a, b兲., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 14.8, , LAGRANGE MULTIPLIERS, 2, , 2. (a) Find the first- and second-degree Taylor polynomials L and Q of f 共x, y兲 苷 e⫺x ⫺y, , ;, , 981, 2, , at (0, 0)., (b) Graph f , L , and Q. Comment on how well L and Q approximate f ., 3. (a) Find the first- and second-degree Taylor polynomials L and Q for f 共x, y兲 苷 xe y, , ;, , at (1, 0)., (b) Compare the values of L , Q, and f at (0.9, 0.1)., (c) Graph f , L , and Q. Comment on how well L and Q approximate f ., 4. In this problem we analyze the behavior of the polynomial f 共x, y兲 苷 ax 2 ⫹ b x y ⫹ cy 2, , (without using the Second Derivatives Test) by identifying the graph as a paraboloid., (a) By completing the square, show that if a 苷 0, then, f 共x, y兲 苷 ax 2 ⫹ bx y ⫹ cy 2 苷 a, , 冋冉, , x⫹, , 冊 冉, , b, y, 2a, , 2, , ⫹, , 冊册, , 4ac ⫺ b 2 2, y, 4a 2, , (b) Let D 苷 4ac ⫺ b 2. Show that if D ⬎ 0 and a ⬎ 0, then f has a local minimum, at (0, 0)., (c) Show that if D ⬎ 0 and a ⬍ 0, then f has a local maximum at (0, 0)., (d) Show that if D ⬍ 0, then (0, 0) is a saddle point., 5. (a) Suppose f is any function with continuous second-order partial derivatives such that, , f 共0, 0兲 苷 0 and (0, 0) is a critical point of f . Write an expression for the seconddegree Taylor polynomial, Q, of f at (0, 0)., (b) What can you conclude about Q from Problem 4?, (c) In view of the quadratic approximation f 共x, y兲 ⬇ Q共x, y兲, what does part (b) suggest, about f ?, , ;, , 14.8, , Graphing calculator or computer required, , Lagrange Multipliers, , y, , g(x, y)=k, , f(x, y)=11, f(x, y)=10, f(x, y)=9, f(x, y)=8, f(x, y)=7, , 0, , FIGURE 1, , TEC Visual 14.8 animates Figure 1 for both, level curves and level surfaces., , x, , In Example 6 in Section 14.7 we maximized a volume function V 苷 xyz subject to the constraint 2xz ⫹ 2yz ⫹ xy 苷 12, which expressed the side condition that the surface area was, 12 m2. In this section we present Lagrange’s method for maximizing or minimizing a general function f 共x, y, z兲 subject to a constraint (or side condition) of the form t共x, y, z兲 苷 k., It’s easier to explain the geometric basis of Lagrange’s method for functions of two variables. So we start by trying to find the extreme values of f 共x, y兲 subject to a constraint of, the form t共x, y兲 苷 k. In other words, we seek the extreme values of f 共x, y兲 when the point, 共x, y兲 is restricted to lie on the level curve t共x, y兲 苷 k. Figure 1 shows this curve together, with several level curves of f . These have the equations f 共x, y兲 苷 c, where c 苷 7, 8, 9, 10,, 11. To maximize f 共x, y兲 subject to t共x, y兲 苷 k is to find the largest value of c such that the, level curve f 共x, y兲 苷 c intersects t共x, y兲 苷 k. It appears from Figure 1 that this happens, when these curves just touch each other, that is, when they have a common tangent line., (Otherwise, the value of c could be increased further.) This means that the normal lines at, the point 共x 0 , y0 兲 where they touch are identical. So the gradient vectors are parallel; that is,, ⵜ f 共x 0 , y0 兲 苷 ␭ ⵜt共x 0 , y0 兲 for some scalar ␭., This kind of argument also applies to the problem of finding the extreme values of, f 共x, y, z兲 subject to the constraint t共x, y, z兲 苷 k. Thus the point 共x, y, z兲 is restricted to lie, on the level surface S with equation t共x, y, z兲 苷 k. Instead of the level curves in Figure 1,, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p980-989.qk_97817_14_ch14_p980-989 11/8/10 1:32 PM Page 987, , SECTION 14.8, , 14.8, , LAGRANGE MULTIPLIERS, , 987, , Exercises, , 1. Pictured are a contour map of f and a curve with equation, , t共x, y兲 苷 8. Estimate the maximum and minimum values, of f subject to the constraint that t共x, y兲 苷 8. Explain your, reasoning., , 16. f 共x, y, z兲 苷 3x ⫺ y ⫺ 3z;, , x ⫹ y ⫺ z 苷 0,, , x 2 ⫹ 2z 2 苷 1, , 17. f 共x, y, z兲 苷 yz ⫹ x y ;, , x y 苷 1,, , 18. f 共x, y, z兲 苷 x 2 ⫹ y 2 ⫹ z 2;, , y, , y 2 ⫹ z2 苷 1, , x ⫺ y 苷 1, y 2 ⫺ z 2 苷 1, , g(x, y)=8, , 19–21 Find the extreme values of f on the region described by, the inequality., , 40, 50, 70, , 19. f 共x, y兲 苷 x 2 ⫹ y 2 ⫹ 4x ⫺ 4y,, , 60, , 0, , 20. f 共x, y兲 苷 2x 2 ⫹ 3y 2 ⫺ 4x ⫺ 5,, 30, , x, , 20, 10, , 21. f 共x, y兲 苷 e ⫺xy,, , x2 ⫹ y2 艋 9, x 2 ⫹ y 2 艋 16, , x 2 ⫹ 4y 2 艋 1, , 22. Consider the problem of maximizing the function, , ; 2. (a) Use a graphing calculator or computer to graph the circle, , f 共x, y兲 苷 2x ⫹ 3y subject to the constraint sx ⫹ sy 苷 5., (a) Try using Lagrange multipliers to solve the problem., (b) Does f 共25, 0兲 give a larger value than the one in part (a)?, (c) Solve the problem by graphing the constraint equation, and several level curves of f., (d) Explain why the method of Lagrange multipliers fails to, solve the problem., (e) What is the significance of f 共9, 4兲?, , x 2 ⫹ y 2 苷 1. On the same screen, graph several curves of, ;, the form x 2 ⫹ y 苷 c until you find two that just touch the, circle. What is the significance of the values of c for these, two curves?, (b) Use Lagrange multipliers to find the extreme values of, f 共x, y兲 苷 x 2 ⫹ y subject to the constraint x 2 ⫹ y 2 苷 1., 23. Consider the problem of minimizing the function f 共x, y兲 苷 x, Compare your answers with those in part (a)., on the curve y 2 ⫹ x 4 ⫺ x 3 苷 0 (a piriform)., (a) Try using Lagrange multipliers to solve the problem., 3–14 Use Lagrange multipliers to find the maximum and mini(b) Show that the minimum value is f 共0, 0兲 苷 0 but the, mum values of the function subject to the given constraint., Lagrange condition ⵜ f 共0, 0兲 苷 ␭ⵜt共0, 0兲 is not satisfied, 3. f 共x, y兲 苷 x 2 ⫹ y 2 ; x y 苷 1, for any value of ␭., (c) Explain why Lagrange multipliers fail to find the mini4. f 共x, y兲 苷 3x ⫹ y; x 2 ⫹ y 2 苷 10, mum value in this case., 1, 5. f 共x, y兲 苷 y 2 ⫺ x 2; 4 x 2 ⫹ y 2 苷 1, CAS 24. (a) If your computer algebra system plots implicitly defined, 6. f 共x, y兲 苷 e xy; x 3 ⫹ y 3 苷 16, curves, use it to estimate the minimum and maximum, 7. f 共x, y, z兲 苷 2x ⫹ 2y ⫹ z; x 2 ⫹ y 2 ⫹ z 2 苷 9, values of f 共x, y兲 苷 x 3 ⫹ y 3 ⫹ 3xy subject to the constraint 共x ⫺ 3兲2 ⫹ 共 y ⫺ 3兲2 苷 9 by graphical methods., 8. f 共x, y, z兲 苷 x 2 ⫹ y 2 ⫹ z 2; x ⫹ y ⫹ z 苷 12, (b) Solve the problem in part (a) with the aid of Lagrange, multipliers. Use your CAS to solve the equations numeri9. f 共x, y, z兲 苷 x yz ; x 2 ⫹ 2y 2 ⫹ 3z 2 苷 6, cally. Compare your answers with those in part (a)., 10. f 共x, y, z兲 苷 x 2 y 2z 2 ; x 2 ⫹ y 2 ⫹ z 2 苷 1, 25. The total production P of a certain product depends on the, 11. f 共x, y, z兲 苷 x 2 ⫹ y 2 ⫹ z 2 ; x 4 ⫹ y 4 ⫹ z 4 苷 1, amount L of labor used and the amount K of capital investment. In Sections 14.1 and 14.3 we discussed how the Cobb12. f 共x, y, z兲 苷 x 4 ⫹ y 4 ⫹ z 4 ; x 2 ⫹ y 2 ⫹ z 2 苷 1, Douglas model P 苷 bL␣K 1⫺␣ follows from certain economic, 2, 2, 2, 2, assumptions, where b and ␣ are positive constants and, 13. f 共x, y, z, t兲 苷 x ⫹ y ⫹ z ⫹ t ; x ⫹ y ⫹ z ⫹ t 苷 1, ␣ ⬍ 1. If the cost of a unit of labor is m and the cost of a unit, 14. f 共x 1, x 2 , . . . , x n兲 苷 x 1 ⫹ x 2 ⫹ ⭈ ⭈ ⭈ ⫹ x n ;, of capital is n, and the company can spend only p dollars as, its total budget, then maximizing the production P is subject, x 12 ⫹ x 22 ⫹ ⭈ ⭈ ⭈ ⫹ x n2 苷 1, to the constraint mL ⫹ nK 苷 p. Show that the maximum, production occurs when, 15–18 Find the extreme values of f subject to both constraints., ␣p, 共1 ⫺ ␣兲p, L苷, and, K苷, 15. f 共x, y, z兲 苷 x ⫹ 2y ; x ⫹ y ⫹ z 苷 1, y 2 ⫹ z 2 苷 4, m, n, , ;, , Graphing calculator or computer required, , CAS Computer algebra system required, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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988, , CHAPTER 14, , PARTIAL DERIVATIVES, , 26. Referring to Exercise 25, we now suppose that the pro-, , duction is fixed at bL␣K 1⫺␣ 苷 Q, where Q is a constant., What values of L and K minimize the cost function, C共L, K 兲 苷 mL ⫹ nK ?, 27. Use Lagrange multipliers to prove that the rectangle with, , maximum area that has a given perimeter p is a square., 28. Use Lagrange multipliers to prove that the triangle with, , maximum area that has a given perimeter p is equilateral., Hint: Use Heron’s formula for the area:, , (b) Use Lagrange multipliers to find the highest and lowest, points on the ellipse., CAS, , 45 – 46 Find the maximum and minimum values of f subject to, the given constraints. Use a computer algebra system to solve, the system of equations that arises in using Lagrange multipliers., (If your CAS finds only one solution, you may need to use additional commands.), 45. f 共x, y, z兲 苷 ye x⫺z ;, , 9x 2 ⫹ 4y 2 ⫹ 36z 2 苷 36, x y ⫹ yz 苷 1, , 46. f 共x, y, z兲 苷 x ⫹ y ⫹ z;, , x 2 ⫺ y 2 苷 z, x 2 ⫹ z 2 苷 4, , A 苷 ss共s ⫺ x兲共s ⫺ y兲共s ⫺ z兲, where s 苷 p兾2 and x, y, z are the lengths of the sides., 29– 41 Use Lagrange multipliers to give an alternate solution to, , the indicated exercise in Section 14.7., 29. Exercise 39, , 30. Exercise 40, , 31. Exercise 41, , 32. Exercise 42, , 33. Exercise 43, , 34. Exercise 44, , 35. Exercise 45, , 36. Exercise 46, , 37. Exercise 47, , 38. Exercise 48, , 39. Exercise 49, , 40. Exercise 50, , 41. Exercise 53, , 47. (a) Find the maximum value of, n, f 共x1 , x 2 , . . . , x n 兲 苷 s, x1 x 2 ⭈ ⭈ ⭈ x n, , given that x1 , x 2 , . . . , x n are positive numbers and, x1 ⫹ x 2 ⫹ ⭈ ⭈ ⭈ ⫹ x n 苷 c, where c is a constant., (b) Deduce from part (a) that if x1 , x 2 , . . . , x n are positive, numbers, then, n, x1 x 2 ⭈ ⭈ ⭈ x n 艋, s, , x1 ⫹ x 2 ⫹ ⭈ ⭈ ⭈ ⫹ x n, n, , This inequality says that the geometric mean of n, numbers is no larger than the arithmetic mean of the, numbers. Under what circumstances are these two means, equal?, 48. (a) Maximize 冘ni苷1 x i yi subject to the constraints 冘ni苷1 x i2 苷 1, , 42. Find the maximum and minimum volumes of a rectangular, , box whose surface area is 1500 cm2 and whose total edge, length is 200 cm., 43. The plane x ⫹ y ⫹ 2z 苷 2 intersects the paraboloid, , z 苷 x 2 ⫹ y 2 in an ellipse. Find the points on this ellipse, that are nearest to and farthest from the origin., , and 冘ni苷1 y i2 苷 1., (b) Put, ai, xi 苷, s 冘 aj2, , and yi 苷, , bi, s冘 bj2, , to show that, , 兺ab, i, , i, , 艋 s冘 aj2 s冘 bj2, , 44. The plane 4x ⫺ 3y ⫹ 8z 苷 5 intersects the cone, , ;, , z 2 苷 x 2 ⫹ y 2 in an ellipse., (a) Graph the cone, the plane, and the ellipse., , APPLIED PROJECT, , for any numbers a1, . . . , an, b1, . . . , bn. This inequality is, known as the Cauchy-Schwarz Inequality., , ROCKET SCIENCE, Many rockets, such as the Pegasus XL currently used to launch satellites and the Saturn V that first, put men on the moon, are designed to use three stages in their ascent into space. A large first stage, initially propels the rocket until its fuel is consumed, at which point the stage is jettisoned to, reduce the mass of the rocket. The smaller second and third stages function similarly in order to, place the rocket’s payload into orbit about the earth. (With this design, at least two stages are, required in order to reach the necessary velocities, and using three stages has proven to be a good, compromise between cost and performance.) Our goal here is to determine the individual masses, of the three stages, which are to be designed in such a way as to minimize the total mass of the, rocket while enabling it to reach a desired velocity., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p980-989.qk_97817_14_ch14_p980-989 11/8/10 1:33 PM Page 989, , APPLIED PROJECT, , ROCKET SCIENCE, , 989, , For a single-stage rocket consuming fuel at a constant rate, the change in velocity resulting, from the acceleration of the rocket vehicle has been modeled by, , 冉, , ⌬V 苷 ⫺c ln 1 ⫺, , 共1 ⫺ S兲Mr, P ⫹ Mr, , 冊, , where Mr is the mass of the rocket engine including initial fuel, P is the mass of the payload,, S is a structural factor determined by the design of the rocket (specifically, it is the ratio of the, mass of the rocket vehicle without fuel to the total mass of the rocket with payload), and c is the, (constant) speed of exhaust relative to the rocket., Now consider a rocket with three stages and a payload of mass A. Assume that outside forces, are negligible and that c and S remain constant for each stage. If Mi is the mass of the ith stage,, we can initially consider the rocket engine to have mass M1 and its payload to have mass, M2 ⫹ M3 ⫹ A; the second and third stages can be handled similarly., 1. Show that the velocity attained after all three stages have been jettisoned is given by, , 冋冉, , Courtesy of Orbital Sciences Corporation, , vf 苷 c ln, , M1 ⫹ M2 ⫹ M3 ⫹ A, SM1 ⫹ M2 ⫹ M3 ⫹ A, , 冊 冉, ⫹ ln, , M2 ⫹ M3 ⫹ A, SM2 ⫹ M3 ⫹ A, , 冊 冉, ⫹ ln, , M3 ⫹ A, SM3 ⫹ A, , 冊册, , 2. We wish to minimize the total mass M 苷 M1 ⫹ M2 ⫹ M3 of the rocket engine subject, to the constraint that the desired velocity vf from Problem 1 is attained. The method of, , Lagrange multipliers is appropriate here, but difficult to implement using the current expressions. To simplify, we define variables Ni so that the constraint equation may be expressed as, vf 苷 c共ln N1 ⫹ ln N2 ⫹ ln N3 兲. Since M is now difficult to express in terms of the Ni’s, we, wish to use a simpler function that will be minimized at the same place as M. Show that, M1 ⫹ M2 ⫹ M3 ⫹ A, 共1 ⫺ S 兲N1, 苷, M2 ⫹ M3 ⫹ A, 1 ⫺ SN1, M2 ⫹ M3 ⫹ A, 共1 ⫺ S 兲N2, 苷, M3 ⫹ A, 1 ⫺ SN2, M3 ⫹ A, 共1 ⫺ S 兲N3, 苷, A, 1 ⫺ SN3, and conclude that, M⫹A, 共1 ⫺ S 兲3N1 N2 N3, 苷, A, 共1 ⫺ SN1兲共1 ⫺ SN2 兲共1 ⫺ SN3 兲, 3. Verify that ln共共M ⫹ A兲兾A兲 is minimized at the same location as M ; use Lagrange multipliers, , and the results of Problem 2 to find expressions for the values of Ni where the minimum, occurs subject to the constraint vf 苷 c共ln N1 ⫹ ln N2 ⫹ ln N3 兲. [Hint: Use properties of, logarithms to help simplify the expressions.], 4. Find an expression for the minimum value of M as a function of vf ., 5. If we want to put a three-stage rocket into orbit 100 miles above the earth’s surface, a final, , velocity of approximately 17,500 mi兾h is required. Suppose that each stage is built with a, structural factor S 苷 0.2 and an exhaust speed of c 苷 6000 mi兾h., (a) Find the minimum total mass M of the rocket engines as a function of A., (b) Find the mass of each individual stage as a function of A. (They are not equally sized!), 6. The same rocket would require a final velocity of approximately 24,700 mi兾h in order to, , escape earth’s gravity. Find the mass of each individual stage that would minimize the total, mass of the rocket engines and allow the rocket to propel a 500-pound probe into deep space., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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990, , CHAPTER 14, , PARTIAL DERIVATIVES, , APPLIED PROJECT, , HYDRO-TURBINE OPTIMIZATION, The Katahdin Paper Company in Millinocket, Maine, operates a hydroelectric generating station, on the Penobscot River. Water is piped from a dam to the power station. The rate at which the, water flows through the pipe varies, depending on external conditions., The power station has three different hydroelectric turbines, each with a known (and unique), power function that gives the amount of electric power generated as a function of the water flow, arriving at the turbine. The incoming water can be apportioned in different volumes to each, turbine, so the goal is to determine how to distribute water among the turbines to give the maximum total energy production for any rate of flow., Using experimental evidence and Bernoulli’s equation, the following quadratic models were, determined for the power output of each turbine, along with the allowable flows of operation:, KW1 苷 共⫺18.89 ⫹ 0.1277Q1 ⫺ 4.08 ⴢ 10⫺5Q12 兲共170 ⫺ 1.6 ⴢ 10⫺6QT2 兲, KW2 苷 共⫺24.51 ⫹ 0.1358Q2 ⫺ 4.69 ⴢ 10⫺5Q22 兲共170 ⫺ 1.6 ⴢ 10⫺6QT2 兲, KW3 苷 共⫺27.02 ⫹ 0.1380Q3 ⫺ 3.84 ⴢ 10⫺5Q32 兲共170 ⫺ 1.6 ⴢ 10⫺6QT2 兲, 250 艋 Q1 艋 1110 ,, , 250 艋 Q2 艋 1110 ,, , 250 艋 Q3 艋 1225, , where, Qi 苷 flow through turbine i in cubic feet per second, KWi 苷 power generated by turbine i in kilowatts, QT 苷 total flow through the station in cubic feet per second, 1. If all three turbines are being used, we wish to determine the flow Qi to each turbine that will, , give the maximum total energy production. Our limitations are that the flows must sum to, the total incoming flow and the given domain restrictions must be observed. Consequently,, use Lagrange multipliers to find the values for the individual flows (as functions of QT ) that, maximize the total energy production KW1 ⫹ KW2 ⫹ KW3 subject to the constraints, Q1 ⫹ Q2 ⫹ Q3 苷 QT and the domain restrictions on each Qi ., 2. For which values of QT is your result valid?, 3. For an incoming flow of 2500 ft3兾s, determine the distribution to the turbines and verify, , (by trying some nearby distributions) that your result is indeed a maximum., 4. Until now we have assumed that all three turbines are operating; is it possible in some situa-, , tions that more power could be produced by using only one turbine? Make a graph of the, three power functions and use it to help decide if an incoming flow of 1000 ft3兾s should be, distributed to all three turbines or routed to just one. (If you determine that only one turbine, should be used, which one would it be?) What if the flow is only 600 ft3兾s?, 5. Perhaps for some flow levels it would be advantageous to use two turbines. If the incoming, , flow is 1500 ft3兾s, which two turbines would you recommend using? Use Lagrange multipliers to determine how the flow should be distributed between the two turbines to maximize the energy produced. For this flow, is using two turbines more efficient than using all, three?, 6. If the incoming flow is 3400 ft3兾s, what would you recommend to the company?, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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CHAPTER 14, , 14, , REVIEW, , 991, , Review, , Concept Check, 1. (a) What is a function of two variables?, , (b) Describe three methods for visualizing a function of two, variables., 2. What is a function of three variables? How can you visualize, , such a function?, , 11. State the Chain Rule for the case where z 苷 f 共x, y兲 and x and y, , are functions of one variable. What if x and y are functions of, two variables?, 12. If z is defined implicitly as a function of x and y by an equation, , of the form F共x, y, z兲 苷 0, how do you find ⭸z兾⭸x and ⭸z兾⭸y ?, 13. (a) Write an expression as a limit for the directional derivative, , 3. What does, , lim, , 共x, y兲 l 共a, b兲, , f 共x, y兲 苷 L, , mean? How can you show that such a limit does not exist?, 4. (a) What does it mean to say that f is continuous at 共a, b兲?, , (b) If f is continuous on ⺢2, what can you say about its graph?, , 5. (a) Write expressions for the partial derivatives fx 共a, b兲 and, , fy 共a, b兲 as limits., (b) How do you interpret fx 共a, b兲 and fy 共a, b兲 geometrically?, How do you interpret them as rates of change?, (c) If f 共x, y兲 is given by a formula, how do you calculate fx, and fy ?, , 6. What does Clairaut’s Theorem say?, 7. How do you find a tangent plane to each of the following types, , of surfaces?, (a) A graph of a function of two variables, z 苷 f 共x, y兲, (b) A level surface of a function of three variables,, F共x, y, z兲 苷 k, 8. Define the linearization of f at 共a, b兲. What is the correspond-, , ing linear approximation? What is the geometric interpretation, of the linear approximation?, 9. (a) What does it mean to say that f is differentiable at 共a, b兲?, , (b) How do you usually verify that f is differentiable?, 10. If z 苷 f 共x, y兲, what are the differentials dx, dy, and dz ?, , of f at 共x 0 , y0 兲 in the direction of a unit vector u 苷 具a, b典 ., How do you interpret it as a rate? How do you interpret it, geometrically?, (b) If f is differentiable, write an expression for Du f 共x 0 , y0 兲 in, terms of fx and fy ., 14. (a) Define the gradient vector ⵜ f for a function f of two or, , three variables., (b) Express Du f in terms of ⵜ f ., (c) Explain the geometric significance of the gradient., 15. What do the following statements mean?, , (a), (b), (c), (d), (e), , f, f, f, f, f, , has a local maximum at 共a, b兲., has an absolute maximum at 共a, b兲., has a local minimum at 共a, b兲., has an absolute minimum at 共a, b兲., has a saddle point at 共a, b兲., , 16. (a) If f has a local maximum at 共a, b兲, what can you say about, , its partial derivatives at 共a, b兲?, (b) What is a critical point of f ?, , 17. State the Second Derivatives Test., 18. (a) What is a closed set in ⺢ 2 ? What is a bounded set?, , (b) State the Extreme Value Theorem for functions of two, variables., (c) How do you find the values that the Extreme Value, Theorem guarantees?, 19. Explain how the method of Lagrange multipliers works, , in finding the extreme values of f 共x, y, z兲 subject to the, constraint t共x, y, z兲 苷 k. What if there is a second constraint, h共x, y, z兲 苷 c ?, , True-False Quiz, Determine whether the statement is true or false. If it is true, explain why., If it is false, explain why or give an example that disproves the statement., 1. fy 共a, b兲 苷 lim, , ylb, , f 共a, y兲 ⫺ f 共a, b兲, y⫺b, , 2. There exists a function f with continuous second-order, , partial derivatives such that fx 共x, y兲 苷 x ⫹ y 2 and, fy 共x, y兲 苷 x ⫺ y 2., ⭸2 f, 3. fxy 苷, ⭸x ⭸y, 4. Dk f 共x, y, z兲 苷 fz共x, y, z兲, , 5. If f 共x, y兲 l L as 共x, y兲 l 共a, b兲 along every straight line, , through 共a, b兲, then lim 共x, y兲 l 共a, b兲 f 共x, y兲 苷 L., , 6. If fx 共a, b兲 and fy 共a, b兲 both exist, then f is differentiable, , at 共a, b兲., , 7. If f has a local minimum at 共a, b兲 and f is differentiable at, , 共a, b兲, then ⵜ f 共a, b兲 苷 0., , 8. If f is a function, then, , lim, , 共x, y兲 l 共2, 5兲, , f 共x, y兲 苷 f 共2, 5兲, , 9. If f 共x, y兲 苷 ln y, then ⵜ f 共x, y兲 苷 1兾y., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p990-996.qk_97817_14_ch14_p990-996 11/8/10 1:33 PM Page 992, , 992, , CHAPTER 14, , PARTIAL DERIVATIVES, , 10. If 共2, 1兲 is a critical point of f and, , 11. If f 共x, y兲 苷 sin x ⫹ sin y, then ⫺s2 艋 Du f 共x, y兲 艋 s2 ., , fxx 共2, 1兲 fyy 共2, 1兲 ⬍ 关 fx y 共2, 1兲兴, , 2, , 12. If f 共x, y兲 has two local maxima, then f must have a local, , then f has a saddle point at 共2, 1兲., , minimum., , Exercises, 1–2 Find and sketch the domain of the function., , at equally spaced points were measured and recorded in the, table., (a) Estimate the values of the partial derivatives Tx 共6, 4兲, and Ty共6, 4兲. What are the units?, (b) Estimate the value of Du T 共6, 4兲, where u 苷 共i ⫹ j兲兾s2 ., Interpret your result., (c) Estimate the value of Txy 共6, 4兲., , 1. f 共x, y兲 苷 ln共x ⫹ y ⫹ 1兲, 2. f 共x, y兲 苷 s4 ⫺ x 2 ⫺ y 2 ⫹ s1 ⫺ x 2, 3– 4 Sketch the graph of the function., 3. f 共x, y兲 苷 1 ⫺ y 2, 4. f 共x, y兲 苷 x 2 ⫹ 共 y ⫺ 2兲2, , y, , 0, , 2, , 4, , 6, , 8, , 0, , 30, , 38, , 45, , 51, , 55, , 2, , 52, , 56, , 60, , 62, , 61, , 4, , 78, , 74, , 72, , 68, , 66, , 6, , 98, , 87, , 80, , 75, , 71, , 8, , 96, , 90, , 86, , 80, , 75, , 10, , 92, , 92, , 91, , 87, , 78, , x, , 5–6 Sketch several level curves of the function., 5. f 共x, y兲 苷 s4x 2 ⫹ y 2, , 6. f 共x, y兲 苷 e x ⫹ y, , 7. Make a rough sketch of a contour map for the function whose, , graph is shown., z, , x, , 2, , 2, , y, , 12. Find a linear approximation to the temperature function T 共x, y兲, , in Exercise 11 near the point (6, 4). Then use it to estimate the, temperature at the point (5, 3.8)., 13–17 Find the first partial derivatives., , 8. A contour map of a function f is shown. Use it to make a, , rough sketch of the graph of f ., y, 1, , u ⫹ 2v, u2 ⫹ v2, , 13. f 共x, y兲 苷 共5y 3 ⫹ 2x 2 y兲8, , 14. t共u, v兲 苷, , 15. F 共␣, ␤ 兲 苷 ␣ 2 ln共␣ 2 ⫹ ␤ 2 兲, , 16. G共x, y, z兲 苷 e xz sin共 y兾z兲, , 17. S共u, v, w兲 苷 u arctan(v sw ), , 1.5, , 18. The speed of sound traveling through ocean water is a function, , 2, , of temperature, salinity, and pressure. It has been modeled by, the function, 4, , C 苷 1449.2 ⫹ 4.6T ⫺ 0.055T 2 ⫹ 0.00029T 3, , x, , 9–10 Evaluate the limit or show that it does not exist., 9., , lim, , 共x, y兲 l 共1, 1兲, , 2xy, x 2 ⫹ 2y 2, , 10., , lim, , 共x, y兲 l 共0, 0兲, , 2xy, x 2 ⫹ 2y 2, , 11. A metal plate is situated in the xy-plane and occupies the, , rectangle 0 艋 x 艋 10, 0 艋 y 艋 8, where x and y are measured, in meters. The temperature at the point 共x, y兲 in the plate is, T 共x, y兲, where T is measured in degrees Celsius. Temperatures, , ;, , ⫹ 共1.34 ⫺ 0.01T 兲共S ⫺ 35兲 ⫹ 0.016D, where C is the speed of sound (in meters per second), T is the, temperature (in degrees Celsius), S is the salinity (the concentration of salts in parts per thousand, which means the number, of grams of dissolved solids per 1000 g of water), and D is the, depth below the ocean surface (in meters). Compute ⭸C兾⭸T ,, ⭸C兾⭸S, and ⭸C兾⭸D when T 苷 10⬚C, S 苷 35 parts per thousand,, and D 苷 100 m. Explain the physical significance of these, partial derivatives., , Graphing calculator or computer required, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p990-996.qk_97817_14_ch14_p990-996 11/8/10 1:33 PM Page 994, , 994, , CHAPTER 14, , PARTIAL DERIVATIVES, , 50. Find parametric equations of the tangent line at the point, , 共⫺2, 2, 4兲 to the curve of intersection of the surface, z 苷 2x 2 ⫺ y 2 and the plane z 苷 4., , 60. f 共x, y兲 苷, , 1, 1, ⫹ ;, x, y, , 61. f 共x, y, z兲 苷 xyz;, , 1, 1, ⫹ 2 苷1, x2, y, x2 ⫹ y2 ⫹ z2 苷 3, , 51–54 Find the local maximum and minimum values and saddle, , points of the function. If you have three-dimensional graphing, software, graph the function with a domain and viewpoint that, reveal all the important aspects of the function., , 62. f 共x, y, z兲 苷 x 2 ⫹ 2y 2 ⫹ 3z 2;, , 51. f 共x, y兲 苷 x 2 ⫺ xy ⫹ y 2 ⫹ 9x ⫺ 6y ⫹ 10, , 63. Find the points on the surface xy 2z 3 苷 2 that are closest to, , the origin., , 3, , 64. A package in the shape of a rectangular box can be mailed by, , 53. f 共x, y兲 苷 3xy ⫺ x 2 y ⫺ xy 2, , the US Postal Service if the sum of its length and girth (the, perimeter of a cross-section perpendicular to the length) is at, most 108 in. Find the dimensions of the package with largest, volume that can be mailed., , 54. f 共x, y兲 苷 共x 2 ⫹ y兲e y兾2, , 55–56 Find the absolute maximum and minimum values of f on, , the set D., 55. f 共x, y兲 苷 4xy 2 ⫺ x 2 y 2 ⫺ xy 3;, , D is the closed triangular, region in the xy-plane with vertices 共0, 0兲, 共0, 6兲, and 共6, 0兲, 2, , 2, , 56. f 共x, y兲 苷 e⫺x ⫺y 共x 2 ⫹ 2y 2 兲;, , x ⫺ y ⫹ 2z 苷 2, , 65. A pentagon is formed by placing an isosceles triangle on a, , rectangle, as shown in the figure. If the pentagon has fixed, perimeter P, find the lengths of the sides of the pentagon that, maximize the area of the pentagon., , D is the disk x 2 ⫹ y 2 艋 4, , =, , =, , 52. f 共x, y兲 苷 x ⫺ 6xy ⫹ 8y, 3, , x ⫹ y ⫹ z 苷 1,, , ¨, , ; 57. Use a graph or level curves or both to estimate the local, maximum and minimum values and saddle points of, f 共x, y兲 苷 x 3 ⫺ 3x ⫹ y 4 ⫺ 2y 2. Then use calculus to find, these values precisely., , ; 58. Use a graphing calculator or computer (or Newton’s method, or a computer algebra system) to find the critical points of, f 共x, y兲 苷 12 ⫹ 10y ⫺ 2x 2 ⫺ 8xy ⫺ y 4 correct to three, decimal places. Then classify the critical points and find, the highest point on the graph., 59–62 Use Lagrange multipliers to find the maximum and mini-, , mum values of f subject to the given constraint(s)., 59. f 共x, y兲 苷 x 2 y ;, , x2 ⫹ y2 苷 1, , 66. A particle of mass m moves on the surface z 苷 f 共x, y兲. Let, , x 苷 x共t兲 and y 苷 y共t兲 be the x- and y-coordinates of the, particle at time t., (a) Find the velocity vector v and the kinetic energy, K 苷 12 m v 2 of the particle., (b) Determine the acceleration vector a., (c) Let z 苷 x 2 ⫹ y 2 and x共t兲 苷 t cos t, y共t兲 苷 t sin t. Find, the velocity vector, the kinetic energy, and the acceleration vector., , ⱍ ⱍ, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Problems Plus, 1. A rectangle with length L and width W is cut into four smaller rectangles by two lines paral-, , lel to the sides. Find the maximum and minimum values of the sum of the squares of the, areas of the smaller rectangles., 2. Marine biologists have determined that when a shark detects the presence of blood in the, , water, it will swim in the direction in which the concentration of the blood increases most, rapidly. Based on certain tests, the concentration of blood (in parts per million) at a point, P共x, y兲 on the surface of seawater is approximated by, 2, , C共x, y兲 苷 e⫺共x ⫹2y, , 2, , 兲兾10 4, , where x and y are measured in meters in a rectangular coordinate system with the blood, source at the origin., (a) Identify the level curves of the concentration function and sketch several members of this, family together with a path that a shark will follow to the source., (b) Suppose a shark is at the point 共x 0 , y0 兲 when it first detects the presence of blood in, the water. Find an equation of the shark’s path by setting up and solving a differential, equation., 3. A long piece of galvanized sheet metal with width w is to be bent into a symmetric form with, , three straight sides to make a rain gutter. A cross-section is shown in the figure., (a) Determine the dimensions that allow the maximum possible flow; that is, find the dimensions that give the maximum possible cross-sectional area., (b) Would it be better to bend the metal into a gutter with a semicircular cross-section?, , x, , ¨, , ¨, , x, , w-2x, 4. For what values of the number r is the function, , 再, , 共x ⫹ y ⫹ z兲r, f 共x, y, z兲 苷 x 2 ⫹ y 2 ⫹ z 2, 0, , if 共x, y, z兲 苷 共0, 0, 0兲, if 共x, y, z兲 苷 共0, 0, 0兲, , continuous on ⺢ 3 ?, 5. Suppose f is a differentiable function of one variable. Show that all tangent planes to the, , surface z 苷 x f 共 y兾x兲 intersect in a common point., 6. (a) Newton’s method for approximating a root of an equation f 共x兲 苷 0 (see Section 4.8), , can be adapted to approximating a solution of a system of equations f 共x, y兲 苷 0 and, t共x, y兲 苷 0. The surfaces z 苷 f 共x, y兲 and z 苷 t共x, y兲 intersect in a curve that intersects, the xy-plane at the point 共r, s兲, which is the solution of the system. If an initial approximation 共x 1, y1兲 is close to this point, then the tangent planes to the surfaces at 共x 1, y1兲, intersect in a straight line that intersects the xy-plane in a point 共x 2 , y2 兲, which should be, closer to 共r, s兲. (Compare with Figure 2 in Section 3.8.) Show that, x2 苷 x1 ⫺, , fty ⫺ fy t, fx ty ⫺ fy tx, , and, , y2 苷 y1 ⫺, , fx t ⫺ ftx, fx ty ⫺ fy tx, , where f , t, and their partial derivatives are evaluated at 共x 1, y1兲. If we continue this procedure, we obtain successive approximations 共x n , yn 兲., , 995, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_14_ch14_p990-996.qk_97817_14_ch14_p990-996 11/8/10 1:33 PM Page 996, , (b) It was Thomas Simpson (1710–1761) who formulated Newton’s method as we know it, today and who extended it to functions of two variables as in part (a). (See the biography, of Simpson on page 537.) The example that he gave to illustrate the method was to solve, the system of equations, x x ⫹ y y 苷 1000, , x y ⫹ y x 苷 100, , In other words, he found the points of intersection of the curves in the figure. Use the, method of part (a) to find the coordinates of the points of intersection correct to six decimal places., y, , x x+y y=1000, 4, , x y+y x=100, 2, , 0, , 2, , 4, , x, , 7. If the ellipse x 2兾a 2 ⫹ y 2兾b 2 苷 1 is to enclose the circle x 2 ⫹ y 2 苷 2y, what values of a and b, , minimize the area of the ellipse?, 8. Among all planes that are tangent to the surface xy 2z 2 苷 1, find the ones that are farthest, , from the origin., , 996, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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15, , Multiple Integrals, , FPO, New Art to, come, , Geologists study how mountain ranges were formed, and estimate the work required to lift them from sea, level. In Section 15.8 you are asked to use a triple, integral to compute the work done in the formation, of Mount Fuji in Japan., , © S.R. Lee Photo Traveller / Shutterstock, , In this chapter we extend the idea of a definite integral to double and triple integrals of functions of two, or three variables. These ideas are then used to compute volumes, masses, and centroids of more general, regions than we were able to consider in Chapters 5 and 8. We also use double integrals to calculate, probabilities when two random variables are involved., We will see that polar coordinates are useful in computing double integrals over some types of regions., In a similar way, we will introduce two new coordinate systems in three-dimensional space––cylindrical, coordinates and spherical coordinates––that greatly simplify the computation of triple integrals over, certain commonly occurring solid regions., , 997, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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998, , CHAPTER 15, , MULTIPLE INTEGRALS, , Double Integrals over Rectangles, , 15.1, , In much the same way that our attempt to solve the area problem led to the definition of a, definite integral, we now seek to find the volume of a solid and in the process we arrive at, the definition of a double integral., , Review of the Definite Integral, First let’s recall the basic facts concerning definite integrals of functions of a single variable. If f 共x兲 is defined for a 艋 x 艋 b, we start by dividing the interval 关a, b兴 into n subintervals 关x i⫺1, x i 兴 of equal width ⌬x 苷 共b ⫺ a兲兾n and we choose sample points x*i in these, subintervals. Then we form the Riemann sum, n, , 兺 f 共x*兲 ⌬x, , 1, , i, , i苷1, , and take the limit of such sums as n l ⬁ to obtain the definite integral of f from a to b :, , y, , 2, , b, , a, , n, , f 共x兲 dx 苷 lim, , 兺 f 共x*兲 ⌬x, i, , n l ⬁ i苷1, , In the special case where f 共x兲 艌 0, the Riemann sum can be interpreted as the sum of the, areas of the approximating rectangles in Figure 1, and xab f 共x兲 dx represents the area under, the curve y 苷 f 共x兲 from a to b., y, , Îx, , f(x *), i, , 0, , FIGURE 1, , z, , a, x*¡, , ⁄, , ¤, x™*, , ‹, , xi-1, , x£*, , x *i, , 0, , x, , b, , FIGURE 2, , x, , x n*, , Volumes and Double Integrals, z=f(x, y), , In a similar manner we consider a function f of two variables defined on a closed rectangle, R 苷 关a, b兴 ⫻ 关c, d兴 苷 兵共x, y兲 僆 ⺢ 2, , a, , b, , xn-1, , xi, , c, , R, , d, y, , ⱍ a 艋 x 艋 b,, , c 艋 y 艋 d其, , and we first suppose that f 共x, y兲 艌 0. The graph of f is a surface with equation z 苷 f 共x, y兲., Let S be the solid that lies above R and under the graph of f, that is,, S 苷 兵共x, y, z兲 僆 ⺢ 3, , ⱍ 0 艋 z 艋 f 共x, y兲,, , 共x, y兲 僆 R 其, , (See Figure 2.) Our goal is to find the volume of S., The first step is to divide the rectangle R into subrectangles. We accomplish this by, dividing the interval 关a, b兴 into m subintervals 关x i⫺1, x i 兴 of equal width ⌬x 苷 共b ⫺ a兲兾m, and dividing 关c, d兴 into n subintervals 关yj⫺1, yj 兴 of equal width ⌬y 苷 共d ⫺ c兲兾n. By drawing lines parallel to the coordinate axes through the endpoints of these subintervals, as in, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 999, , SECTION 15.1, , DOUBLE INTEGRALS OVER RECTANGLES, , 999, , Figure 3, we form the subrectangles, Rij 苷 关x i⫺1, x i 兴 ⫻ 关yj⫺1, yj 兴 苷 兵共x, y兲, , ⱍx, , i⫺1, , 艋 x 艋 x i , yj⫺1 艋 y 艋 yj 其, , each with area ⌬A 苷 ⌬x ⌬y., y, , R ij, , d, , (xi, yj), , (x *ij , y ij* ), , yj, yj-1, , Îy, , ›, c, *, (x *£™, y£™), , FIGURE 3, , 0, , a, , ⁄, , ¤, , x i-1 x i, , Dividing R into subrectangles, , b, , x, , Îx, , If we choose a sample point 共x ij*, y ij*兲 in each Rij , then we can approximate the part of, S that lies above each Rij by a thin rectangular box (or “column”) with base Rij and height, f 共x ij*, yij*兲 as shown in Figure 4. (Compare with Figure 1.) The volume of this box is the, height of the box times the area of the base rectangle:, f 共x ij*, yij*兲 ⌬A, If we follow this procedure for all the rectangles and add the volumes of the corresponding, boxes, we get an approximation to the total volume of S:, m, , V⬇, , 3, , n, , 兺 兺 f 共x *, y *兲 ⌬A, ij, , ij, , i苷1 j苷1, , (See Figure 5.) This double sum means that for each subrectangle we evaluate f at the chosen point and multiply by the area of the subrectangle, and then we add the results., z, , z, , f(x *ij , y*ij ), , 0, , 0, , c, , a, , d, , y, , y, x, , b, , x, , R ij, FIGURE 4, , FIGURE 5, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 1000, , 1000, , CHAPTER 15, , MULTIPLE INTEGRALS, , Our intuition tells us that the approximation given in 3 becomes better as m and n, become larger and so we would expect that, The meaning of the double limit in Equation 4 is, that we can make the double sum as close as we, like to the number V [for any choice of 共x ij*, yij*兲, in Rij ] by taking m and n sufficiently large., , m, , n, , 兺 兺 f 共x *, y *兲 ⌬A, , V 苷 lim, , 4, , ij, , m, n l ⬁ i苷1 j苷1, , ij, , We use the expression in Equation 4 to define the volume of the solid S that lies under the, graph of f and above the rectangle R. (It can be shown that this definition is consistent with, our formula for volume in Section 5.2.), Limits of the type that appear in Equation 4 occur frequently, not just in finding volumes but in a variety of other situations as well—as we will see in Section 15.5—even, when f is not a positive function. So we make the following definition., 5, , Definition The double integral of f over the rectangle R is, , Notice the similarity between Definition 5, and the definition of a single integral in, Equation 2., , m, , yy f 共x, y兲 dA 苷, R, , n, , 兺 兺 f 共x *, y *兲 ⌬A, , lim, , ij, , m, n l ⬁ i苷1 j苷1, , ij, , if this limit exists., , Although we have defined the double integral by, dividing R into equal-sized subrectangles, we, could have used subrectangles Rij of unequal, size. But then we would have to ensure that all, of their dimensions approach 0 in the limiting, process., , The precise meaning of the limit in Definition 5 is that for every number ␧ ⬎ 0 there is, an integer N such that, , 冟 yy, R, , m, , f 共x, y兲 dA ⫺, , n, , 兺 兺 f 共x *, y *兲 ⌬A, ij, , ij, , i苷1 j苷1, , 冟, , ⬍␧, , for all integers m and n greater than N and for any choice of sample points 共x ij*, yij*兲 in Rij., A function f is called integrable if the limit in Definition 5 exists. It is shown in courses, on advanced calculus that all continuous functions are integrable. In fact, the double integral of f exists provided that f is “not too discontinuous.” In particular, if f is bounded, [that is, there is a constant M such that f 共x, y兲 艋 M for all 共x, y兲 in R ], and f is continuous there, except on a finite number of smooth curves, then f is integrable over R ., The sample point 共x ij*, yij*兲 can be chosen to be any point in the subrectangle Rij , but if, we choose it to be the upper right-hand corner of Rij [namely 共x i, yj 兲, see Figure 3], then, the expression for the double integral looks simpler:, , ⱍ, , ⱍ, , m, , 6, , yy, R, , f 共x, y兲 dA 苷 lim, , n, , 兺 兺 f 共x , y 兲 ⌬A, , m, n l ⬁ i苷1 j苷1, , i, , j, , By comparing Definitions 4 and 5, we see that a volume can be written as a double, integral:, If f 共x, y兲 艌 0, then the volume V of the solid that lies above the rectangle R and, below the surface z 苷 f 共x, y兲 is, V 苷 yy f 共x, y兲 dA, R, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 1001, , DOUBLE INTEGRALS OVER RECTANGLES, , SECTION 15.1, , 1001, , The sum in Definition 5,, m, , n, , 兺 兺 f 共x *, y *兲 ⌬A, ij, , ij, , i苷1 j苷1, , is called a double Riemann sum and is used as an approximation to the value of the, double integral. [Notice how similar it is to the Riemann sum in 1 for a function of a, single variable.] If f happens to be a positive function, then the double Riemann sum, represents the sum of volumes of columns, as in Figure 5, and is an approximation to the, volume under the graph of f ., y, , v EXAMPLE 1 Estimate the volume of the solid that lies above the square, R 苷 关0, 2兴 ⫻ 关0, 2兴 and below the elliptic paraboloid z 苷 16 ⫺ x 2 ⫺ 2y 2. Divide R into, four equal squares and choose the sample point to be the upper right corner of each, square Rij . Sketch the solid and the approximating rectangular boxes., , (1, 2), (2, 2), , 2, , R¡™, 1, , R™™, (2, 1), , (1, 1), , SOLUTION The squares are shown in Figure 6. The paraboloid is the graph of, , R¡¡, , f 共x, y兲 苷 16 ⫺ x 2 ⫺ 2y 2 and the area of each square is ⌬A 苷 1. Approximating the, volume by the Riemann sum with m 苷 n 苷 2, we have, , R™¡, , 0, , 1, , x, , 2, , 2, , V⬇, , FIGURE 6, , 2, , 兺 兺 f 共x , y 兲 ⌬A, i, , j, , i苷1 j苷1, , z, 16, , 苷 f 共1, 1兲 ⌬A ⫹ f 共1, 2兲 ⌬A ⫹ f 共2, 1兲 ⌬A ⫹ f 共2, 2兲 ⌬A, , z=16-≈-2¥, , 苷 13共1兲 ⫹ 7共1兲 ⫹ 10共1兲 ⫹ 4共1兲 苷 34, This is the volume of the approximating rectangular boxes shown in Figure 7., We get better approximations to the volume in Example 1 if we increase the number of, squares. Figure 8 shows how the columns start to look more like the actual solid and the, corresponding approximations become more accurate when we use 16, 64, and 256, squares. In the next section we will be able to show that the exact volume is 48., 2, , 2, , y, , x, , FIGURE 7, , FIGURE 8, , The Riemann sum approximations to, the volume under z=16-≈-2¥, become more accurate as m and, n increase., , (a) m=n=4, VÅ41.5, , v, , EXAMPLE 2 If R 苷 兵共x, y兲, , (b) m=n=8, VÅ44.875, , ⱍ, , (c) m=n=16, VÅ46.46875, , ⫺1 艋 x 艋 1, ⫺2 艋 y 艋 2其, evaluate the integral, , yy s1 ⫺ x, , 2, , dA, , R, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 1002, , 1002, , MULTIPLE INTEGRALS, , CHAPTER 15, z, , SOLUTION It would be very difficult to evaluate this integral directly from Definition 5, (0, 0, 1), , but, because s1 ⫺ x 2 艌 0, we can compute the integral by interpreting it as a volume. If, z 苷 s1 ⫺ x 2 , then x 2 ⫹ z 2 苷 1 and z 艌 0, so the given double integral represents the, volume of the solid S that lies below the circular cylinder x 2 ⫹ z 2 苷 1 and above the, rectangle R. (See Figure 9.) The volume of S is the area of a semicircle with radius 1 times, the length of the cylinder. Thus, , S, , x, , (1, 0, 0), , (0, 2, 0), , y, , yy s1 ⫺ x, , FIGURE 9, , 2, , dA 苷 12 ␲ 共1兲2 ⫻ 4 苷 2␲, , R, , The Midpoint Rule, The methods that we used for approximating single integrals (the Midpoint Rule, the, Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals. Here we, consider only the Midpoint Rule for double integrals. This means that we use a double Riemann sum to approximate the double integral, where the sample point 共x ij*, yij*兲 in Rij is, chosen to be the center 共xi , yj兲 of Rij . In other words, xi is the midpoint of 关x i⫺1, x i 兴 and yj, is the midpoint of 关yj⫺1, yj 兴., Midpoint Rule for Double Integrals, m, , n, , yy f 共x, y兲 dA ⬇ 兺 兺 f 共 x , y 兲 ⌬A, i, , j, , i苷1 j苷1, , R, , where xi is the midpoint of 关x i⫺1, x i 兴 and yj is the midpoint of 关yj⫺1, yj 兴., , v EXAMPLE 3 Use the Midpoint Rule with m 苷 n 苷 2 to estimate the value of the, integral xxR 共x ⫺ 3y 2 兲 dA, where R 苷 兵共x, y兲 0 艋 x 艋 2, 1 艋 y 艋 2其., , ⱍ, , SOLUTION In using the Midpoint Rule with m 苷 n 苷 2, we evaluate f 共x, y兲 苷 x ⫺ 3y 2 at, , y, (2, 2), , 2, 3, 2, , the centers of the four subrectangles shown in Figure 10. So x1 苷 12 , x2 苷 32 , y1 苷 54 , and, y2 苷 74 . The area of each subrectangle is ⌬A 苷 12 . Thus, , R¡™, , R™™, , R¡¡, , 2, , yy, , R™¡, , 1, , 共x ⫺ 3y 2 兲 dA ⬇, , R, , 2, , 兺 兺 f 共x , y 兲 ⌬A, i, , j, , i苷1 j苷1, , 苷 f 共x1, y1兲 ⌬A ⫹ f 共x1, y2 兲 ⌬A ⫹ f 共x2 , y1 兲 ⌬A ⫹ f 共x2 , y2 兲 ⌬A, 0, , 1, , 2, , 苷 f ( 12 , 54 ) ⌬A ⫹ f ( 12 , 74 ) ⌬A ⫹ f ( 32 , 54 ) ⌬A ⫹ f ( 32 , 74 ) ⌬A, , x, , 苷 (⫺ 16 ) 2 ⫹ (⫺ 16 ) 2 ⫹ (⫺ 16) 2 ⫹ (⫺ 16 ) 2, 67 1, , FIGURE 10, , 139 1, , 51 1, , 123 1, , 苷 ⫺ 958 苷 ⫺11.875, Thus we have, , yy 共x ⫺ 3y, , 2, , 兲 dA ⬇ ⫺11.875, , R, , NOTE In the next section we will develop an efficient method for computing double, integrals and then we will see that the exact value of the double integral in Example 3 is, ⫺12. (Remember that the interpretation of a double integral as a volume is valid only when, the integrand f is a positive function. The integrand in Example 3 is not a positive function, so its integral is not a volume. In Examples 2 and 3 in Section 15.2 we will discuss, how to interpret integrals of functions that are not always positive in terms of volumes.) If, we keep dividing each subrectangle in Figure 10 into four smaller ones with similar shape,, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 1003, , SECTION 15.1, , Number of, subrectangles, , Midpoint Rule, approximation, , 1, 4, 16, 64, 256, 1024, , ⫺11.5000, ⫺11.8750, ⫺11.9687, ⫺11.9922, ⫺11.9980, ⫺11.9995, , DOUBLE INTEGRALS OVER RECTANGLES, , 1003, , we get the Midpoint Rule approximations displayed in the chart in the margin. Notice how, these approximations approach the exact value of the double integral, ⫺12., , Average Value, Recall from Section 5.5 that the average value of a function f of one variable defined on, an interval 关a, b兴 is, fave 苷, , 1, b⫺a, , y, , b, , a, , f 共x兲 dx, , In a similar fashion we define the average value of a function f of two variables defined, on a rectangle R to be, 1, fave 苷, yy f 共x, y兲 dA, A共R兲 R, where A共R兲 is the area of R., If f 共x, y兲 艌 0, the equation, A共R兲 ⫻ fave 苷, , yy f 共x, y兲 dA, R, , says that the box with base R and height fave has the same volume as the solid that lies, under the graph of f . [If z 苷 f 共x, y兲 describes a mountainous region and you chop off the, tops of the mountains at height fave , then you can use them to fill in the valleys so that the, region becomes completely flat. See Figure 11.], , FIGURE 11, , EXAMPLE 4 The contour map in Figure 12 shows the snowfall, in inches, that fell on the, state of Colorado on December 20 and 21, 2006. (The state is in the shape of a rectangle, that measures 388 mi west to east and 276 mi south to north.) Use the contour map to, estimate the average snowfall for the entire state of Colorado on those days., , 12, 40 36, 44, , 20, , 12, , 16, , 32, 28, , 16, , 24, , 40, , 36, 32, , 12, , 28, 24, , 32, 28, , 8, , 24, 0, , 4, , 8 12 16, , 20, , FIGURE 12, SOLUTION Let’s place the origin at the southwest corner of the state. Then 0 艋 x 艋 388,, , 0 艋 y 艋 276, and f 共x, y兲 is the snowfall, in inches, at a location x miles to the east and, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 1004, , 1004, , CHAPTER 15, , MULTIPLE INTEGRALS, , y miles to the north of the origin. If R is the rectangle that represents Colorado, then the, average snowfall for the state on December 20–21 was, 1, A共R兲, , fave 苷, , yy f 共x, y兲 dA, R, , where A共R兲 苷 388 ⴢ 276. To estimate the value of this double integral, let’s use the Midpoint Rule with m 苷 n 苷 4. In other words, we divide R into 16 subrectangles of equal, size, as in Figure 13. The area of each subrectangle is, ⌬A 苷 161 共388兲共276兲 苷 6693 mi2, y, 276, 12, 40 36, 44, , 20, , 12, , 16, , 32, 28, , 16, , 24, , 40, , 36, , 32, , 12, , 28, 24, , 0, , 4, , 32, 28, , 16 20, 8 12, , 8, , 24, , 0, , 388 x, , FIGURE 13, , Using the contour map to estimate the value of f at the center of each subrectangle,, we get, 4, , 4, , yy f 共x, y兲 dA ⬇ 兺 兺 f 共x , y 兲 ⌬A, i, , R, , j, , i苷1 j苷1, , ⬇ ⌬A关0 ⫹ 15 ⫹ 8 ⫹ 7 ⫹ 2 ⫹ 25 ⫹ 18.5 ⫹ 11, ⫹ 4.5 ⫹ 28 ⫹ 17 ⫹ 13.5 ⫹ 12 ⫹ 15 ⫹ 17.5 ⫹ 13兴, 苷 共6693兲共207兲, Therefore, , fave ⬇, , 共6693兲共207兲, ⬇ 12.9, 共388兲共276兲, , On December 20–21, 2006, Colorado received an average of approximately 13 inches of, snow., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p997-1005.qk_97817_15_ch15_p997-1005 11/8/10 3:33 PM Page 1005, , SECTION 15.1, , DOUBLE INTEGRALS OVER RECTANGLES, , 1005, , Properties of Double Integrals, We list here three properties of double integrals that can be proved in the same manner as, in Section 4.2. We assume that all of the integrals exist. Properties 7 and 8 are referred to, as the linearity of the integral., , yy 关 f 共x, y兲 ⫹ t共x, y兲兴 dA 苷 yy f 共x, y兲 dA ⫹ yy t共x, y兲 dA, , 7, Double integrals behave this way because the, double sums that define them behave this way., , R, , 8, , R, , R, , yy c f 共x, y兲 dA 苷 c yy f 共x, y兲 dA, R, , where c is a constant, , R, , If f 共x, y兲 艌 t共x, y兲 for all 共x, y兲 in R, then, 9, , yy, , f 共x, y兲 dA 艌 yy t共x, y兲 dA, , R, , 15.1, , R, , Exercises, , 1. (a) Estimate the volume of the solid that lies below the surface, , z 苷 xy and above the rectangle, R 苷 兵共x, y兲, , ⱍ, , (b) Estimate the double integral with m 苷 n 苷 4 by choosing, the sample points to be the points closest to the origin., , 0 艋 x 艋 6, 0 艋 y 艋 4其, , Use a Riemann sum with m 苷 3, n 苷 2, and take the, sample point to be the upper right corner of each square., (b) Use the Midpoint Rule to estimate the volume of the solid, in part (a)., 2. If R 苷 关0, 4兴 ⫻ 关⫺1, 2兴, use a Riemann sum with m 苷 2,, , n 苷 3 to estimate the value of xxR 共1 ⫺ x y 2 兲 dA. Take the, sample points to be (a) the lower right corners and (b) the, upper left corners of the rectangles., , 3. (a) Use a Riemann sum with m 苷 n 苷 2 to estimate the value, , of xxR xe ⫺xy dA, where R 苷 关0, 2兴 ⫻ 关0, 1兴. Take the sample, points to be upper right corners., (b) Use the Midpoint Rule to estimate the integral in part (a)., , 4. (a) Estimate the volume of the solid that lies below the surface, , z 苷 1 ⫹ x 2 ⫹ 3y and above the rectangle, R 苷 关1, 2兴 ⫻ 关0, 3兴. Use a Riemann sum with m 苷 n 苷 2, and choose the sample points to be lower left corners., (b) Use the Midpoint Rule to estimate the volume in part (a)., , y, , 2.0, , 2.5, , 3.0, , 3.5, , 4.0, , 0, , ⫺3, , ⫺5, , ⫺6, , ⫺4, , ⫺1, , 1, , ⫺1, , ⫺2, , ⫺3, , ⫺1, , 1, , 2, , 1, , 0, , ⫺1, , 1, , 4, , 3, , 2, , 2, , 1, , 3, , 7, , 4, , 3, , 4, , 2, , 5, , 9, , x, , 6. A 20-ft-by-30-ft swimming pool is filled with water. The depth, , is measured at 5-ft intervals, starting at one corner of the pool,, and the values are recorded in the table. Estimate the volume of, water in the pool., 0, , 5, , 10, , 15, , 20, , 25, , 30, , 0, , 2, , 3, , 4, , 6, , 7, , 8, , 8, , 5, , 2, , 3, , 4, , 7, , 8, , 10, , 8, , 10, , 2, , 4, , 6, , 8, , 10, , 12, , 10, , 15, , 2, , 3, , 4, , 5, , 6, , 8, , 7, , 20, , 2, , 2, , 2, , 2, , 3, , 4, , 4, , 5. A table of values is given for a function f 共x, y兲 defined on, , R 苷 关0, 4兴 ⫻ 关2, 4兴., (a) Estimate xxR f 共x, y兲 dA using the Midpoint Rule with, m 苷 n 苷 2., , 7. Let V be the volume of the solid that lies under the graph of, , f 共x, y兲 苷 s52 ⫺ x 2 ⫺ y 2 and above the rectangle given by, 2 艋 x 艋 4, 2 艋 y 艋 6. We use the lines x 苷 3 and y 苷 4 to, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1006, , MULTIPLE INTEGRALS, , CHAPTER 15, , divide R into subrectangles. Let L and U be the Riemann sums, computed using lower left corners and upper right corners,, respectively. Without calculating the numbers V, L , and U,, arrange them in increasing order and explain your reasoning., , 24, , 20, , 32 4444, , 16, , 28, 24, , 8. The figure shows level curves of a function f in the square, , 32, , 40, 3236, , 16, , R 苷 关0, 2兴  关0, 2兴. Use the Midpoint Rule with m 苷 n 苷 2, to estimate xxR f 共x, y兲 dA. How could you improve your, estimate?, , 44, , 48, , 28, 56, 52, , y, 2, 5, , 4, 3, , 1, , 32, 36, 40, 44, , 20, 24, 28, , 6 7, , 48, 5256, , 2, , 11–13 Evaluate the double integral by first identifying it as the, volume of a solid., , 1, , 0, , 1, , 2, , x, , 11., 12., 13., , xxR 3 dA, R 苷 兵共x, y兲 ⱍ 2  x  2, 1  y  6其, xxR 共5  x兲 dA, R 苷 兵共x, y兲 ⱍ 0  x  5, 0  y  3其, xxR 共4  2y兲 dA, R 苷 关0, 1兴  关0, 1兴, , 9. A contour map is shown for a function f on the square, , R 苷 关0, 4兴  关0, 4兴., (a) Use the Midpoint Rule with m 苷 n 苷 2 to estimate the, value of xxR f 共x, y兲 dA., (b) Estimate the average value of f ., , 14. The integral xxR s9  y 2 dA, where R 苷 关0, 4兴  关0, 2兴,, , represents the volume of a solid. Sketch the solid., 15. Use a programmable calculator or computer (or the sum, , command on a CAS) to estimate, , yy, , y, 4, , s1  xey dA, , R, , 10, , 0, , 0, , 2, , 10 20 30, , where R 苷 关0, 1兴  关0, 1兴. Use the Midpoint Rule with the, following numbers of squares of equal size: 1, 4, 16, 64, 256,, and 1024., , 10, , 16. Repeat Exercise 15 for the integral xxR sin( x  sy ) dA., , 20, , 17. If f is a constant function, f 共x, y兲 苷 k, and, , R 苷 关a, b兴  关c, d兴, show that, , 30, , 0, , 2, , 4 x, , yy k dA 苷 k共b  a兲共d  c兲, R, , 10. The contour map shows the temperature, in degrees Fahrenheit,, , at 4:00 PM on February 26, 2007, in Colorado. (The state measures 388 mi west to east and 276 mi south to north.) Use, the Midpoint Rule with m 苷 n 苷 4 to estimate the average, temperature in Colorado at that time., , 15.2, , 18. Use the result of Exercise 17 to show that, , 0  yy sin  x cos  y dA , R, , 1, 32, , [ ] [ , ]., , where R 苷 0, 14 , , 1 1, 4 2, , Iterated Integrals, Recall that it is usually difficult to evaluate single integrals directly from the definition of, an integral, but the Fundamental Theorem of Calculus provides a much easier method. The, evaluation of double integrals from first principles is even more difficult, but in this sec-, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1006-1015.qk_97817_15_ch15_p1006-1015 11/8/10 3:34 PM Page 1007, , SECTION 15.2, , ITERATED INTEGRALS, , 1007, , tion we see how to express a double integral as an iterated integral, which can then be evaluated by calculating two single integrals., Suppose that f is a function of two variables that is integrable on the rectangle, R 苷 关a, b兴  关c, d 兴. We use the notation xcd f 共x, y兲 dy to mean that x is held fixed and, f 共x, y兲 is integrated with respect to y from y 苷 c to y 苷 d. This procedure is called partial integration with respect to y. (Notice its similarity to partial differentiation.) Now, xcd f 共x, y兲 dy is a number that depends on the value of x, so it defines a function of x :, A共x兲 苷, , y, , d, , c, , f 共x, y兲 dy, , If we now integrate the function A with respect to x from x 苷 a to x 苷 b, we get, , y, , 1, , b, , a, , A共x兲 dx 苷 y, , 冋y, , b, , a, , 册, , d, , f 共x, y兲 dy dx, , c, , The integral on the right side of Equation 1 is called an iterated integral. Usually the, brackets are omitted. Thus, b, , y y, , 2, , a, , d, , c, , f 共x, y兲 dy dx 苷 y, , 冋y, , b, , a, , d, , c, , 册, , f 共x, y兲 dy dx, , means that we first integrate with respect to y from c to d and then with respect to x from, a to b., Similarly, the iterated integral, d, , y y, , 3, , c, , b, , a, , f 共x, y兲 dx dy 苷 y, , 冋y, , d, , c, , b, , a, , 册, , f 共x, y兲 dx dy, , means that we first integrate with respect to x (holding y fixed) from x 苷 a to x 苷 b and, then we integrate the resulting function of y with respect to y from y 苷 c to y 苷 d. Notice, that in both Equations 2 and 3 we work from the inside out., EXAMPLE 1 Evaluate the iterated integrals., , (a), , 3, , 2, , 0, , 1, , y y, , x 2y dy dx, , 2, , y y, , (b), , 1, , 3, , 0, , x 2 y dx dy, , SOLUTION, , (a) Regarding x as a constant, we obtain, , y, , 2, , 1, , 冋 册, , y2, x y dy 苷 x, 2, 2, , 冉冊 冉冊, , y苷2, , 22, 2, , 苷 x2, , 2, , y苷1, ,  x2, , 12, 2, , 苷 2 x2, 3, , Thus the function A in the preceding discussion is given by A共x兲 苷 32 x 2 in this example., We now integrate this function of x from 0 to 3:, 3, , yy, 0, , 2, , 1, , x 2 y dy dx 苷 y, , 3, , 0, , 冋y, , 3 3, 2, 0, , 苷y, , 2, , 1, , 册, 册, , x 2 y dy dx, , x 2 dx 苷, , x3, 2, , 3, , 苷, 0, , 27, 2, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1006-1015.qk_97817_15_ch15_p1006-1015 11/8/10 3:34 PM Page 1008, , 1008, , CHAPTER 15, , MULTIPLE INTEGRALS, , (b) Here we first integrate with respect to x :, 2, , y y, 1, , 3, , 0, , x 2 y dx dy 苷 y, , 冋y, , 2, , 1, , 苷, , y, , 3, , 0, , 2, , 册, , x 2 y dx dy 苷, y2, 2, , 9y dy 苷 9, , 1, , y, , 2, , 1, , 册, , 2, , 冋 册, x3, y, 3, , x苷3, , dy, , x苷0, , 27, 2, , 苷, , 1, , Notice that in Example 1 we obtained the same answer whether we integrated with, respect to y or x first. In general, it turns out (see Theorem 4) that the two iterated integrals, in Equations 2 and 3 are always equal; that is, the order of integration does not matter., (This is similar to Clairaut’s Theorem on the equality of the mixed partial derivatives.), The following theorem gives a practical method for evaluating a double integral by, expressing it as an iterated integral (in either order)., , Theorem 4 is named after the Italian mathematician Guido Fubini (1879–1943), who proved a, very general version of this theorem in 1907. But, the version for continuous functions was known, to the French mathematician Augustin-Louis, Cauchy almost a century earlier., , 4, , Fubini’s Theorem If f is continuous on the rectangle, , R 苷 兵共x, y兲, , ⱍ, , a  x  b, c  y  d 其, then, b, , yy f 共x, y兲 dA 苷 y y, a, , d, , c, , f 共x, y兲 dy dx 苷 y, , d, , y, , c, , b, , a, , f 共x, y兲 dx dy, , R, , More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a finite number of smooth curves, and the iterated integrals exist., z, , The proof of Fubini’s Theorem is too difficult to include in this book, but we can at least, give an intuitive indication of why it is true for the case where f 共x, y兲  0. Recall that if, f is positive, then we can interpret the double integral xxR f 共x, y兲 dA as the volume V of, the solid S that lies above R and under the surface z 苷 f 共x, y兲. But we have another formula that we used for volume in Chapter 5, namely,, , C, , x, x, , a, , 0, , A(x), y, , b, , V 苷 y A共x兲 dx, , b, , a, , FIGURE 1, , where A共x兲 is the area of a cross-section of S in the plane through x perpendicular to the, x-axis. From Figure 1 you can see that A共x兲 is the area under the curve C whose equation, is z 苷 f 共x, y兲, where x is held constant and c  y  d. Therefore, , TEC Visual 15.2 illustrates Fubini’s, Theorem by showing an animation of, Figures 1 and 2., , d, , A共x兲 苷 y f 共x, y兲 dy, , z, , c, , and we have, , yy f 共x, y兲 dA 苷 V 苷 y, , b, , a, , 0, , c, , FIGURE 2, , b, , a, , y, , d, , c, , f 共x, y兲 dy dx, , R, , y, , d, y, , x, , A共x兲 dx 苷 y, , A similar argument, using cross-sections perpendicular to the y-axis as in Figure 2, shows, that, d, , yy f 共x, y兲 dA 苷 y y, c, , b, , a, , f 共x, y兲 dx dy, , R, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1006-1015.qk_97817_15_ch15_p1006-1015 11/8/10 3:34 PM Page 1013, , DOUBLE INTEGRALS OVER GENERAL REGIONS, , SECTION 15.3, z, , 1013, , If F is integrable over R, then we define the double integral of f over D by, graph of f, , 2, , 0, , yy f 共x, y兲 dA 苷 yy F共x, y兲 dA, D, , where F is given by Equation 1, , R, , y, , D, , Definition 2 makes sense because R is a rectangle and so xxR F共x, y兲 dA has been previously defined in Section 15.1. The procedure that we have used is reasonable because the, values of F共x, y兲 are 0 when 共x, y兲 lies outside D and so they contribute nothing to the integral. This means that it doesn’t matter what rectangle R we use as long as it contains D., In the case where f 共x, y兲  0, we can still interpret xxD f 共x, y兲 dA as the volume of the, solid that lies above D and under the surface z 苷 f 共x, y兲 (the graph of f ). You can see that, this is reasonable by comparing the graphs of f and F in Figures 3 and 4 and remembering that xxR F共x, y兲 dA is the volume under the graph of F., Figure 4 also shows that F is likely to have discontinuities at the boundary points, of D. Nonetheless, if f is continuous on D and the boundary curve of D is “well behaved”, (in a sense outside the scope of this book), then it can be shown that xxR F共x, y兲 dA exists, and therefore xxD f 共x, y兲 dA exists. In particular, this is the case for the following two types, of regions., A plane region D is said to be of type I if it lies between the graphs of two continuous, functions of x, that is,, , x, , FIGURE 3, , z, , graph of F, 0, y, , D, x, , FIGURE 4, , D 苷 兵共x, y兲, , ⱍ a  x  b,, , t1共x兲  y  t 2共x兲其, , where t1 and t 2 are continuous on 关a, b兴. Some examples of type I regions are shown in, Figure 5., y, , y, , y=g™(x), , y, , y=g™(x), , y=g™(x), D, , D, , D, , y=g¡(x), y=g¡(x), 0, , a, , y=g¡(x), b, , x, , 0, , a, , x, , b, , 0, , a, , b, , x, , FIGURE 5 Some type I regions, , y, , In order to evaluate xxD f 共x, y兲 dA when D is a region of type I, we choose a rectangle, R 苷 关a, b兴  关c, d 兴 that contains D, as in Figure 6, and we let F be the function given by, Equation 1; that is, F agrees with f on D and F is 0 outside D. Then, by Fubini’s Theorem,, , y=g™(x), , d, b, , yy f 共x, y兲 dA 苷 yy F共x, y兲 dA 苷 y y, a, , D, , D, , y=g¡(x), 0, , FIGURE 6, , a, , x, , b, , x, , F共x, y兲 dy dx, , R, , Observe that F共x, y兲 苷 0 if y, Therefore, , c, , d, , c, , y, , d, , c, , t1共x兲 or y, , F共x, y兲 dy 苷 y, , t 2共x兲, , t1共x兲, , t 2共x兲 because 共x, y兲 then lies outside D., , F共x, y兲 dy 苷, , y, , t 2共x兲, , t1共x兲, , f 共x, y兲 dy, , because F共x, y兲 苷 f 共x, y兲 when t1共x兲  y  t 2共x兲. Thus we have the following formula, that enables us to evaluate the double integral as an iterated integral., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1006-1015.qk_97817_15_ch15_p1006-1015 11/8/10 3:34 PM Page 1015, , DOUBLE INTEGRALS OVER GENERAL REGIONS, , SECTION 15.3, , 1015, , NOTE When we set up a double integral as in Example 1, it is essential to draw a, diagram. Often it is helpful to draw a vertical arrow as in Figure 8. Then the limits of, integration for the inner integral can be read from the diagram as follows: The arrow, starts at the lower boundary y 苷 t1共x兲, which gives the lower limit in the integral, and, the arrow ends at the upper boundary y 苷 t 2共x兲, which gives the upper limit of integration., For a type II region the arrow is drawn horizontally from the left boundary to the right, boundary., y, , EXAMPLE 2 Find the volume of the solid that lies under the paraboloid z 苷 x 2  y 2 and, , (2, 4), , above the region D in the xy-plane bounded by the line y 苷 2x and the parabola y 苷 x 2., , y=2x, , SOLUTION 1 From Figure 9 we see that D is a type I region and, y=≈, , D 苷 兵共x, y兲, , D, 0, , 1, , Therefore the volume under z 苷 x 2  y 2 and above D is, , x, , 2, , 0  x  2, x 2  y  2x其, , ⱍ, , V 苷 yy 共x 2  y 2 兲 dA 苷 y, , FIGURE 9, , D as a type I region, , y, , 2x, , x2, , 共x 2  y 2 兲 dy dx, , D, , y, 4, , 2, , 0, , 苷, , (2, 4), , 冋, y冋, y冉, y, , 2, , 0, , x= 12 y, , 苷, , 2, , D, , 苷, , 2, , , , 0, , dx, , y苷x 2, , 共2x兲3, 共x 2 兲3,  x 2x 2 , 3, 3, , x 2共2x兲 , , 0, , x=œ„, y, , 册, , y苷2x, , y3, 3, , x2y , , x6, 14x 3,  x4 , 3, 3, , x, , 0, , 苷, , FIGURE 10, , x7, x5, 7x 4, , , 21, 5, 6, , 册, , 冊, , 2, , 苷, , 0, , 册, , dx, , dx, , 216, 35, , D as a type II region, SOLUTION 2 From Figure 10 we see that D can also be written as a type II region:, Figure 11 shows the solid whose volume, is calculated in Example 2. It lies above the, xy-plane, below the paraboloid z 苷 x 2  y 2,, and between the plane y 苷 2x and the, parabolic cylinder y 苷 x 2., , 0  y  4, 2 y  x  sy 其, , ⱍ, , 1, , Therefore another expression for V is, , z, , y=≈, , D 苷 兵共x, y兲, , V 苷 yy 共x 2  y 2 兲 dA 苷 y, , z=≈+¥, , 4, , y, , 0, , D, , 苷, , y冋, 4, , 0, , x, , FIGURE 11, , y=2x, , 册, , x3,  y 2x, 3, , sy, 1, 2, , y, , 共x 2  y 2 兲 dx dy, , x苷sy, , dy 苷, x苷2 y, 1, , y冉, 4, , 0, , y 3兾2, y3, y3,  y 5兾2 , , 3, 24, 2, , 冊, , dy, , y, , 苷 15 y 5兾2  7 y 7兾2  96 y 4, 2, , 2, , 13, , ], , 4, 0, , 苷, , 216, 35, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1016-1025.qk_97817_15_ch15_p1016-1025 11/8/10 3:34 PM Page 1016, , 1016, , CHAPTER 15, , MULTIPLE INTEGRALS, , v EXAMPLE 3 Evaluate xxD xy dA, where D is the region bounded by the line y 苷 x  1, and the parabola y 2 苷 2x  6., SOLUTION The region D is shown in Figure 12. Again D is both type I and type II, but the, , description of D as a type I region is more complicated because the lower boundary consists of two parts. Therefore we prefer to express D as a type II region:, D 苷 兵(x, y), , ⱍ 2  y  4,, , 1, 2, , y 2  3  x  y  1其, , y, , y, (5, 4), , (5, 4), , ¥, , x= 2 -3, , y=œ„„„„„, 2x+6, y=x-1, , x=y+1, x, , 0, , _3, , (_1, _2), , _2, , (_1, _2), , y=_œ„„„„„, 2x+6, FIGURE 12, , x, , 0, , (a) D as a type I region, , (b) D as a type II region, , Then 5 gives, 4, , yy xy dA 苷 y y, 2, , D, , y1, , 1, 2, , 4, , y 23, , xy dx dy 苷, , y, , 4, , 2, , 冋 册, , x苷y1, , x2, y, 2, , [, , dy, , x苷2 y 23, 1, , ], , 苷 12 y y 共y  1兲 2  ( 12 y 2  3) 2 dy, 2, , z, , 苷2, 1, , (0, 0, 2), , 苷, x+2y+z=2, , x=2y, T, , 冉, , 2, , 冋, , , , 冊, 册, , y5,  4y 3  2y 2  8y dy, 4, , 1, y6, y3, ,  y4  2,  4y 2, 2, 24, 3, , 4, , 苷 36, , 2, , If we had expressed D as a type I region using Figure 12(a), then we would have, obtained, , y, , 1, , (0, 1, 0), , 0, , y, , 4, , yy xy dA 苷 y y, D, , 1, , ”1, 2 , 0’, , s2x6, , s2x6, , 3, , xy dy dx  y, , 5, , 1, , y, , s2x6, , x1, , xy dy dx, , but this would have involved more work than the other method., , x, , EXAMPLE 4 Find the volume of the tetrahedron bounded by the planes x  2y  z 苷 2,, x 苷 2y, x 苷 0, and z 苷 0., , FIGURE 13, , SOLUTION In a question such as this, it’s wise to draw two diagrams: one of the three-, , y, 1, , x+2y=2, (or y=1-x/2), , ”1, 21 ’, , D, y=x/2, 0, , FIGURE 14, , 1, , x, , dimensional solid and another of the plane region D over which it lies. Figure 13 shows, the tetrahedron T bounded by the coordinate planes x 苷 0, z 苷 0, the vertical plane, x 苷 2y, and the plane x  2y  z 苷 2. Since the plane x  2y  z 苷 2 intersects the, xy-plane (whose equation is z 苷 0) in the line x  2y 苷 2, we see that T lies above the, triangular region D in the xy-plane bounded by the lines x 苷 2y, x  2y 苷 2, and x 苷 0., (See Figure 14.), The plane x  2y  z 苷 2 can be written as z 苷 2  x  2y, so the required volume, lies under the graph of the function z 苷 2  x  2y and above, D 苷 兵共x, y兲, , ⱍ, , 0  x  1, x兾2  y  1  x兾2其, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1016-1025.qk_97817_15_ch15_p1016-1025 11/8/10 3:35 PM Page 1018, , 1018, , CHAPTER 15, , MULTIPLE INTEGRALS, , If f 共x, y兲  t共x, y兲 for all 共x, y兲 in D, then, , yy f 共x, y兲 dA  yy t共x, y兲 dA, , 8, , D, , The next property of double integrals is similar to the property of single integrals given, by the equation xab f 共x兲 dx 苷 xac f 共x兲 dx  xcb f 共x兲 dx., If D 苷 D1 傼 D2 , where D1 and D2 don’t overlap except perhaps on their boundaries, (see Figure 17), then, , y, , D, D¡, , D, , D™, , x, , 0, , yy f 共x, y兲 dA 苷 yy f 共x, y兲 dA  yy f 共x, y兲 dA, , 9, , D1, , D, , D2, , FIGURE 17, , Property 9 can be used to evaluate double integrals over regions D that are neither type I, nor type II but can be expressed as a union of regions of type I or type II. Figure 18 illustrates this procedure. (See Exercises 55 and 56.), y, , y, , D™, D, , D¡, , 0, , FIGURE 18, , x, , 0, , x, , (b) D=D¡ 傼 D™, D¡ is type I, D™ is type II., , (a) D is neither type I nor type II., , The next property of integrals says that if we integrate the constant function f 共x, y兲 苷 1, over a region D, we get the area of D :, , 10, , yy 1 dA 苷 A共D兲, D, , Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is D and whose, height is 1 has volume A共D兲 ⴢ 1 苷 A共D兲, but we know that we can also write its volume, as xxD 1 dA., Finally, we can combine Properties 7, 8, and 10 to prove the following property. (See, Exercise 61.), , z, , z=1, , 0, , D, , y, , 11, , If m  f 共x, y兲  M for all 共x, y兲 in D, then, , x, , FIGURE 19, , mA共D兲  yy f 共x, y兲 dA  MA共D兲, D, , Cylinder with base D and height 1, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1016-1025.qk_97817_15_ch15_p1016-1025 11/8/10 3:35 PM Page 1020, , 1020, , CHAPTER 15, , MULTIPLE INTEGRALS, , 40. Between the paraboloids z 苷 2x 2  y 2 and, , 23–32 Find the volume of the given solid., , z 苷 8  x 2  2y 2 and inside the cylinder x 2  y 2 苷 1, , 23. Under the plane x  2y  z 苷 1 and above the region, , bounded by x  y 苷 1 and x 2  y 苷 1, , 41. Enclosed by z 苷 1  x 2  y 2 and z 苷 0, , 24. Under the surface z 苷 1  x 2 y 2 and above the region, , enclosed by x 苷 y and x 苷 4, 2, , 25. Under the surface z 苷 xy and above the triangle with, , vertices 共1, 1兲, 共4, 1兲, and 共1, 2兲, , 42. Enclosed by z 苷 x 2  y 2 and z 苷 2y, , 43– 48 Sketch the region of integration and change the order of, integration., , 26. Enclosed by the paraboloid z 苷 x 2  3y 2 and the planes, , x 苷 0, y 苷 1, y 苷 x, z 苷 0, , 1, , 45., , y y, , 0, , 47., , 0, , 兾2, , 0, , 28. Bounded by the planes z 苷 x, y 苷 x, x  y 苷 2, and z 苷 0, 29. Enclosed by the cylinders z 苷 x 2, y 苷 x 2 and the planes, , f 共x, y兲 dx dy, , yy, , 27. Bounded by the coordinate planes and the plane, , 3x  2y  z 苷 6, , y, , 43., , 2, , yy, 1, , cos x, , f 共x, y兲 dy dx, , 0, , ln x, , f 共x, y兲 dy dx, , 0, , 2, , 4, , 44., , yy, , 46., , y y, , 48., , 0, , x2, , 2, , 2, , s4y 2, , 0, , 1, , yy, 0, , f 共x, y兲 dy dx, , 兾4, , arctan x, , f 共x, y兲 dx dy, , f 共x, y兲 dy dx, , z 苷 0, y 苷 4, , 30. Bounded by the cylinder y 2  z 2 苷 4 and the planes x 苷 2y,, , 49–54 Evaluate the integral by reversing the order of integration., , 31. Bounded by the cylinder x 2  y 2 苷 1 and the planes y 苷 z,, , 49., , yy, , 32. Bounded by the cylinders x 2  y 2 苷 r 2 and y 2  z 2 苷 r 2, , 51., , yy, , x 苷 0, z 苷 0 in the first octant, , x 苷 0, z 苷 0 in the first octant, , ; 34. Find the approximate volume of the solid in the first octant, that is bounded by the planes y 苷 x, z 苷 0, and z 苷 x and, the cylinder y 苷 cos x. (Use a graphing device to estimate, the points of intersection.), , 4, , 0, , 1, , 54., , yy, , 1, dy dx, y3  1, , 2, , 兾2, , arcsin y, , 8, , 0, , 2, , e x dx dy, , sx, , yy, 0, , 3, , 3y, , 53., , ; 33. Use a graphing calculator or computer to estimate the, x-coordinates of the points of intersection of the curves, y 苷 x 4 and y 苷 3x  x 2. If D is the region bounded by, these curves, estimate xxD x dA., , 1, , 0, , 2, , s, , s, , 0, , y, , 50., , y y, , 52., , yy, , 1, , 0, , 1, , x, , cos共x 2 兲 dx dy, , e x兾y dy dx, , cos x s1  cos 2 x dx dy, 4, , 3, sy, , e x dx dy, , 55–56 Express D as a union of regions of type I or type II and, evaluate the integral., 55., , yy x, , 2, , dA, , 56., , yy y dA, , D, , D, , 35–36 Find the volume of the solid by subtracting two volumes., , y, , 35. The solid enclosed by the parabolic cylinders y 苷 1  x 2,, , 1, , y 苷 x 2  1 and the planes x  y  z 苷 2,, 2x  2y  z  10 苷 0, , y, , 1, (1, 1), , x=y-Á, , y=(x+1)@, , D, , _1, _1, , 36. The solid enclosed by the parabolic cylinder y 苷 x and the, 2, , 0, , 1, , x, , 0, , x, , planes z 苷 3y, z 苷 2  y, , _1, , _1, 37–38 Sketch the solid whose volume is given by the iterated, , integral., 37., , 1, , yy, 0, , 1x, , 0, , 共1  x  y兲 dy dx, , 38., , 1, , y y, 0, , 1x 2, , 0, , 共1  x兲 dy dx, , 57–58 Use Property 11 to estimate the value of the integral., 57., , yy e, , 共x 2 y 2 兲2, , dA,, , Q is the quarter-circle with center the, , Q, , CAS, , origin and radius 12 in the first quadrant, , 39– 42 Use a computer algebra system to find the exact volume, , of the solid., , 58., , 39. Under the surface z 苷 x y  xy and above the region, 3, , 4, , 2, , bounded by the curves y 苷 x 3  x and y 苷 x 2  x for x  0, , yy sin 共x  y兲 dA,, 4, , T is the triangle enclosed by the lines, , T, , y 苷 0, y 苷 2x, and x 苷 1, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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DOUBLE INTEGRALS IN POLAR COORDINATES, , SECTION 15.4, , 59–60 Find the average value of f over the region D., 59. f 共x, y兲 苷 xy,, , 64., , D is the triangle with vertices 共0, 0兲, 共1, 0兲,, , and 共1, 3兲, , 60. f 共x, y兲 苷 x sin y,, , y 苷 x 2, and x 苷 1, , D is enclosed by the curves y 苷 0,, , 65., , yy 共2x  3y兲 dA,, D, , D is the rectangle 0  x  a, 0  y  b, 66., , yy f 共x, y兲 dA 苷 y y, , 2y, , 0, , f 共x, y兲 dx dy , , 3, , y y, 1, , 3y, , 0, , yy 共2  x, D, , 2, , y 3  y 2 sin x兲 dA,, , D 苷 兵共x, y兲, , iterated integrals was obtained as follows:, f 共x, y兲 dx dy, , 67., , yy (ax, , 3, , ⱍ ⱍ x ⱍ  ⱍ y ⱍ  1其, ,  by 3  sa 2  x 2 ) dA,, , D, , D, , Sketch the region D and express the double integral as an, iterated integral with reversed order of integration., , D 苷 关a, a兴  关b, b兴, , CAS, , 63–67 Use geometry or symmetry, or both, to evaluate the, , yy 共x  2兲 dA,, , D 苷 兵共x, y兲, , D, , 15.4, , 0  y  s9  x 2 其, , ⱍ, , 68. Graph the solid bounded by the plane x  y  z 苷 1 and, , the paraboloid z 苷 4  x 2  y 2 and find its exact volume., (Use your CAS to do the graphing, to find the equations of, the boundary curves of the region of integration, and to evaluate the double integral.), , double integral., 63., ,  x 2  y 2 dA,, , D is the disk with center the origin and radius R, , 62. In evaluating a double integral over a region D, a sum of, 1, , 2, , D, , 61. Prove Property 11., , 0, , yy sR, , 1021, , Double Integrals in Polar Coordinates, Suppose that we want to evaluate a double integral xxR f 共x, y兲 dA, where R is one of the, regions shown in Figure 1. In either case the description of R in terms of rectangular coordinates is rather complicated, but R is easily described using polar coordinates., y, , y, , ≈+¥=4, , ≈+¥=1, R, , R, 0, , x, 0, , FIGURE 1, y, , P (r, ¨ ) =P (x, y), r, , (a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd, , x, , (b) R=s(r, ¨) | 1¯r¯2, 0¯¨¯πd, , Recall from Figure 2 that the polar coordinates 共r,  兲 of a point are related to the rectangular coordinates 共x, y兲 by the equations, x 苷 r cos , , r2 苷 x2  y2, , y, , ≈+¥=1, , y 苷 r sin , , ¨, O, , FIGURE 2, , x, , x, , (See Section 10.3.), The regions in Figure 1 are special cases of a polar rectangle, R 苷 兵共r, 兲, , ⱍ, , a  r  b,, , , , 其, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1016-1025.qk_97817_15_ch15_p1016-1025 11/8/10 3:35 PM Page 1023, , DOUBLE INTEGRALS IN POLAR COORDINATES, , SECTION 15.4, , 1023, , which is a Riemann sum for the double integral, , y y, , b, , a, , t共r,  兲 dr d, , Therefore we have, m, , yy f 共x, y兲 dA 苷, , i, , m, nl, , m, , i, , Ai, , j, , n, , 兺 兺 t共r*,  * 兲, , 苷 lim, , i, , m, nl, , y, , j, , i苷1 j苷1, , R, , 苷y, , n, , 兺 兺 f 共r* cos  *, r* sin  * 兲, , lim, , j, , r 苷y, , i苷1 j苷1, b, , a, , y, , b, , a, , t共r,  兲 dr d, , f 共r cos , r sin  兲 r dr d, , Change to Polar Coordinates in a Double Integral If f is continuous on a polar, , 2, ,    , where 0 , , rectangle R given by 0  a  r  b,, , yy f 共x, y兲 dA 苷 y y, , b, , a, , , ,  2, then, , f 共r cos , r sin  兲 r dr d, , R, , |, , dA, , The formula in 2 says that we convert from rectangular to polar coordinates in a, double integral by writing x 苷 r cos  and y 苷 r sin , using the appropriate limits of integration for r and , and replacing dA by r dr d. Be careful not to forget the additional, factor r on the right side of Formula 2. A classical method for remembering this is shown, in Figure 5, where the “infinitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions r d and dr and therefore has “area” dA 苷 r dr d., , d¨, dr, r, , r d¨, , EXAMPLE 1 Evaluate xxR 共3x  4y 2 兲 dA, where R is the region in the upper half-plane, , bounded by the circles x 2  y 2 苷 1 and x 2  y 2 苷 4., SOLUTION The region R can be described as, , R 苷 兵共x, y兲, , O, , ⱍ y  0,, , 1  x 2  y 2  4其, , It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by 1  r  2,, 0    . Therefore, by Formula 2,, , FIGURE 5, , yy 共3x  4y, , 2, , 兲 dA 苷 y, , , , 0, , y, , 2, , 1, , 共3r cos   4r 2 sin 2 兲 r dr d, , R, , 苷y, , , , 0, , 苷y, , , , 0, , Here we use the trigonometric identity, sin  苷 共1  cos 2 兲, 2, , 1, 2, , See Section 7.2 for advice on integrating, trigonometric functions., , 苷y, , , , 0, , y, , 2, , 1, , [r, , 共3r 2 cos   4r 3 sin 2 兲 dr d, , 3, , ], , cos   r 4 sin 2, , [7 cos  , , 苷 7 sin  , , 15, 2, , r苷2, r苷1, , , , d 苷 y 共7 cos   15 sin 2 兲 d, 0, , ], , 共1  cos 2 兲 d, , 册, , 15, 15, , sin 2, 2, 4, , , , 苷, 0, , 15, 2, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1026, , 1026, , MULTIPLE INTEGRALS, , CHAPTER 15, , Exercises, , 15.4, , 1– 4 A region R is shown. Decide whether to use polar coordinates, or rectangular coordinates and write xxR f 共x, y兲 dA as an iterated, integral, where f is an arbitrary continuous function on R., y, 4, , 1., , y, , 2., , 1, , 14., , xxD x dA, where D is the region in the first quadrant that lies, between the circles x 2  y 2 苷 4 and x 2  y 2 苷 2x, , 15–18 Use a double integral to find the area of the region., , y=1-≈, , 15. One loop of the rose r 苷 cos 3, 16. The region enclosed by both of the cardioids r 苷 1  cos , , 0, , x, , 4, , 0, , _1, , x, , 1, , and r 苷 1  cos , , 17. The region inside the circle 共x  1兲2  y 2 苷 1 and outside the, , circle x 2  y 2 苷 1, , 18. The region inside the cardioid r 苷 1  cos  and outside the, 3., , 4., , y, , circle r 苷 3 cos , , y, 6, , 1, , 19–27 Use polar coordinates to find the volume of the given solid., , 3, , 19. Under the cone z 苷 sx 2  y 2 and above the disk x 2  y 2  4, 0, , _1, , 1, , 0, , x, , x, , 20. Below the paraboloid z 苷 18  2x 2  2y 2 and above the, , xy-plane, 21. Enclosed by the hyperboloid x 2  y 2  z 2 苷 1 and the, , plane z 苷 2, , 5 –6 Sketch the region whose area is given by the integral and eval-, , uate the integral., 5., , 3兾4, , y y, 兾4, , 2, , 1, , r dr d, , 6., , , , y y, 兾2, , 2 sin , , 0, , 22. Inside the sphere x 2  y 2  z 2 苷 16 and outside the, , cylinder x 2  y 2 苷 4, , 23. A sphere of radius a, , r dr d, , 24. Bounded by the paraboloid z 苷 1  2x 2  2y 2 and the, , plane z 苷 7 in the first octant, , 7–14 Evaluate the given integral by changing to polar coordinates., 7., , 8., , 9., , xxD x 2 y dA,, , 25. Above the cone z 苷 sx 2  y 2 and below the sphere, , x 2  y 2  z2 苷 1, , where D is the top half of the disk with center the, origin and radius 5, , 26. Bounded by the paraboloids z 苷 3x 2  3y 2 and, , xxR 共2x  y兲 dA, where R is the region in the first quadrant, enclosed by the circle x 2  y 2 苷 4 and the lines x 苷 0 and, y苷x, , 27. Inside both the cylinder x 2  y 2 苷 4 and the ellipsoid, , xxR sin共x 2  y 2 兲 dA,, , 28. (a) A cylindrical drill with radius r 1 is used to bore a hole, , where R is the region in the first quadrant, between the circles with center the origin and radii 1 and 3, , z 苷 4  x2  y2, , 4x 2  4y 2  z 2 苷 64, , through the center of a sphere of radius r 2 . Find the volume, of the ring-shaped solid that remains., (b) Express the volume in part (a) in terms of the height h of, the ring. Notice that the volume depends only on h, not, on r 1 or r 2 ., , y2, dA, where R is the region that lies between the, x  y2, circles x 2  y 2 苷 a 2 and x 2  y 2 苷 b 2 with 0  a  b, , 10., , xxR, , 11., , xxD ex y, , dA, where D is the region bounded by the, semicircle x 苷 s4  y 2 and the y-axis, , 29–32 Evaluate the iterated integral by converting to polar, coordinates., , 12., , xxD cos sx 2  y 2, , dA, where D is the disk with center the, origin and radius 2, , 29., , y y, , 13., , xxR arctan共 y兾x兲 dA,, , 31., , yy, , 2, , 2, , 2, , where R 苷 兵共x, y兲, , ⱍ1x, , 2, ,  y  4, 0  y  x其, 2, , 3, , s9x 2, , 3, , 0, , 1, , s2y 2, , 0, , y, , a, , sin共x 2  y 2 兲 dy dx, , 30., , y y, , 共x  y兲 dx dy, , 32., , y y, , 0, , 2, , 0, , 0, , sa 2 y 2, s2xx 2, , 0, , x 2 y dx dy, sx 2  y 2 dy dx, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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APPLICATIONS OF DOUBLE INTEGRALS, , SECTION 15.5, , 33–34 Express the double integral in terms of a single integral with, , 40. (a) We define the improper integral (over the entire plane ⺢ 2 兲, , respect to r. Then use your calculator to evaluate the integral correct to four decimal places., 33., , xxD e 共x y 兲, 2, , 2 2, , I 苷 yy e共x y 兲 dA 苷, 2, , 共x 2y 2 兲, , 苷 lim, , yy e, , al, , xxD xys1  x 2  y 2, , dA, where D is the portion of the disk, x 2  y 2  1 that lies in the first quadrant, , , , , , , , , , y y, , on the annular region a 2  x 2  y 2  b 2, where 0  a  b., , 共x 2y 2 兲, , ⺢, , 1兾s2, , s1x, , 2, , xy dy dx , , s2, , y y, 1, , x, , 0, , y, , , , dA 苷 lim, , 2, , , , ex dx y, , , , 共x 2y 2 兲, , yy e, , dA, , Sa, , a兲. Use this to, , 2, , , , ey dy 苷 , , (c) Deduce that, , y, , , , 2, , , , ex dx 苷 s, , (d) By making the change of variable t 苷 s2 x, show that, , y, , , , 2, , , , ex 兾2 dx 苷 s2, , (This is a fundamental result for probability and statistics.), , xy dy dx  y, , 2, , s2, , y, , s4x 2, , 0, , xy dy dx, , into one double integral. Then evaluate the double integral., , 15.5, , 2, , where Sa is the square with vertices 共 a,, show that, , 39. Use polar coordinates to combine the sum, x, , 2, , e共x y 兲 dA 苷 , , al, , the average distance from points in D to the origin?, , 1, , 2, , dA, , 2, , 38. Let D be the disk with center the origin and radius a. What is, , y y, , 2, , e共x y 兲 dy dx, , (b) An equivalent definition of the improper integral in part (a), is, , yy e, , 37. Find the average value of the function f 共x, y兲 苷 1兾sx 2  y 2, , , , , , where Da is the disk with radius a and center the origin., Show that, , 36. An agricultural sprinkler distributes water in a circular pattern, , of radius 100 ft. It supplies water to a depth of er feet per hour, at a distance of r feet from the sprinkler., (a) If 0  R  100, what is the total amount of water supplied, per hour to the region inside the circle of radius R centered, at the sprinkler?, (b) Determine an expression for the average amount of water, per hour per square foot supplied to the region inside the, circle of radius R., , , , , , y y, , Da, , 35. A swimming pool is circular with a 40-ft diameter. The depth is, , constant along east-west lines and increases linearly from 2 ft, at the south end to 7 ft at the north end. Find the volume of, water in the pool., , 2, , ⺢2, , dA, where D is the disk with center the origin and, , radius 1, 34., , 1027, , 41. Use the result of Exercise 40 part (c) to evaluate the following, , integrals., (a), , y, , , , 0, , 2, , x 2ex dx, , (b), , y, , , , 0, , sx ex dx, , Applications of Double Integrals, We have already seen one application of double integrals: computing volumes. Another, geometric application is finding areas of surfaces and this will be done in the next section., In this section we explore physical applications such as computing mass, electric charge,, center of mass, and moment of inertia. We will see that these physical ideas are also important when applied to probability density functions of two random variables., , Density and Mass, In Section 8.3 we were able to use single integrals to compute moments and the center of, mass of a thin plate or lamina with constant density. But now, equipped with the double, integral, we can consider a lamina with variable density. Suppose the lamina occupies a, region D of the xy-plane and its density (in units of mass per unit area) at a point 共x, y兲 in, D is given by 共x, y兲, where is a continuous function on D. This means that, 共x, y兲 苷 lim, , m, A, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1028, , 1028, , CHAPTER 15, , MULTIPLE INTEGRALS, , where m and A are the mass and area of a small rectangle that contains 共x, y兲 and the limit, is taken as the dimensions of the rectangle approach 0. (See Figure 1.), To find the total mass m of the lamina we divide a rectangle R containing D into subrectangles Rij of the same size (as in Figure 2) and consider 共x, y兲 to be 0 outside D. If we, choose a point 共x ij*, yij* 兲 in Rij , then the mass of the part of the lamina that occupies Rij is, approximately 共x ij*, yij* 兲 A, where A is the area of Rij . If we add all such masses, we get, an approximation to the total mass:, , y, (x, y), , D, , 0, , x, , k, , FIGURE 1, , l, , 兺兺, , m⬇, , 共x *ij , y*ij 兲 A, , i苷1 j苷1, , y, , (xij* , yij* ), , If we now increase the number of subrectangles, we obtain the total mass m of the lamina, as the limiting value of the approximations:, , R ij, , k, , 1, , 0, , x, , FIGURE 2, , m 苷 lim, , l, , 兺兺, , k, l l  i苷1 j苷1, , 共x *ij , y*ij 兲 A 苷 yy 共x, y兲 dA, D, , Physicists also consider other types of density that can be treated in the same manner. For, example, if an electric charge is distributed over a region D and the charge density (in units, of charge per unit area) is given by 共x, y兲 at a point 共x, y兲 in D, then the total charge Q is, given by, Q 苷 yy, , 2, , 共x, y兲 dA, , D, , y, , y=1, , 1, , EXAMPLE 1 Charge is distributed over the triangular region D in Figure 3 so that the, charge density at 共x, y兲 is 共x, y兲 苷 xy, measured in coulombs per square meter (C兾m 2 )., Find the total charge., , (1, 1), , D, , SOLUTION From Equation 2 and Figure 3 we have, , Q 苷 yy, , y=1-x, , 共x, y兲 dA 苷 y, , D, , 0, , x, , 苷, , y, , 冋 册, y2, x, 2, , 1, , 0, , FIGURE 3, , 苷, Thus the total charge is, , 1, 2, , y, , 5, 24, , 1, , 0, , 1, , 0, , y, , 1, , y苷1, , dx 苷 y, y苷1x, , 3, , 1, , 0, , 1, 共2x  x 兲 dx 苷, 2, 2, , xy dy dx, , 1x, , x 2, 关1  共1  x兲2 兴 dx, 2, , 冋, , 2x 3, x4, , 3, 4, , 册, , 1, , 0, , 苷, , 5, 24, , C., , Moments and Centers of Mass, In Section 8.3 we found the center of mass of a lamina with constant density; here we consider a lamina with variable density. Suppose the lamina occupies a region D and has density function 共x, y兲. Recall from Chapter 8 that we defined the moment of a particle about, an axis as the product of its mass and its directed distance from the axis. We divide D into, small rectangles as in Figure 2. Then the mass of Rij is approximately 共x *ij , y*ij 兲 A, so we, can approximate the moment of Rij with respect to the x-axis by, 关 共x *ij , y*ij 兲 A兴 y*ij, If we now add these quantities and take the limit as the number of subrectangles becomes, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1030, , 1030, , MULTIPLE INTEGRALS, , CHAPTER 15, , y苷, 苷, , 1, m, 3, 8, , 1, 苷, 4, , 共x, y兲 dA 苷, , yy y, , 3, 8, , 1, , y y, 0, , D, , y冋, 冋, 1, , 0, , 22x, , 共y  3xy  y 2 兲 dy dx, , 0, , y2, y2, y3,  3x, , 2, 2, 3, , 册, , y苷22x, 1, , dx 苷 14 y 共7  9x  3x 2  5x 3 兲 dx, 0, , y苷0, , 册, , 1, , x2, x4, 7x  9,  x3  5, 2, 4, , 苷, 0, , 11, 16, , The center of mass is at the point ( 8 , 16 )., 3 11, , v EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the, distance from the center of the circle. Find the center of mass of the lamina., SOLUTION Let’s place the lamina as the upper half of the circle x 2  y 2 苷 a 2. (See Fig-, , y, a, , D, , _a, , ure 6.) Then the distance from a point 共x, y兲 to the center of the circle (the origin) is, sx 2  y 2 . Therefore the density function is, , ≈+¥=a@, 3a, , ”0, 2π ’, 0, , 共x, y兲 苷 Ksx 2  y 2, a, , x, , where K is some constant. Both the density function and the shape of the lamina suggest, that we convert to polar coordinates. Then sx 2  y 2 苷 r and the region D is given by, 0  r  a, 0    . Thus the mass of the lamina is, , FIGURE 6, , m 苷 yy 共x, y兲 dA 苷 yy Ksx 2  y 2 dA, D, , 苷y, , , , 0, , D, , y, , 苷 K, , a, , 0, , , , a, , 共Kr兲 r dr d 苷 K y d y r 2 dr, , r3, 3, , 0, , 册, , a, , 苷, , 0, , 0, , K a 3, 3, , Both the lamina and the density function are symmetric with respect to the y-axis, so the, center of mass must lie on the y-axis, that is, x 苷 0. The y-coordinate is given by, y苷, , Compare the location of the center of mass in, Example 3 with Example 4 in Section 8.3,, where we found that the center of mass of, a lamina with the same shape but uniform, density is located at the point 共0, 4a兾共3兲兲., , 1, m, , yy y, D, , 3, 苷, a3, 苷, , 3, K a 3, , 共x, y兲 dA 苷, , y, , , , 0, , sin  d, , y, , a, , 0, , , , yy, 0, , a, , 0, , r sin  共 r兲 r dr d, , 3, r dr 苷, cos , a3, 3, , [, , 冋册, , , 0, , ], , r4, 4, , a, , 0, , 3 2a 4, 3a, 苷, a3 4, 2, , Therefore the center of mass is located at the point 共0, 3a兾共2兲兲., , Moment of Inertia, The moment of inertia (also called the second moment) of a particle of mass m about an, axis is defined to be mr 2, where r is the distance from the particle to the axis. We extend this, concept to a lamina with density function 共x, y兲 and occupying a region D by proceeding, as we did for ordinary moments. We divide D into small rectangles, approximate the, moment of inertia of each subrectangle about the x-axis, and take the limit of the sum, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1032, , 1032, , CHAPTER 15, , MULTIPLE INTEGRALS, , that mass plays in linear motion. The moment of inertia of a wheel is what makes it difficult to start or stop the rotation of the wheel, just as the mass of a car is what makes it difficult to start or stop the motion of the car., The radius of gyration of a lamina about an axis is the number R such that, mR 2 苷 I, , 9, , where m is the mass of the lamina and I is the moment of inertia about the given axis., Equation 9 says that if the mass of the lamina were concentrated at a distance R from the, axis, then the moment of inertia of this “point mass” would be the same as the moment of, inertia of the lamina., In particular, the radius of gyration y with respect to the x-axis and the radius of gyration x with respect to the y-axis are given by the equations, my 2 苷 I x, , 10, , mx 2 苷 I y, , Thus 共x, y兲 is the point at which the mass of the lamina can be concentrated without changing the moments of inertia with respect to the coordinate axes. (Note the analogy with the, center of mass.), , v, , EXAMPLE 5 Find the radius of gyration about the x-axis of the disk in Example 4., , SOLUTION As noted, the mass of the disk is m 苷, , y2 苷, , a 2, so from Equations 10 we have, , 1, Ix,  a4, a2, 苷 4, 苷, m, a 2, 4, , Therefore the radius of gyration about the x-axis is y 苷 2 a , which is half the radius of, the disk., 1, , Probability, In Section 8.5 we considered the probability density function f of a continuous random vari, able X. This means that f 共x兲  0 for all x, x, f 共x兲 dx 苷 1, and the probability that X lies, between a and b is found by integrating f from a to b:, b, , P共a  X  b兲 苷 y f 共x兲 dx, a, , Now we consider a pair of continuous random variables X and Y, such as the lifetimes, of two components of a machine or the height and weight of an adult female chosen at random. The joint density function of X and Y is a function f of two variables such that the, probability that 共X, Y兲 lies in a region D is, P (共X, Y 兲 僆 D) 苷 yy f 共x, y兲 dA, D, , In particular, if the region is a rectangle, the probability that X lies between a and b and, Y lies between c and d is, P共a  X  b, c  Y  d 兲 苷 y, , b, , a, , y, , d, , c, , f 共x, y兲 dy dx, , (See Figure 7.), , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1034, , 1034, , CHAPTER 15, , MULTIPLE INTEGRALS, , Suppose X is a random variable with probability density function f1共x兲 and Y is a random, variable with density function f2共y兲. Then X and Y are called independent random variables if their joint density function is the product of their individual density functions:, f 共x, y兲 苷 f1共x兲 f2共 y兲, In Section 8.5 we modeled waiting times by using exponential density functions, f 共t兲 苷, , 再, , 0, if t  0, 1et兾 if t  0, , where  is the mean waiting time. In the next example we consider a situation with two independent waiting times., EXAMPLE 7 The manager of a movie theater determines that the average time moviegoers wait in line to buy a ticket for this week’s film is 10 minutes and the average time, they wait to buy popcorn is 5 minutes. Assuming that the waiting times are independent,, find the probability that a moviegoer waits a total of less than 20 minutes before taking, his or her seat., SOLUTION Assuming that both the waiting time X for the ticket purchase and the waiting, , time Y in the refreshment line are modeled by exponential probability density functions,, we can write the individual density functions as, f1共x兲 苷, , 再, , 0, 1 x兾10, 10, , e, , if x  0, if x  0, , f2共y兲 苷, , 再, , if y  0, if y  0, , 0, 1 y兾5, 5, , e, , Since X and Y are independent, the joint density function is the product:, f 共x, y兲 苷 f1共x兲 f2共y兲 苷, , 再, , 1 x兾10 y兾5, 50, , e, , if x  0, y  0, otherwise, , e, , 0, , We are asked for the probability that X  Y  20:, P共X  Y  20兲 苷 P (共X, Y兲 僆 D), where D is the triangular region shown in Figure 8. Thus, , y, 20, , P共X  Y  20兲 苷 yy f 共x, y兲 dA 苷 y, x+y=20, , 1, , 20, , 0, , 0, , 20 x, , y, , 20x 1 x兾10 y兾5, 50, , e, , 0, , e, , dy dx, , D, , 苷 50 y, , D, , 20, , 0, , [e, , x兾10, , ], , 共5兲ey兾5, , y苷20x, y苷0, , dx, , 20, , 苷 10 y ex兾10共1  e 共x20兲兾5 兲 dx, 1, , 0, , FIGURE 8, , 20, , 苷 10 y 共ex兾10  e4e x兾10 兲 dx, 1, , 0, , 苷 1  e4  2e2 ⬇ 0.7476, This means that about 75% of the moviegoers wait less than 20 minutes before taking, their seats., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1035, , SECTION 15.5, , APPLICATIONS OF DOUBLE INTEGRALS, , 1035, , Expected Values, Recall from Section 8.5 that if X is a random variable with probability density function f,, then its mean is, , ,  苷 y x f 共x兲 dx, , , Now if X and Y are random variables with joint density function f, we define the X-mean, and Y-mean, also called the expected values of X and Y, to be, , 1 苷 yy x f 共x, y兲 dA, , 11, ,  2 苷 yy yf 共x, y兲 dA, , ⺢2, , ⺢2, , Notice how closely the expressions for 1 and  2 in 11 resemble the moments Mx and My, of a lamina with density function in Equations 3 and 4. In fact, we can think of probability as being like continuously distributed mass. We calculate probability the way we calculate mass—by integrating a density function. And because the total “probability mass” is 1,, the expressions for x and y in 5 show that we can think of the expected values of X and Y,, 1 and  2 , as the coordinates of the “center of mass” of the probability distribution., In the next example we deal with normal distributions. As in Section 8.5, a single random variable is normally distributed if its probability density function is of the form, f 共x兲 苷, where  is the mean and, , 1, 2, e共x兲 兾共2, s2, , 2, , 兲, , is the standard deviation., , EXAMPLE 8 A factory produces (cylindrically shaped) roller bearings that are sold as, having diameter 4.0 cm and length 6.0 cm. In fact, the diameters X are normally distributed with mean 4.0 cm and standard deviation 0.01 cm while the lengths Y are normally, distributed with mean 6.0 cm and standard deviation 0.01 cm. Assuming that X and Y are, independent, write the joint density function and graph it. Find the probability that a bearing randomly chosen from the production line has either length or diameter that differs, from the mean by more than 0.02 cm., , 1 苷 4.0,  2 苷 6.0,, 苷 0.01. So the individual density functions for X and Y are, , SOLUTION We are given that X and Y are normally distributed with, , and, , 1, , 苷, , 2, , f1共x兲 苷, , 1, 2, e共x4兲 兾0.0002, 0.01s2, , f2共y兲 苷, , 1, 2, e共 y6兲 兾0.0002, 0.01s2, , Since X and Y are independent, the joint density function is the product:, 1500, , f 共x, y兲 苷 f1共x兲 f2共y兲, , 1000, 500, 0, 5.95, , 3.95, y, , 4, , 6, 6.05, , 苷, , 1, 2, 2, e共x4兲 兾0.0002e共y6兲 兾0.0002, 0.0002, , 苷, , 5000 5000关共x4兲2共 y6兲2兴, e, , , x, , 4.05, , FIGURE 9, , Graph of the bivariate normal joint, density function in Example 8, , A graph of this function is shown in Figure 9., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1036-1045.qk_97817_15_ch15_p1036-1045 11/8/10 3:37 PM Page 1036, , 1036, , CHAPTER 15, , MULTIPLE INTEGRALS, , Let’s first calculate the probability that both X and Y differ from their means by less, than 0.02 cm. Using a calculator or computer to estimate the integral, we have, P共3.98 ⬍ X ⬍ 4.02, 5.98 ⬍ Y ⬍ 6.02兲 苷 y, , 4.02, , 3.98, , 苷, , y, , 5000, ␲, , 6.02, , 5.98, , f 共x, y兲 dy dx, , 4.02, , y y, 3.98, , 6.02, , 5.98, , 2, , 2, , e⫺5000关共x⫺4兲 ⫹共 y⫺6兲 兴 dy dx, , ⬇ 0.91, Then the probability that either X or Y differs from its mean by more than 0.02 cm is, approximately, 1 ⫺ 0.91 苷 0.09, , 15.5, , Exercises, , 1. Electric charge is distributed over the rectangle 0 艋 x 艋 5,, , 2 艋 y 艋 5 so that the charge density at 共x, y兲 is, ␴ 共x, y兲 苷 2x ⫹ 4y (measured in coulombs per square meter)., Find the total charge on the rectangle., , 2. Electric charge is distributed over the disk x 2 ⫹ y 2 艋 1 so, , that the charge density at 共x, y兲 is ␴ 共x, y兲 苷 sx 2 ⫹ y 2, (measured in coulombs per square meter). Find the total charge, on the disk., , 3–10 Find the mass and center of mass of the lamina that occupies, the region D and has the given density function ␳., , ⱍ 1 艋 x 艋 3, 1 艋 y 艋 4其 ; ␳ 共x, y兲 苷 ky, D 苷 兵共x, y兲 ⱍ 0 艋 x 艋 a, 0 艋 y 艋 b其 ; ␳ 共x, y兲 苷 1 ⫹ x, , 3. D 苷 兵共x, y兲, 4., , 2, , 2, , ⫹ y2, , 5. D is the triangular region with vertices 共0, 0兲, 共2, 1兲, 共0, 3兲;, , ␳ 共x, y兲 苷 x ⫹ y, , 6. D is the triangular region enclosed by the lines x 苷 0, y 苷 x,, , and 2x ⫹ y 苷 6; ␳ 共x, y兲 苷 x 2, , 7. D is bounded by y 苷 1 ⫺ x and y 苷 0; ␳ 共x, y兲 苷 ky, 8. D is bounded by y 苷 x 2 and y 苷 x ⫹ 2; ␳ 共x, y兲 苷 kx, , ⱍ 0 艋 y 艋 sin共␲ x兾L兲, 0 艋 x 艋 L其 ;, , ␳ 共x, y兲 苷 y, , 10. D is bounded by the parabolas y 苷 x and x 苷 y ;, 2, , 2, , ␳ 共x, y兲 苷 sx, , 11. A lamina occupies the part of the disk x 2 ⫹ y 2 艋 1 in the first, , quadrant. Find its center of mass if the density at any point is, proportional to its distance from the x-axis., 12. Find the center of mass of the lamina in Exercise 11 if the, , density at any point is proportional to the square of its, distance from the origin., CAS Computer algebra system required, , y 苷 s1 ⫺ x 2 and y 苷 s4 ⫺ x 2 together with the portions, of the x-axis that join them. Find the center of mass of the lamina if the density at any point is proportional to its distance, from the origin., 14. Find the center of mass of the lamina in Exercise 13 if the den-, , sity at any point is inversely proportional to its distance from, the origin., 15. Find the center of mass of a lamina in the shape of an isos-, , celes right triangle with equal sides of length a if the density at, any point is proportional to the square of the distance from the, vertex opposite the hypotenuse., 16. A lamina occupies the region inside the circle x 2 ⫹ y 2 苷 2y, , but outside the circle x 2 ⫹ y 2 苷 1. Find the center of mass, if the density at any point is inversely proportional to its, distance from the origin., , 17. Find the moments of inertia I x , I y , I 0 for the lamina of, , Exercise 7., , 2, , 9. D 苷 兵共x, y兲, , 13. The boundary of a lamina consists of the semicircles, , 18. Find the moments of inertia I x , I y , I 0 for the lamina of, , Exercise 12., 19. Find the moments of inertia I x , I y , I 0 for the lamina of, , Exercise 15., 20. Consider a square fan blade with sides of length 2 and the, , lower left corner placed at the origin. If the density of the blade, is ␳ 共x, y兲 苷 1 ⫹ 0.1x, is it more difficult to rotate the blade, about the x-axis or the y-axis?, 21–24 A lamina with constant density ␳ 共x, y兲 苷 ␳ occupies the, , given region. Find the moments of inertia I x and I y and the radii of, gyration x and y., 21. The rectangle 0 艋 x 艋 b, 0 艋 y 艋 h, 22. The triangle with vertices 共0, 0兲, 共b, 0兲, and 共0, h兲, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 15.6, , 23. The part of the disk x 2 ⫹ y 2 艋 a 2 in the first quadrant, , 25–26 Use a computer algebra system to find the mass, center, of mass, and moments of inertia of the lamina that occupies the, region D and has the given density function., 25. D is enclosed by the right loop of the four-leaved rose, , r 苷 cos 2␪ ; ␳ 共x, y兲 苷 x 2 ⫹ y 2, 26. D 苷 兵共x, y兲, , ⱍ 0 艋 y 艋 xe, , ⫺x, , , 0 艋 x 艋 2 其 ; ␳ 共x, y兲 苷 x 2 y 2, , 27. The joint density function for a pair of random variables X, , and Y is, f 共x, y兲 苷, , 再, , Cx共1 ⫹ y兲 if 0 艋 x 艋 1, 0 艋 y 艋 2, 0, otherwise, , (a) Find the value of the constant C., (b) Find P共X 艋 1, Y 艋 1兲., (c) Find P共X ⫹ Y 艋 1兲., 28. (a) Verify that, , f 共x, y兲 苷, , 再, , 4xy if 0 艋 x 艋 1, 0 艋 y 艋 1, 0, otherwise, , is a joint density function., (b) If X and Y are random variables whose joint density, function is the function f in part (a), find, (i) P (X 艌 12 ), (ii) P (X 艌 12 , Y 艋 12 ), (c) Find the expected values of X and Y., 29. Suppose X and Y are random variables with joint density, , function, f 共x, y兲 苷, , 再, , 0.1e⫺共0.5x⫹0.2y兲 if x 艌 0, y 艌 0, 0, otherwise, , (a) Verify that f is indeed a joint density function., (b) Find the following probabilities., (i) P共Y 艌 1兲, (ii) P共X 艋 2, Y 艋 4兲, (c) Find the expected values of X and Y., 30. (a) A lamp has two bulbs of a type with an average lifetime, , of 1000 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density, function with mean ␮ 苷 1000, find the probability that, both of the lamp’s bulbs fail within 1000 hours., , 15.6, , 1037, , (b) Another lamp has just one bulb of the same type as in, part (a). If one bulb burns out and is replaced by a bulb, of the same type, find the probability that the two bulbs, fail within a total of 1000 hours., , 24. The region under the curve y 苷 sin x from x 苷 0 to x 苷 ␲, , CAS, , SURFACE AREA, , CAS, , 31. Suppose that X and Y are independent random variables,, , where X is normally distributed with mean 45 and standard, deviation 0.5 and Y is normally distributed with mean 20, and standard deviation 0.1., (a) Find P共40 艋 X 艋 50, 20 艋 Y 艋 25兲., (b) Find P (4共X ⫺ 45兲2 ⫹ 100共Y ⫺ 20兲2 艋 2)., 32. Xavier and Yolanda both have classes that end at noon and, , they agree to meet every day after class. They arrive at the, coffee shop independently. Xavier’s arrival time is X and, Yolanda’s arrival time is Y, where X and Y are measured in, minutes after noon. The individual density functions are, f1共x兲 苷, , 再, , e⫺x, 0, , if x 艌 0, if x ⬍ 0, , f2共 y兲 苷, , 再, , 1, 50, , y, , 0, , if 0 艋 y 艋 10, otherwise, , (Xavier arrives sometime after noon and is more likely, to arrive promptly than late. Yolanda always arrives by, 12:10 PM and is more likely to arrive late than promptly.), After Yolanda arrives, she’ll wait for up to half an hour for, Xavier, but he won’t wait for her. Find the probability that, they meet., 33. When studying the spread of an epidemic, we assume that, , the probability that an infected individual will spread the, disease to an uninfected individual is a function of the distance between them. Consider a circular city of radius, 10 miles in which the population is uniformly distributed., For an uninfected individual at a fixed point A共x 0 , y0 兲,, assume that the probability function is given by, f 共P兲 苷 20 关20 ⫺ d共P, A兲兴, 1, , where d共P, A兲 denotes the distance between points P and A., (a) Suppose the exposure of a person to the disease is the, sum of the probabilities of catching the disease from all, members of the population. Assume that the infected, people are uniformly distributed throughout the city,, with k infected individuals per square mile. Find a, double integral that represents the exposure of a person, residing at A., (b) Evaluate the integral for the case in which A is the center, of the city and for the case in which A is located on the, edge of the city. Where would you prefer to live?, , Surface Area, , In Section 16.6 we will deal with areas of more, general surfaces, called parametric surfaces, and, so this section need not be covered if that later, section will be covered., , In this section we apply double integrals to the problem of computing the area of a surface., In Section 8.2 we found the area of a very special type of surface––a surface of revolution––by the methods of single-variable calculus. Here we compute the area of a surface, with equation z 苷 f 共x, y兲, the graph of a function of two variables., Let S be a surface with equation z 苷 f 共x, y兲, where f has continuous partial derivatives., For simplicity in deriving the surface area formula, we assume that f 共x, y兲 艌 0 and the, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1036-1045.qk_97817_15_ch15_p1036-1045 11/8/10 3:37 PM Page 1040, , 1040, , CHAPTER 15, , 15.6, , MULTIPLE INTEGRALS, , Exercises, , 1–12 Find the area of the surface., , 16. (a) Use the Midpoint Rule for double integrals with, , CAS, , m 苷 n 苷 2 to estimate the area of the surface, z 苷 xy ⫹ x 2 ⫹ y 2, 0 艋 x 艋 2, 0 艋 y 艋 2., (b) Use a computer algebra system to approximate the surface area in part (a) to four decimal places. Compare, with the answer to part (a)., , CAS, , 17. Find the exact area of the surface z 苷 1 ⫹ 2x ⫹ 3y ⫹ 4y 2,, , 1. The part of the plane z 苷 2 ⫹ 3x ⫹ 4y that lies above the, , rectangle 关0, 5兴 ⫻ 关1, 4兴, , 2. The part of the plane 2x ⫹ 5y ⫹ z 苷 10 that lies inside the, , cylinder x 2 ⫹ y 2 苷 9, , 3. The part of the plane 3x ⫹ 2y ⫹ z 苷 6 that lies in the, , 1 艋 x 艋 4, 0 艋 y 艋 1., , first octant, 4. The part of the surface z 苷 1 ⫹ 3x ⫹ 2y that lies above, 2, , the triangle with vertices 共0, 0兲, 共0, 1兲, and 共2, 1兲, , CAS, , z 苷 1 ⫹ x ⫹ y ⫹ x2, , 5. The part of the cylinder y 2 ⫹ z 2 苷 9 that lies above the rect-, , angle with vertices 共0, 0兲, 共4, 0兲, 共0, 2兲, and 共4, 2兲, , CAS, , 19. Find, to four decimal places, the area of the part of the sur-, , face z 苷 1 ⫹ x 2 y 2 that lies above the disk x 2 ⫹ y 2 艋 1., , 7. The part of the hyperbolic paraboloid z 苷 y ⫺ x that lies, 2, , 2, , between the cylinders x 2 ⫹ y 2 苷 1 and x 2 ⫹ y 2 苷 4, , 8. The surface z 苷 共x, 2, 3, , 3兾2, , ⫹y, , 3兾2, , 兲, 0 艋 x 艋 1, 0 艋 y 艋 1, , 9. The part of the surface z 苷 xy that lies within the cylinder, , x2 ⫹ y2 苷 1, , CAS, , 20. Find, to four decimal places, the area of the part of the, , surface z 苷 共1 ⫹ x 2 兲兾共1 ⫹ y 2 兲 that lies above the square, x ⫹ y 艋 1. Illustrate by graphing this part of the, surface., , ⱍ ⱍ ⱍ ⱍ, , 21. Show that the area of the part of the plane z 苷 ax ⫹ by ⫹ c, , 10. The part of the sphere x ⫹ y ⫹ z 苷 4 that lies above the, 2, , 2, , 2, , plane z 苷 1, , 11. The part of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 a 2 that lies within the, , cylinder x 2 ⫹ y 2 苷 ax and above the xy-plane, , 12. The part of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 4z that lies inside the, , paraboloid z 苷 x 2 ⫹ y 2, , 13–14 Find the area of the surface correct to four decimal places, , by expressing the area in terms of a single integral and using, your calculator to estimate the integral., ⫺x 2⫺y 2, , 13. The part of the surface z 苷 e, , that lies above the disk, , x ⫹y 艋4, 2, , 14. The part of the surface z 苷 cos共x 2 ⫹ y 2 兲 that lies inside the, , cylinder x 2 ⫹ y 2 苷 1, , that projects onto a region D in the xy-plane with area A共D兲, is sa 2 ⫹ b 2 ⫹ 1 A共D兲., 22. If you attempt to use Formula 2 to find the area of the top, , half of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 a 2, you have a slight, problem because the double integral is improper. In fact, the, integrand has an infinite discontinuity at every point of the, boundary circle x 2 ⫹ y 2 苷 a 2. However, the integral can, be computed as the limit of the integral over the disk, x 2 ⫹ y 2 艋 t 2 as t l a ⫺. Use this method to show that the, area of a sphere of radius a is 4␲ a 2., 23. Find the area of the finite part of the paraboloid y 苷 x 2 ⫹ z 2, , cut off by the plane y 苷 25. [Hint: Project the surface onto, the xz-plane.], , 24. The figure shows the surface created when the cylinder, , y 2 ⫹ z 2 苷 1 intersects the cylinder x 2 ⫹ z 2 苷 1. Find the, area of this surface., z, , 15. (a) Use the Midpoint Rule for double integrals (see Sec-, , CAS, , ⫺1 艋 y 艋 1, , 2, , the xy-plane, , 2, , ⫺2 艋 x 艋 1, , Illustrate by graphing the surface., , 6. The part of the paraboloid z 苷 4 ⫺ x ⫺ y that lies above, 2, , 18. Find the exact area of the surface, , tion 15.1) with four squares to estimate the surface area, of the portion of the paraboloid z 苷 x 2 ⫹ y 2 that lies, above the square 关0, 1兴 ⫻ 关0, 1兴., (b) Use a computer algebra system to approximate the surface area in part (a) to four decimal places. Compare, with the answer to part (a)., , CAS Computer algebra system required, , x, , y, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1036-1045.qk_97817_15_ch15_p1036-1045 11/11/10 9:37 AM Page 1042, , 1042, , CHAPTER 15, , MULTIPLE INTEGRALS, , which we can integrate, all of which give the same value. For instance, if we integrate with, respect to y, then z, and then x, we have, b, , s, , yyy f 共x, y, z兲 dV 苷 y y y, a, , r, , d, , c, , f 共x, y, z兲 dy dz dx, , B, , v, , EXAMPLE 1 Evaluate the triple integral xxxB xyz 2 dV, where B is the rectangular box, , given by, B 苷 兵共x, y, z兲, , ⱍ, , 0 艋 x 艋 1, ⫺1 艋 y 艋 2, 0 艋 z 艋 3其, , SOLUTION We could use any of the six possible orders of integration. If we choose to, , integrate with respect to x, then y, and then z, we obtain, , yyy xyz, , 2, , dV 苷 y, , 3, , 0, , B, , 苷y, , 3, , 0, , 苷y, , 3, , 0, , 2, , y y, ⫺1, , y, , 2, , ⫺1, , 1, , 0, , xyz dx dy dz 苷, 2, , yz 2, dy dz 苷, 2, , 3z 2, z3, dz 苷, 4, 4, , 册, , y, , 3, , 0, , 3, , 苷, , 0, , 3, , 2, , 0, , ⫺1, , yy, , 冋 册, y 2z 2, 4, , 冋 册, x 2 yz 2, 2, , x苷1, , dy dz, , x苷0, , y苷2, , dz, , y苷⫺1, , 27, 4, , Now we define the triple integral over a general bounded region E in threedimensional space (a solid) by much the same procedure that we used for double integrals, (15.3.2). We enclose E in a box B of the type given by Equation 1. Then we define F so, that it agrees with f on E but is 0 for points in B that are outside E. By definition,, , yyy f 共x, y, z兲 dV 苷 yyy F共x, y, z兲 dV, E, , z, , z=u™(x, y), E, z=u¡(x, y), , 0, x, , D, , FIGURE 2, , A type 1 solid region, , y, , B, , This integral exists if f is continuous and the boundary of E is “reasonably smooth.” The, triple integral has essentially the same properties as the double integral (Properties 6–9 in, Section 15.3)., We restrict our attention to continuous functions f and to certain simple types of regions., A solid region E is said to be of type 1 if it lies between the graphs of two continuous functions of x and y, that is,, 5, , E 苷 兵共x, y, z兲, , u 1共x, y兲 艋 z 艋 u 2共x, y兲其, , ⱍ 共x, y兲 僆 D,, , where D is the projection of E onto the xy-plane as shown in Figure 2. Notice that the, upper boundary of the solid E is the surface with equation z 苷 u 2共x, y兲, while the lower, boundary is the surface z 苷 u1共x, y兲., By the same sort of argument that led to (15.3.3), it can be shown that if E is a type 1, region given by Equation 5, then, , 6, , 冋, , yyy f 共x, y, z兲 dV 苷 yy y, E, , D, , u 2共x, y兲, , u1共x, y兲, , 册, , f 共x, y, z兲 dz dA, , The meaning of the inner integral on the right side of Equation 6 is that x and y are held, fixed, and therefore u1共x, y兲 and u 2共x, y兲 are regarded as constants, while f 共x, y, z兲 is integrated with respect to z., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1036-1045.qk_97817_15_ch15_p1036-1045 11/8/10 3:38 PM Page 1044, , 1044, , MULTIPLE INTEGRALS, , CHAPTER 15, z, , where, this time, D is the projection of E onto the yz-plane (see Figure 7). The back surface is x 苷 u1共y, z兲, the front surface is x 苷 u 2共y, z兲, and we have, D, , 0, , E, , y, , E, , x, , x=u¡(y, z), , D, , u 2共 y, z兲, , u1共 y, z兲, , 册, , f 共x, y, z兲 dx dA, , Finally, a type 3 region is of the form, , x=u™(y, z), , E 苷 兵共x, y, z兲 共x, z兲 僆 D, u1共x, z兲 艋 y 艋 u 2共x, z兲其, , ⱍ, , FIGURE 7, , A type 2 region, , where D is the projection of E onto the xz-plane, y 苷 u1共x, z兲 is the left surface, and, y 苷 u 2共x, z兲 is the right surface (see Figure 8). For this type of region we have, , z, , y=u™(x, z), , 冋, , yyy f 共x, y, z兲 dV 苷 yy y, , 11, , E, , D, , 冋, , yyy f 共x, y, z兲 dV 苷 yy y, , 10, , D, , u 2共x, z兲, , u1共x, z兲, , 册, , f 共x, y, z兲 dy dA, , E, 0, , y=u¡(x, z), x, , y, , FIGURE 8, , A type 3 region, , In each of Equations 10 and 11 there may be two possible expressions for the integral, depending on whether D is a type I or type II plane region (and corresponding to Equations 7 and 8)., , v, , EXAMPLE 3 Evaluate xxxE sx 2 ⫹ z 2 dV, where E is the region bounded by the parabo-, , loid y 苷 x 2 ⫹ z 2 and the plane y 苷 4., , SOLUTION The solid E is shown in Figure 9. If we regard it as a type 1 region, then we, , need to consider its projection D1 onto the xy-plane, which is the parabolic region in, Figure 10. (The trace of y 苷 x 2 ⫹ z 2 in the plane z 苷 0 is the parabola y 苷 x 2.), y, , z, , y=≈+z@, , TEC Visual 15.7 illustrates how solid regions, (including the one in Figure 9) project onto, coordinate planes., , y=4, D¡, , E, , 0, 4, x, , y=≈, y, , 0, , FIGURE 9, , FIGURE 10, , Region of integration, , Projection onto xy-plane, , x, , From y 苷 x 2 ⫹ z 2 we obtain z 苷 ⫾sy ⫺ x 2 , so the lower boundary surface of E is, z 苷 ⫺sy ⫺ x 2 and the upper surface is z 苷 sy ⫺ x 2 . Therefore the description of E as, a type 1 region is, E 苷 兵共x, y, z兲, , ⱍ ⫺2 艋 x 艋 2,, , x 2 艋 y 艋 4, ⫺sy ⫺ x 2 艋 z 艋 sy ⫺ x 2 其, , and so we obtain, , yyy sx, E, , 2, , ⫹ z 2 dV 苷 y, , 2, , ⫺2, , 4, , sy⫺x 2, , x2, , ⫺sy⫺x 2, , y y, , sx 2 ⫹ z 2 dz dy dx, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1046-1055.qk_97817_15_ch15_p1046-1055 11/8/10 3:39 PM Page 1046, , 1046, , CHAPTER 15, , MULTIPLE INTEGRALS, , Applications of Triple Integrals, Recall that if f 共x兲  0, then the single integral xab f 共x兲 dx represents the area under the, curve y 苷 f 共x兲 from a to b, and if f 共x, y兲  0, then the double integral xxD f 共x, y兲 dA represents the volume under the surface z 苷 f 共x, y兲 and above D. The corresponding interpretation of a triple integral xxxE f 共x, y, z兲 dV, where f 共x, y, z兲  0, is not very useful, because it would be the “hypervolume” of a four-dimensional object and, of course, that, is very difficult to visualize. (Remember that E is just the domain of the function f ; the, graph of f lies in four-dimensional space.) Nonetheless, the triple integral xxxE f 共x, y, z兲 dV, can be interpreted in different ways in different physical situations, depending on the physical interpretations of x, y, z, and f 共x, y, z兲., Let’s begin with the special case where f 共x, y, z兲 苷 1 for all points in E. Then the triple, integral does represent the volume of E:, V共E兲 苷, , 12, , yyy dV, E, , For example, you can see this in the case of a type 1 region by putting f 共x, y, z兲 苷 1 in, Formula 6:, , 冋, , yyy 1 dV 苷 yy y, E, , D, , u 2共x, y兲, , u1共x, y兲, , 册, , dz dA 苷 yy 关u 2共x, y兲  u1共x, y兲兴 dA, D, , and from Section 15.3 we know this represents the volume that lies between the surfaces, z 苷 u1共x, y兲 and z 苷 u 2共x, y兲., EXAMPLE 5 Use a triple integral to find the volume of the tetrahedron T bounded by the, planes x  2y  z 苷 2, x 苷 2y, x 苷 0, and z 苷 0., SOLUTION The tetrahedron T and its projection D onto the xy-plane are shown in Fig-, , ures 14 and 15. The lower boundary of T is the plane z 苷 0 and the upper boundary is the, plane x  2y  z 苷 2, that is, z 苷 2  x  2y., z, (0, 0, 2), , y, , x+2y+z=2, , x=2y, T, , x+2y=2, (or y=1- x/2), , 1, y, , ”1, 21 ’, , D, , (0, 1, 0), , 0, , y=x/2, , 1, , ”1, 2 , 0’, , 0, , 1, , x, , x, , FIGURE 15, , FIGURE 14, , Therefore we have, V共T兲 苷 yyy dV 苷 y, , 1, , 0, , y, , 1x兾2, , x兾2, , y, , 2x2y, , 0, , dz dy dx, , T, , 苷y, , 1, , 0, , y, , 1x兾2, , x兾2, , 共2  x  2y兲 dy dx 苷 13, , by the same calculation as in Example 4 in Section 15.3., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1046-1055.qk_97817_15_ch15_p1046-1055 11/8/10 3:39 PM Page 1050, , 1050, , MULTIPLE INTEGRALS, , CHAPTER 15, , x 苷 2,, , 30. y 2  z 2 苷 9,, 31. y 苷 x 2,, 32. x 苷 2,, , z 苷 0,, , x苷2, , 38., , x2  y2  z2  1, , y  2z 苷 4, , y 苷 2,, , z 苷 0,, , x  y  2z 苷 2, , 39– 42 Find the mass and center of mass of the solid E with the, given density function ., , 33. The figure shows the region of integration for the integral, 1, , 1, , y y y, 0, , 1y, , 0, , sx, , xxxB 共z 3  sin y  3兲 dV, where B is the unit ball, , 39. E is the solid of Exercise 13;, ,  共x, y, z兲 苷 2, , 40. E is bounded by the parabolic cylinder z 苷 1  y 2 and the, , f 共x, y, z兲 dz dy dx, , planes x  z 苷 1, x 苷 0, and z 苷 0;, , Rewrite this integral as an equivalent iterated integral in the, five other orders., ,  共x, y, z兲 苷 4, , 41. E is the cube given by 0  x  a, 0  y  a, 0  z  a ;, ,  共x, y, z兲 苷 x 2  y 2  z 2, , z, , 42. E is the tetrahedron bounded by the planes x 苷 0, y 苷 0,, , z 苷 0, x  y  z 苷 1;, , 1, ,  共x, y, z兲 苷 y, , z=1-y, 43– 46 Assume that the solid has constant density k., , y=œ„, x, , 43. Find the moments of inertia for a cube with side length L if, , 0, 1, , one vertex is located at the origin and three edges lie along, the coordinate axes., , y, , x, , 44. Find the moments of inertia for a rectangular brick with, , dimensions a, b, and c and mass M if the center of the brick, is situated at the origin and the edges are parallel to the coordinate axes., , 34. The figure shows the region of integration for the integral, 1, , yy, 0, , 1x 2, , 0, , y, , 1x, , 0, , 45. Find the moment of inertia about the z-axis of the solid, , f 共x, y, z兲 dy dz dx, , cylinder x 2  y 2  a 2, 0  z  h., 46. Find the moment of inertia about the z-axis of the solid cone, , Rewrite this integral as an equivalent iterated integral in the, five other orders., , sx 2  y 2  z  h., , z, , 47– 48 Set up, but do not evaluate, integral expressions for, (a) the mass, (b) the center of mass, and (c) the moment of, inertia about the z-axis., , 1, , z=1-≈, , 47. The solid of Exercise 21;, 0, 1, x, , 1, , y=1-x, , 48. The hemisphere x 2  y 2  z 2  1, z  0;, , y, ,  共x, y, z兲 苷 sx 2  y 2  z 2, CAS, , iterated integral., 1, , 1, , yyy, , 36., , y y y, , 0, , 1, , 0, , y, , 1, , y, , y, , 0, z, , 0, , f 共x, y, z兲 dz dx dy, f 共x, y, z兲 dx dz dy, CAS, , 37–38 Evaluate the triple integral using only geometric interpreta-, , tion and symmetry., 37., , xxxC 共4  5x 2 yz 2 兲 dV, where C is the cylindrical region, x 2  y 2  4, 2  z  2, , 49. Let E be the solid in the first octant bounded by the cylinder, , x 2  y 2 苷 1 and the planes y 苷 z, x 苷 0, and z 苷 0 with the, density function  共x, y, z兲 苷 1  x  y  z. Use a computer algebra system to find the exact values of the, following quantities for E., (a) The mass, (b) The center of mass, (c) The moment of inertia about the z-axis, , 35–36 Write five other iterated integrals that are equal to the given, , 35., ,  共x, y, z兲 苷 sx 2  y 2, , 50. If E is the solid of Exercise 18 with density function, ,  共x, y, z兲 苷 x 2  y 2, find the following quantities, correct, to three decimal places., (a) The mass, (b) The center of mass, (c) The moment of inertia about the z-axis, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 15.8, , TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES, , where V共E 兲 is the volume of E. For instance, if  is a density, function, then  ave is the average density of E., , 51. The joint density function for random variables X , Y , and Z is, , f 共x, y, z兲 苷 Cxyz if 0  x  2, 0  y  2, 0  z  2, and, f 共x, y, z兲 苷 0 otherwise., (a) Find the value of the constant C., (b) Find P共X  1, Y  1, Z  1兲., (c) Find P共X  Y  Z  1兲., , 53. Find the average value of the function f 共x, y, z兲 苷 xyz over, , the cube with side length L that lies in the first octant with one, vertex at the origin and edges parallel to the coordinate axes., 54. Find the average value of the function f 共x, y, z兲 苷 x 2 z  y 2 z, , 52. Suppose X , Y , and Z are random variables with joint density, , over the region enclosed by the paraboloid z 苷 1  x 2  y 2, and the plane z 苷 0., , function f 共x, y, z兲 苷 Ce共0.5x0.2y0.1z兲 if x  0, y  0, z  0,, and f 共x, y, z兲 苷 0 otherwise., (a) Find the value of the constant C., (b) Find P共X  1, Y  1兲., (c) Find P共X  1, Y  1, Z  1兲., , 55. (a) Find the region E for which the triple integral, , yyy 共1  x, , 53–54 The average value of a function f 共x, y, z兲 over a solid, , 1, V共E兲, , yyy f 共x, y, z兲 dV, E, , DISCOVERY PROJECT, , 2, ,  2y 2  3z 2 兲 dV, , E, , region E is defined to be, fave 苷, , 1051, , CAS, , is a maximum., (b) Use a computer algebra system to calculate the exact, maximum value of the triple integral in part (a)., , VOLUMES OF HYPERSPHERES, In this project we find formulas for the volume enclosed by a hypersphere in n-dimensional space., 1. Use a double integral and trigonometric substitution, together with Formula 64 in the Table, , of Integrals, to find the area of a circle with radius r., 2. Use a triple integral and trigonometric substitution to find the volume of a sphere with, , radius r., 3. Use a quadruple integral to find the hypervolume enclosed by the hypersphere, x 2  y 2  z 2  w 2 苷 r 2 in ⺢ 4. (Use only trigonometric substitution and the reduction, , formulas for x sin n x dx or x cos n x dx.), 4. Use an n-tuple integral to find the volume enclosed by a hypersphere of radius r in, , n-dimensional space ⺢ n. [Hint: The formulas are different for n even and n odd.], , 15.8, , Triple Integrals in Cylindrical Coordinates, , y, P(r, ¨)=P (x, y), , r, , y, , In plane geometry the polar coordinate system is used to give a convenient description of, certain curves and regions. (See Section 10.3.) Figure 1 enables us to recall the connection, between polar and Cartesian coordinates. If the point P has Cartesian coordinates 共x, y兲 and, polar coordinates 共r, 兲, then, from the figure,, x 苷 r cos, , y 苷 r sin, , r2 苷 x2  y2, , tan 苷, , ¨, O, , FIGURE 1, , x, , x, , y, x, , In three dimensions there is a coordinate system, called cylindrical coordinates, that is, similar to polar coordinates and gives convenient descriptions of some commonly occurring surfaces and solids. As we will see, some triple integrals are much easier to evaluate, in cylindrical coordinates., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1046-1055.qk_97817_15_ch15_p1046-1055 11/8/10 3:39 PM Page 1052, , 1052, , MULTIPLE INTEGRALS, , CHAPTER 15, z, , Cylindrical Coordinates, In the cylindrical coordinate system, a point P in three-dimensional space is represented, by the ordered triple 共r, , z兲, where r and are polar coordinates of the projection of P onto, the xy-plane and z is the directed distance from the xy-plane to P. (See Figure 2.), To convert from cylindrical to rectangular coordinates, we use the equations, , P (r, ¨, z), , z, , O, , r, , ¨, x, , y, , x 苷 r cos, , 1, , (r, ¨, 0), , y 苷 r sin, , z苷z, , FIGURE 2, , The cylindrical coordinates of a point, , whereas to convert from rectangular to cylindrical coordinates, we use, , r2 苷 x2  y2, , 2, , tan, , 苷, , y, x, , z苷z, , EXAMPLE 1, , (a) Plot the point with cylindrical coordinates 共2, 2 兾3, 1兲 and find its rectangular, coordinates., (b) Find cylindrical coordinates of the point with rectangular coordinates 共3, 3, 7兲., SOLUTION, , (a) The point with cylindrical coordinates 共2, 2 兾3, 1兲 is plotted in Figure 3. From, Equations 1, its rectangular coordinates are, , z, , ”2,, , 2π, 3 , 1’, , 1, , 2, 1, 苷2 , 3, 2, , 苷 1, , y 苷 2 sin, , 2, s3, 苷2, 3, 2, , 苷 s3, , 2, 0, 2π, 3, , 冉 冊, 冉 冊, , x 苷 2 cos, , y, , z苷1, , x, , FIGURE 3, , Thus the point is (1, s3 , 1) in rectangular coordinates., (b) From Equations 2 we have, r 苷 s3 2  共3兲2 苷 3s2, tan, , z, , 苷, , 3, 苷 1, 3, , so, , 苷, , 7,  2n, 4, , z 苷 7, 0, (c, 0, 0), x, , FIGURE 4, , r=c, a cylinder, , (0, c, 0), , Therefore one set of cylindrical coordinates is (3s2 , 7 兾4, 7). Another is, (3s2 ,  兾4, 7). As with polar coordinates, there are infinitely many choices., , y, , Cylindrical coordinates are useful in problems that involve symmetry about an axis, and, the z-axis is chosen to coincide with this axis of symmetry. For instance, the axis of the circular cylinder with Cartesian equation x 2  y 2 苷 c 2 is the z-axis. In cylindrical coordinates, this cylinder has the very simple equation r 苷 c. (See Figure 4.) This is the reason for the, name “cylindrical” coordinates., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1046-1055.qk_97817_15_ch15_p1046-1055 11/8/10 3:39 PM Page 1053, , TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES, , SECTION 15.8, , v, , z, , 1053, , EXAMPLE 2 Describe the surface whose equation in cylindrical coordinates is z 苷 r., , SOLUTION The equation says that the z-value, or height, of each point on the surface is, , 0, , y, , the same as r, the distance from the point to the z-axis. Because doesn’t appear, it can, vary. So any horizontal trace in the plane z 苷 k 共k 0兲 is a circle of radius k. These, traces suggest that the surface is a cone. This prediction can be confirmed by converting, the equation into rectangular coordinates. From the first equation in 2 we have, z2 苷 r 2 苷 x 2  y 2, , x, , We recognize the equation z 2 苷 x 2  y 2 (by comparison with Table 1 in Section 12.6) as, being a circular cone whose axis is the z-axis (see Figure 5)., , FIGURE 5, , z=r, a cone, , Evaluating Triple Integrals with Cylindrical Coordinates, Suppose that E is a type 1 region whose projection D onto the xy-plane is conveniently, described in polar coordinates (see Figure 6). In particular, suppose that f is continuous, and, , ⱍ, , E 苷 兵共x, y, z兲 共x, y兲 僆 D, u1共x, y兲  z  u 2共x, y兲其, where D is given in polar coordinates by, , ⱍ, , D 苷 兵共r, 兲, ,  , h1共 兲  r  h 2共 兲其, , , , z, , z=u™(x, y), , z=u¡(x, y), , r=h¡(¨ ) 0, ¨=a, FIGURE 6, , ¨=b, , D, , y, , r=h™(¨ ), , x, , We know from Equation 15.7.6 that, , 冋, , yyy f 共x, y, z兲 dV 苷 yy y, , 3, , E, , D, , u 2共x, y兲, , u1共x, y兲, , 册, , f 共x, y, z兲 dz dA, , But we also know how to evaluate double integrals in polar coordinates. In fact, combining Equation 3 with Equation 15.4.3, we obtain, , 4, , yyy f 共x, y, z兲 dV 苷 y y, E, , h2共 兲, , h1共 兲, , y, , u 2共r cos , r sin 兲, , u1共r cos , r sin 兲, , f 共r cos , r sin , z兲 r dz dr d, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1046-1055.qk_97817_15_ch15_p1046-1055 11/8/10 3:40 PM Page 1054, , 1054, , CHAPTER 15, , MULTIPLE INTEGRALS, , z, , Formula 4 is the formula for triple integration in cylindrical coordinates. It says that, we convert a triple integral from rectangular to cylindrical coordinates by writing, x 苷 r cos , y 苷 r sin , leaving z as it is, using the appropriate limits of integration for z,, r , and , and replacing dV by r dz dr d . (Figure 7 shows how to remember this.) It is, worthwhile to use this formula when E is a solid region easily described in cylindrical, coordinates, and especially when the function f 共x, y, z兲 involves the expression x 2  y 2., , dz, , d¨, r, r d¨, , dr, , v, , FIGURE 7, , EXAMPLE 3 A solid E lies within the cylinder x 2  y 2 苷 1, below the plane z 苷 4,, , and above the paraboloid z 苷 1  x 2  y 2. (See Figure 8.) The density at any point is, proportional to its distance from the axis of the cylinder. Find the mass of E., , Volume element in cylindrical, coordinates: dV=r dz dr d¨, , SOLUTION In cylindrical coordinates the cylinder is r 苷 1 and the paraboloid is, , z 苷 1  r 2, so we can write, , z, , z=4, , ⱍ, , E 苷 兵共r, , z兲 0 , (0, 0, 4), ,  2 , 0  r  1, 1  r 2  z  4 其, , Since the density at 共x, y, z兲 is proportional to the distance from the z-axis, the density, function is, f 共x, y, z兲 苷 K sx 2  y 2 苷 Kr, , (0, 0, 1), , z=1-r @, , where K is the proportionality constant. Therefore, from Formula 15.7.13, the mass of E, is, 2, 1 4, m 苷 yyy Ksx 2  y 2 dV 苷 y y y 2 共Kr兲 r dz dr d, 0, , 0, , 1r, , E, 0, , y, , (1, 0, 0), , x, , 苷y, , 2, , 0, , y, , 1, , 0, , 2, , Kr 2 关4  共1  r 2 兲兴 dr d 苷 K y d, , 冋, , r5, 苷2 K r , 5, , FIGURE 8, , EXAMPLE 4 Evaluate, , 3, , 2, , s4x 2, , 2, , s4x, , y y, , 册, , 1, , 苷, , 0, , y, 2, , y, , 0, , 1, , 0, , 共3r 2  r 4 兲 dr, , 12 K, 5, , 2, , sx 2 y 2, , 共x 2  y 2 兲 dz dy dx ., , SOLUTION This iterated integral is a triple integral over the solid region, , ⱍ, , E 苷 兵共x, y, z兲 2  x  2, s4  x 2  y  s4  x 2 , sx 2  y 2  z  2其, and the projection of E onto the xy-plane is the disk x 2  y 2  4. The lower surface of, E is the cone z 苷 sx 2  y 2 and its upper surface is the plane z 苷 2. (See Figure 9.), This region has a much simpler description in cylindrical coordinates:, , ⱍ, , E 苷 兵共r, , z兲 0 , , z, ,  2 , 0  r  2, r  z  2其, , Therefore we have, , z=2, , 2, , s4x 2, , 2, , s4x 2, , y y, , 2, , z=œ„„„„„, ≈+¥, , y, , 2, , sx 2 y 2, , 共x 2  y 2 兲 dz dy dx 苷 yyy 共x 2  y 2 兲 dV, E, , 苷y, , 2, , 苷y, , 2, , 0, , x, , 2, , FIGURE 9, , 2, , y, , 0, , 苷2, , 2, , y y, , 2, , 0, , r, , d, , y, , [, , 1, 2, , 2, , 0, , r 2 r dz dr d, r 3共2  r兲 dr, , r4  5r5, 1, , ], , 2, 0, , 苷, , 16, 5, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1046-1055.qk_97817_15_ch15_p1046-1055 11/8/10 3:40 PM Page 1055, , TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES, , SECTION 15.8, , 1055, , Exercises, , 15.8, , 1–2 Plot the point whose cylindrical coordinates are given. Then, find the rectangular coordinates of the point., 1. (a) 共4, 兾3, 2兲, , (b) 共2,  兾2, 1兲, , 2. (a) (s2 , 3 兾4, 2), , (b) 共1, 1, 1兲, , 18. Evaluate xxxE z dV, where E is enclosed by the paraboloid, , z 苷 x 2  y 2 and the plane z 苷 4., 19. Evaluate xxxE 共x  y  z兲 dV, where E is the solid in the first, , octant that lies under the paraboloid z 苷 4  x 2  y 2., , 20. Evaluate xxxE x dV, where E is enclosed by the planes z 苷 0, , 3– 4 Change from rectangular to cylindrical coordinates., 3. (a) 共1, 1, 1兲, , (b) (2, 2s3 , 3), , 4. (a) (2 s3, 2, 1), , (b) 共4, 3, 2兲, , and z 苷 x  y  5 and by the cylinders x 2  y 2 苷 4 and, x 2  y 2 苷 9., 21. Evaluate xxxE x 2 dV, where E is the solid that lies within the, , cylinder x 2  y 2 苷 1, above the plane z 苷 0, and below the, cone z 2 苷 4x 2  4y 2., , 5–6 Describe in words the surface whose equation is given., , 苷 兾4, , 5., , 6. r 苷 5, , 22. Find the volume of the solid that lies within both the cylinder, , x 2  y 2 苷 1 and the sphere x 2  y 2  z 2 苷 4., 7–8 Identify the surface whose equation is given., 7. z 苷 4  r 2, , 23. Find the volume of the solid that is enclosed by the cone, , 8. 2r 2  z 2 苷 1, , z 苷 sx 2  y 2 and the sphere x 2  y 2  z 2 苷 2., 24. Find the volume of the solid that lies between the paraboloid, , 9–10 Write the equations in cylindrical coordinates., 9. (a) x  x  y  z 苷 1, 2, , 2, , (b) z 苷 x  y, , 2, , 2, , 10. (a) 3x  2y  z 苷 6, , z 苷 x 2  y 2 and the sphere x 2  y 2  z 2 苷 2., , 2, , (b) x 2  y 2  z 2 苷 1, , 25. (a) Find the volume of the region E bounded by the parabo-, , loids z 苷 x 2  y 2 and z 苷 36  3x 2  3y 2., (b) Find the centroid of E (the center of mass in the case, where the density is constant)., , 11–12 Sketch the solid described by the given inequalities., 11. 0  r  2,, 12. 0 , ,  兾2 , ,  兾2,, ,  兾2,, , 0z1, , 26. (a) Find the volume of the solid that the cylinder r 苷 a cos, , rz2, , cuts out of the sphere of radius a centered at the origin., (b) Illustrate the solid of part (a) by graphing the sphere and, the cylinder on the same screen., , ;, 13. A cylindrical shell is 20 cm long, with inner radius 6 cm and, , outer radius 7 cm. Write inequalities that describe the shell, in an appropriate coordinate system. Explain how you have, positioned the coordinate system with respect to the shell., , 27. Find the mass and center of mass of the solid S bounded by, , the paraboloid z 苷 4x 2  4y 2 and the plane z 苷 a 共a, S has constant density K., , 0兲 if, , ; 14. Use a graphing device to draw the solid enclosed by the, paraboloids z 苷 x 2  y 2 and z 苷 5  x 2  y 2., , 28. Find the mass of a ball B given by x 2  y 2  z 2  a 2 if the, , density at any point is proportional to its distance from the, z-axis., , 15–16 Sketch the solid whose volume is given by the integral, , and evaluate the integral., 15., , y, , 兾2, ,  兾2, , 2, , y y, 0, , r2, , 0, , r dz dr d, , 16., , 2, , 2, , r, , 0, , 0, , 0, , y y y, , r dz d dr, , 29–30 Evaluate the integral by changing to cylindrical coordinates., , inside the cylinder x 2  y 2 苷 16 and between the planes, z 苷 5 and z 苷 4., , ;, , Graphing calculator or computer required, , s4y 2, , 2, , s4y 2, , 3, , s9x 2, , y y, , 30., , y y, , 17–28 Use cylindrical coordinates., 17. Evaluate xxxE sx 2  y 2 dV, where E is the region that lies, , 2, , 29., , 3, , 0, , y, , 2, , sx 2y 2, , y, , 9x 2y 2, , 0, , xz dz dx dy, , sx 2  y 2 dz dy dx, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1056, , CHAPTER 15, , MULTIPLE INTEGRALS, , estimate the amount of work required to lift a mountain, from sea level. Consider a mountain that is essentially in the, shape of a right circular cone. Suppose that the weight density of the material in the vicinity of a point P is t共P兲 and, the height is h共P兲., (a) Find a definite integral that represents the total work, done in forming the mountain., (b) Assume that Mount Fuji in Japan is in the shape of a, right circular cone with radius 62,000 ft, height, 12,400 ft, and density a constant 200 lb兾ft3. How much, work was done in forming Mount Fuji if the land was, initially at sea level?, , © S.R. Lee Photo Traveller / Shutterstock, , 31. When studying the formation of mountain ranges, geologists, , L A B O R AT O R Y P R O J E C T THE INTERSECTION OF THREE CYLINDERS, The figure shows the solid enclosed by three circular cylinders with the same diameter that intersect at right angles. In this project we compute its volume and determine how its shape changes if, the cylinders have different diameters., , 1. Sketch carefully the solid enclosed by the three cylinders x 2  y 2 苷 1, x 2  z 2 苷 1, and, , y 2  z 2 苷 1. Indicate the positions of the coordinate axes and label the faces with the, equations of the corresponding cylinders., , 2. Find the volume of the solid in Problem 1., CAS, , 3. Use a computer algebra system to draw the edges of the solid., 4. What happens to the solid in Problem 1 if the radius of the first cylinder is different, , from 1? Illustrate with a hand-drawn sketch or a computer graph., 5. If the first cylinder is x 2  y 2 苷 a 2, where a  1, set up, but do not evaluate, a double, , integral for the volume of the solid. What if a  1?, , CAS Computer algebra system required, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 15.9, , TRIPLE INTEGRALS IN SPHERICAL COORDINATES, , 1057, , Triple Integrals in Spherical Coordinates, , 15.9, , Another useful coordinate system in three dimensions is the spherical coordinate system., It simplifies the evaluation of triple integrals over regions bounded by spheres or cones., z, , Spherical Coordinates, The spherical coordinates 共 , , 兲 of a point P in space are shown in Figure 1, where,  苷 OP is the distance from the origin to P,  is the same angle as in cylindrical coordinates, and  is the angle between the positive z-axis and the line segment OP. Note that, , P ( ∏, ¨, ˙), , ˙, , ⱍ, , ∏, , ⱍ, , 0, , O, , ¨, , x, , y, , FIGURE 1, , The spherical coordinates of a point, , , , 0, , The spherical coordinate system is especially useful in problems where there is symmetry, about a point, and the origin is placed at this point. For example, the sphere with center the, origin and radius c has the simple equation  苷 c (see Figure 2); this is the reason for, the name “spherical” coordinates. The graph of the equation  苷 c is a vertical half-plane, (see Figure 3), and the equation  苷 c represents a half-cone with the z-axis as its axis (see, Figure 4)., z, , z, , z, , z, , c, 0, , 0, , c, , y, x, , 0, y, , x, , 0, , y, , y, , x, , x, , 0<c<π/2, FIGURE 2 ∏=c , a sphere, , FIGURE 3 ¨=c, a half-plane, , c, , π/2<c<π, , FIGURE 4 ˙=c, a half-cone, , The relationship between rectangular and spherical coordinates can be seen from Figure 5. From triangles OPQ and OPP we have, , z, , Q, , z, ˙, , z 苷  cos , , P(x, y, z), P (∏, ¨, ˙), , ∏, , r 苷  sin , , But x 苷 r cos  and y 苷 r sin , so to convert from spherical to rectangular coordinates, we, use the equations, , ˙, , O, , x, x, , r, , ¨, y, , y, P ª(x, y, 0), , 1, , x 苷  sin  cos , , y 苷  sin  sin , , z 苷  cos , , Also, the distance formula shows that, , FIGURE 5, , 2, , 2 苷 x 2  y 2  z2, , We use this equation in converting from rectangular to spherical coordinates., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1056-1065.qk_97817_15_ch15_p1056-1065 11/8/10 3:43 PM Page 1058, , 1058, , CHAPTER 15, , MULTIPLE INTEGRALS, , v EXAMPLE 1 The point 共2, 兾4, 兾3兲 is given in spherical coordinates. Plot the point, and find its rectangular coordinates., SOLUTION We plot the point in Figure 6. From Equations 1 we have, , z, , π, 3, O, , 3, , cos, , y 苷  sin  sin  苷 2 sin, y, , z 苷  cos  苷 2 cos, , FIGURE 6, , 3, , 3, , sin, , 4, , 冉 冊冉 冊 冑, 冉 冊冉 冊 冑, , 苷2, , 4, , 2, , π, 4, , x, , x 苷  sin  cos  苷 2 sin, , (2, π/4, π/3), , 苷2, , s3, 2, , s3, 2, , 1, s2, , 1, s2, , 苷, , 苷, , 3, 2, , 3, 2, , 苷 2( 12 ) 苷 1, , Thus the point 共2, 兾4, 兾3兲 is (s3兾2 , s3兾2 , 1) in rectangular coordinates., , v EXAMPLE 2 The point (0, 2s3 ,, cal coordinates for this point., , 2) is given in rectangular coordinates. Find spheri-, , SOLUTION From Equation 2 we have, ,  苷 sx 2  y 2  z 2 苷 s0  12  4 苷 4, , |, , WARNING There is not universal, agreement on the notation for spherical coordinates. Most books on physics reverse the, meanings of  and  and use r in place of ., , TEC In Module 15.9 you can investigate, families of surfaces in cylindrical and spherical coordinates., , and so Equations 1 give, cos  苷, , z, 2, 苷, 苷, , 4, , 1, 2, , cos  苷, , x, 苷0,  sin , , 苷, 苷, , 2, 3, 2, , (Note that  苷 3, , 兾2 because y 苷 2s3  0.) Therefore spherical coordinates of the, given point are 共4, 兾2, 2 兾3兲., , Evaluating Triple Integrals with Spherical Coordinates, z, , ∏ i sin ˙ k Ψ, , ˙k, , Î∏, , ⱍ, , E 苷 兵共 , , 兲 a, , Î˙, ∏ i Î˙, , 0, x, , In the spherical coordinate system the counterpart of a rectangular box is a spherical, wedge, , ri=∏ i sin ˙ k, , Ψ, , y, , ri Ψ=∏ i sin ˙ k Ψ, FIGURE 7, , , , , , b,, , , c, , , , d其, , where a  0 and , 2 , and d c, . Although we defined triple integrals by, dividing solids into small boxes, it can be shown that dividing a solid into small spherical, wedges always gives the same result. So we divide E into smaller spherical wedges Eijk by, means of equally spaced spheres  苷  i , half-planes  苷  j , and half-cones  苷  k . Figure 7 shows that Eijk is approximately a rectangular box with dimensions ,  i   (arc of, a circle with radius  i , angle ), and  i sin  k  (arc of a circle with radius  i sin  k,, angle  ). So an approximation to the volume of Eijk is given by, Vijk ⬇ 共兲共  i 兲共  i sin  k  兲 苷  2i sin  k   , In fact, it can be shown, with the aid of the Mean Value Theorem (Exercise 47), that the volume of Eijk is given exactly by, 苲, , Vijk 苷 苲 i2 sin  k   , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1056-1065.qk_97817_15_ch15_p1056-1065 11/8/10 3:43 PM Page 1060, , 1060, , CHAPTER 15, , MULTIPLE INTEGRALS, , v, , 2, , 2, , EXAMPLE 3 Evaluate xxxB e 共x y z, , 2 3兾2, , 兲, , dV, where B is the unit ball:, , ⱍ, , B 苷 兵共x, y, z兲 x 2  y 2  z 2, , 1其, , SOLUTION Since the boundary of B is a sphere, we use spherical coordinates:, , B 苷 兵共 , , 兲, , ⱍ0, , , , , , 1, 0, , 2 , 0, , 其, , , , In addition, spherical coordinates are appropriate because, x 2  y 2  z2 苷 2, Thus 3 gives, , yyy e, , 共x 2y 2z 2 兲3兾2, , 2, , dV 苷 y, , y y, , 0, , 0, , 1, , 0, , e共 , ,  2 sin  d d d, , 兲, , 2 3兾2, , B, , 苷 y sin  d, , y, , 0, , [, , 苷, , 2, , d, , 0, , ], , cos  0 共2 兲, , y, , 1, , 0, , [e ], , 1 3 1, 3, 0, ,  2e  d, 3, , 苷 43 共e, , 1兲, , NOTE It would have been extremely awkward to evaluate the integral in Example 3 without spherical coordinates. In rectangular coordinates the iterated integral would have been, 1, , y y, , s1 x 2, s1 x, , 1, , 2, , y, , s1 x 2 y 2, , e 共x y z, 2, , s1 x 2 y 2, , 2, , 兲, , 2 3兾2, , dz dy dx, , v EXAMPLE 4 Use spherical coordinates to find the volume of the solid that lies above, the cone z 苷 sx 2  y 2 and below the sphere x 2  y 2  z 2 苷 z. (See Figure 9.), z, , (0, 0, 1), , ≈+¥+z@=z, , π, 4, , z=œ„„„„„, ≈+¥, y, , x, , FIGURE 9, Figure 10 gives another look (this time drawn, by Maple) at the solid of Example 4., , (, , 1, , ), , SOLUTION Notice that the sphere passes through the origin and has center 0, 0, 2 . We, , write the equation of the sphere in spherical coordinates as, ,  2 苷  cos , ,  苷 cos , , or, , The equation of the cone can be written as, ,  cos  苷 s 2 sin 2 cos 2   2 sin 2 sin 2 苷  sin , This gives sin  苷 cos , or  苷 兾4. Therefore the description of the solid E in, spherical coordinates is, FIGURE 10, , ⱍ, , E 苷 兵共 , , 兲 0, , , , 2 , 0, , , , 兾4, 0, , , , cos  其, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1056-1065.qk_97817_15_ch15_p1056-1065 11/8/10 3:43 PM Page 1061, , TRIPLE INTEGRALS IN SPHERICAL COORDINATES, , SECTION 15.9, , 1061, , Figure 11 shows how E is swept out if we integrate first with respect to , then , and, then . The volume of E is, V共E兲 苷 yyy dV 苷 y, , 2, , 0, , 兾4, , y y, 0, , E, , 苷y, , 2, , d, , 0, , 苷, , TEC Visual 15.9 shows an animation of, , 2, 3, , y, , 0, , y, , 兾4, , 0, , 兾4, , cos , , 0, , 冋册, , 3, sin , 3, ,  2 sin  d d d, 苷cos , , d, 苷0, , sin  cos 3 d 苷, , Figure 11., , z, , 15.9, , 冋, , cos 4, 4, , 册, , 兾4, , 苷, , 0, , z, , x, , 8, , z, , x, , y, , ∏ varies from 0 to cos ˙, while ˙ and ¨ are constant., , FIGURE 11, , 2, 3, , x, , y, , ˙ varies from 0 to π/4, while ¨ is constant., , y, , ¨ varies from 0 to 2π., , Exercises, , 1–2 Plot the point whose spherical coordinates are given. Then find, the rectangular coordinates of the point., 1. (a) 共6, 兾3, 兾6兲, , (b) 共3, 兾2, 3 兾4兲, , 2. (a) 共2, 兾2, 兾2兲, , (b) 共4,, , 9–10 Write the equation in spherical coordinates., 9. (a) z 2 苷 x 2  y 2, 10. (a) x 2, , (b) x 2  z 2 苷 9, , 2x  y 2  z 2 苷 0, , (b) x  2y  3z 苷 1, , 兾4, 兾3兲, 11–14 Sketch the solid described by the given inequalities., , 3– 4 Change from rectangular to spherical coordinates., 3. (a) 共0,, , 2, 0兲, , 4. (a) (1, 0, s3 ), , ( 1, 1, s2 ), (b) (s3 , 1, 2s3 ), (b), , 11. 2, , , , 4,, , 0, , , , 兾3,, , 12. 1, , , , 2,, , 0, , , , 兾2,, , 13. , , 1,, , 14. , , 2, , , 3 兾4, , , , 0, 兾2, , , , 3 兾2, , , , csc , , 5–6 Describe in words the surface whose equation is given., 5.  苷, , 兾3, , 15. A solid lies above the cone z 苷 sx 2  y 2 and below the, , 6.  苷 3, , sphere x 2  y 2  z 2 苷 z. Write a description of the solid in, terms of inequalities involving spherical coordinates., , 7–8 Identify the surface whose equation is given., 7.  苷 sin  sin , , ;, , 16. (a) Find inequalities that describe a hollow ball with diameter, , 8.  共sin  sin   cos 兲 苷 9, 2, , Graphing calculator or computer required, , 2, , 2, , 2, , 30 cm and thickness 0.5 cm. Explain how you have, positioned the coordinate system that you have chosen., , CAS Computer algebra system required, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1056-1065.qk_97817_15_ch15_p1056-1065 11/8/10 3:43 PM Page 1062, , 1062, , MULTIPLE INTEGRALS, , CHAPTER 15, , (b) Suppose the ball is cut in half. Write inequalities that, describe one of the halves., , 32. Let H be a solid hemisphere of radius a whose density at any, , point is proportional to its distance from the center of the base., (a) Find the mass of H ., (b) Find the center of mass of H ., (c) Find the moment of inertia of H about its axis., , 17–18 Sketch the solid whose volume is given by the integral, , and evaluate the integral., 17., 18., , 兾6, , 兾2, , y y y, 0, , 0, , 2, , y y y, 兾2, , 0, , 3, , 0, , 2, , 1, ,  2 sin  d d d, , 33. (a) Find the centroid of a solid homogeneous hemisphere of, , radius a., (b) Find the moment of inertia of the solid in part (a) about a, diameter of its base., ,  sin  d d d, 2, , 34. Find the mass and center of mass of a solid hemisphere of, 19–20 Set up the triple integral of an arbitrary continuous function, , radius a if the density at any point is proportional to its, distance from the base., , f 共x, y, z兲 in cylindrical or spherical coordinates over the solid, shown., z, , 19., , z, , 20., , 35–38 Use cylindrical or spherical coordinates, whichever seems, more appropriate., , 3, , 35. Find the volume and centroid of the solid E that lies, , above the cone z 苷 sx 2  y 2 and below the sphere, x 2  y 2  z 2 苷 1., , 2, y, , x, , x, , 2, , 1, , y, , 36. Find the volume of the smaller wedge cut from a sphere of, , radius a by two planes that intersect along a diameter at an, angle of 兾6., , 21–34 Use spherical coordinates., , CAS, , 21. Evaluate xxxB 共x 2  y 2  z 2 兲 2 dV, where B is the ball with, , 37. Evaluate xxxE z dV, where E lies above the paraboloid, , z 苷 x 2  y 2 and below the plane z 苷 2y. Use either the, Table of Integrals (on Reference Pages 6–10) or a computer, algebra system to evaluate the integral., , center the origin and radius 5., 22. Evaluate xxxH 共9, , x 2 y 2 兲 dV, where H is the solid, hemisphere x  y  z 2 9, z  0., 2, , 2, , CAS, , 38. (a) Find the volume enclosed by the torus  苷 sin ., , (b) Use a computer to draw the torus., , 23. Evaluate xxxE 共x  y 兲 dV, where E lies between the spheres, 2, , 2, , x 2  y 2  z 2 苷 4 and x 2  y 2  z 2 苷 9., , 39– 41 Evaluate the integral by changing to spherical coordinates., , 24. Evaluate xxxE y 2 dV, where E is the solid hemisphere, , x 2  y 2  z2, , 9, y  0., 2, , 2, , s1 x 2, , yy, , 40., , y y, , 41., , y y, , 0, , 0, , y, , s2 x 2 y 2, sx 2y 2, , xy dz dy dx, , 2, , 25. Evaluate xxxE xe x  y  z dV, where E is the portion of the unit, , ball x 2  y 2  z 2, , 1, , 39., , 1 that lies in the first octant., , 26. Evaluate xxxE xyz dV, where E lies between the spheres, ,  苷 2 and  苷 4 and above the cone  苷 兾3., , 27. Find the volume of the part of the ball , , a, , a, , 2, , 2, , sa 2 y 2, sa 2 y 2, s4 x 2, s4 x 2, , y, y, , sa 2 x 2 y 2, sa 2 x 2 y 2, , 2s4 x 2 y 2, , 2 s4 x 2 y 2, , 共x 2z  y 2z  z 3 兲 dz dx dy, 共x 2  y 2  z 2 兲3兾2 dz dy dx, , a that lies between, , the cones  苷 兾6 and  苷 兾3., , 28. Find the average distance from a point in a ball of radius a to, , its center., 29. (a) Find the volume of the solid that lies above the cone, ,  苷 兾3 and below the sphere  苷 4 cos ., (b) Find the centroid of the solid in part (a)., 30. Find the volume of the solid that lies within the sphere, , x 2  y 2  z 2 苷 4, above the xy-plane, and below the cone, z 苷 sx 2  y 2 ., 31. (a) Find the centroid of the solid in Example 4., , (b) Find the moment of inertia about the z-axis for this solid., , 42. A model for the density  of the earth’s atmosphere near its, , surface is, ,  苷 619.09, , 0.000097, , where  (the distance from the center of the earth) is measured in meters and  is measured in kilograms per cubic, meter. If we take the surface of the earth to be a sphere with, radius 6370 km, then this model is a reasonable one for, 6.370  106  6.375  106. Use this model to estimate, the mass of the atmosphere between the ground and an altitude, of 5 km., , ; 43. Use a graphing device to draw a silo consisting of a cylinder, with radius 3 and height 10 surmounted by a hemisphere., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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APPLIED PROJECT, , 44. The latitude and longitude of a point P in the Northern, , Hemisphere are related to spherical coordinates , ,  as, follows. We take the origin to be the center of the earth and, the positive z-axis to pass through the North Pole. The positive x-axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects, the equator. Then the latitude of P is 苷 90   and the, longitude is  苷 360  . Find the great-circle distance, from Los Angeles (lat. 34.06 N, long. 118.25 W) to Montréal (lat. 45.50 N, long. 73.60 W). Take the radius of the, earth to be 3960 mi. (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.), CAS, , ROLLER DERBY, , 1063, , 46. Show that, , , , , y y y, , , , , , , , sx 2  y 2  z 2 e, , 共x 2y 2z 2 兲, , dx dy dz 苷 2, , (The improper triple integral is defined as the limit of a, triple integral over a solid sphere as the radius of the sphere, increases indefinitely.), 47. (a) Use cylindrical coordinates to show that the volume of, , the solid bounded above by the sphere r 2  z 2 苷 a 2 and, below by the cone z 苷 r cot  0 (or  苷  0 ), where, 0   0  兾2, is, V苷, , 45. The surfaces  苷 1  5 sin m sin n have been used as, 1, , models for tumors. The “bumpy sphere” with m 苷 6 and, n 苷 5 is shown. Use a computer algebra system to find the, volume it encloses., , 2 a3, 共1, 3, , cos  0 兲, , (b) Deduce that the volume of the spherical wedge given by,  1   2 ,  1   2 ,  1   2 is, V 苷, ,  23, ,  13, , 共cos  1, , 3, , cos  2 兲共 2, , 1 兲, , (c) Use the Mean Value Theorem to show that the volume in, part (b) can be written as, 苲, , V 苷 苲 2 sin    , 苲, , where 苲 lies between  1 and  2 ,  lies between  1 and,  2 ,  苷  2  1 ,  苷  2  1 , and  苷  2  1 ., , APPLIED PROJECT, , ROLLER DERBY, , h, å, , Suppose that a solid ball (a marble), a hollow ball (a squash ball), a solid cylinder (a steel bar), and, a hollow cylinder (a lead pipe) roll down a slope. Which of these objects reaches the bottom first?, (Make a guess before proceeding.), To answer this question, we consider a ball or cylinder with mass m, radius r, and moment of, inertia I (about the axis of rotation). If the vertical drop is h, then the potential energy at the top, is mth. Suppose the object reaches the bottom with velocity v and angular velocity , so v 苷 r., The kinetic energy at the bottom consists of two parts: 12 mv 2 from translation (moving down the, slope) and 12 I 2 from rotation. If we assume that energy loss from rolling friction is negligible,, then conservation of energy gives, mth 苷 2 mv 2  2 I 2, 1, , 1, , 1. Show that, v2 苷, , 2th, 1  I*, , where I* 苷, , I, mr 2, , 2. If y共t兲 is the vertical distance traveled at time t, then the same reasoning as used in, Problem 1 shows that v 2 苷 2ty兾共1  I*兲 at any time t. Use this result to show that y, , satisfies the differential equation, dy, 苷, dt, where, , 冑, , 2t, 共sin 兲 sy, 1  I*, , is the angle of inclination of the plane., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1064, , CHAPTER 15, , MULTIPLE INTEGRALS, , 3. By solving the differential equation in Problem 2, show that the total travel time is, , T苷, , 冑, , 2h共1  I*兲, t sin 2, , This shows that the object with the smallest value of I* wins the race., 4. Show that I* 苷 2 for a solid cylinder and I* 苷 1 for a hollow cylinder., 1, , 5. Calculate I* for a partly hollow ball with inner radius a and outer radius r . Express your, , answer in terms of b 苷 a兾r. What happens as a l 0 and as a l r?, 6. Show that I* 苷 5 for a solid ball and I* 苷 3 for a hollow ball. Thus the objects finish in the, 2, , 2, , following order: solid ball, solid cylinder, hollow ball, hollow cylinder., , 15.10 Change of Variables in Multiple Integrals, In one-dimensional calculus we often use a change of variable (a substitution) to simplify, an integral. By reversing the roles of x and u, we can write the Substitution Rule (4.5.5) as, , y, , 1, , b, , d, , f 共x兲 dx 苷 y f 共 t共u兲兲 t 共u兲 du, , a, , c, , where x 苷 t共u兲 and a 苷 t共c兲, b 苷 t共d 兲. Another way of writing Formula 1 is as follows:, , y, , 2, , b, , a, , d, , f 共x兲 dx 苷 y f 共x共u兲兲, c, , dx, du, du, , A change of variables can also be useful in double integrals. We have already seen one, example of this: conversion to polar coordinates. The new variables r and  are related to, the old variables x and y by the equations, x 苷 r cos , , y 苷 r sin , , and the change of variables formula (15.4.2) can be written as, , yy f 共x, y兲 dA 苷 yy f 共r cos , r sin  兲 r dr d, R, , S, , where S is the region in the r -plane that corresponds to the region R in the xy-plane., More generally, we consider a change of variables that is given by a transformation T, from the uv-plane to the xy-plane:, T共u, v兲 苷 共x, y兲, where x and y are related to u and v by the equations, 3, , x 苷 t共u, v兲, , y 苷 h共u, v兲, , x 苷 x共u, v兲, , y 苷 y共u, v兲, , or, as we sometimes write,, We usually assume that T is a C 1 transformation, which means that t and h have continuous first-order partial derivatives., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1056-1065.qk_97817_15_ch15_p1056-1065 11/8/10 3:43 PM Page 1065, , SECTION 15.10, , CHANGE OF VARIABLES IN MULTIPLE INTEGRALS, , 1065, , A transformation T is really just a function whose domain and range are both subsets, of ⺢ 2. If T共u1, v1兲 苷 共x 1, y1兲, then the point 共x 1, y1兲 is called the image of the point 共u1, v1兲., If no two points have the same image, T is called one-to-one. Figure 1 shows the effect of, a transformation T on a region S in the uv-plane. T transforms S into a region R in the, xy-plane called the image of S, consisting of the images of all points in S., √, , y, , T, , S, (u¡, √¡), , R, , T –!, , (x¡, y¡), , u, , 0, , x, , 0, , FIGURE 1, , If T is a one-to-one transformation, then it has an inverse transformation T 1 from the, xy-plane to the uv-plane and it may be possible to solve Equations 3 for u and v in terms, of x and y :, u 苷 G共x, y兲, v 苷 H共x, y兲, , v, , EXAMPLE 1 A transformation is defined by the equations, , x 苷 u2, Find the image of the square S 苷 兵共u, v兲, , S£, , (0, 1), , S¢, , u, , 1, 0, , v, , 1其., , we begin by finding the images of the sides of S. The first side, S1 , is given by v 苷 0, 共0 u 1兲. (See Figure 2.) From the given equations we have x 苷 u 2, y 苷 0, and so, 0 x 1. Thus S1 is mapped into the line segment from 共0, 0兲 to 共1, 0兲 in the xy-plane., The second side, S 2, is u 苷 1 共0 v 1兲 and, putting u 苷 1 in the given equations, we, get, x 苷 1 v2, y 苷 2v, Eliminating v, we obtain, , (1, 1), , S, , 0, , S™, , S¡ (1, 0), , u, , T, , 4, , (0, 2), ¥, x= 4 -1, , x苷1, , ¥, , x=1- 4, , 5, , R, 0, , y2, 4, , 0, , x, , 1, , which is part of a parabola. Similarly, S 3 is given by v 苷 1 共0, the parabolic arc, , y, , FIGURE 2, , ⱍ0, , SOLUTION The transformation maps the boundary of S into the boundary of the image. So, , √, , (_1, 0), , y 苷 2uv, , v2, , (1, 0), , x, , x苷, , y2, 4, , 1, , 1, , x, , u, , 1兲, whose image is, , 0, , Finally, S4 is given by u 苷 0 共0 v 1兲 whose image is x 苷 v 2, y 苷 0, that is,, 1 x 0. (Notice that as we move around the square in the counterclockwise direction, we also move around the parabolic region in the counterclockwise direction.) The, image of S is the region R (shown in Figure 2) bounded by the x-axis and the parabolas, given by Equations 4 and 5., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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CHAPTER 15, , REVIEW, , 1073, , Review, , 15, , Concept Check, 1. Suppose f is a continuous function defined on a rectangle, , R 苷 关a, b兴  关c, d 兴., (a) Write an expression for a double Riemann sum of f ., If f 共x, y兲 0, what does the sum represent?, (b) Write the definition of xxR f 共x, y兲 dA as a limit., (c) What is the geometric interpretation of xxR f 共x, y兲 dA if, f 共x, y兲 0? What if f takes on both positive and negative, values?, (d) How do you evaluate xxR f 共x, y兲 dA?, (e) What does the Midpoint Rule for double integrals say?, (f ) Write an expression for the average value of f ., , (b) What properties does f possess?, (c) What are the expected values of X and Y ?, 6. Write an expression for the area of a surface with equation, , z 苷 f 共x, y兲, 共x, y兲 僆 D., 7. (a) Write the definition of the triple integral of f over a, , rectangular box B., (b) How do you evaluate xxxB f 共x, y, z兲 dV ?, (c) How do you define xxxE f 共x, y, z兲 dV if E is a bounded solid, region that is not a box?, (d) What is a type 1 solid region? How do you evaluate, xxxE f 共x, y, z兲 dV if E is such a region?, (e) What is a type 2 solid region? How do you evaluate, xxxE f 共x, y, z兲 dV if E is such a region?, (f ) What is a type 3 solid region? How do you evaluate, xxxE f 共x, y, z兲 dV if E is such a region?, , 2. (a) How do you define xxD f 共x, y兲 dA if D is a bounded region, , that is not a rectangle?, (b) What is a type I region? How do you evaluate, xxD f 共x, y兲 dA if D is a type I region?, (c) What is a type II region? How do you evaluate, xxD f 共x, y兲 dA if D is a type II region?, (d) What properties do double integrals have?, , 8. Suppose a solid object occupies the region E and has density, , function 共x, y, z兲. Write expressions for each of the following., (a) The mass, (b) The moments about the coordinate planes, (c) The coordinates of the center of mass, (d) The moments of inertia about the axes, , 3. How do you change from rectangular coordinates to polar coor-, , dinates in a double integral? Why would you want to make the, change?, 4. If a lamina occupies a plane region D and has density function, , 共x, y兲, write expressions for each of the following in terms of, double integrals., (a) The mass, (b) The moments about the axes, (c) The center of mass, (d) The moments of inertia about the axes and the origin, 5. Let f be a joint density function of a pair of continuous, , random variables X and Y., (a) Write a double integral for the probability that X lies, between a and b and Y lies between c and d., , 9. (a) How do you change from rectangular coordinates to cylin-, , drical coordinates in a triple integral?, (b) How do you change from rectangular coordinates to, spherical coordinates in a triple integral?, (c) In what situations would you change to cylindrical or, spherical coordinates?, 10. (a) If a transformation T is given by x 苷 t共u, v兲,, y 苷 h共u, v兲, what is the Jacobian of T ?, , (b) How do you change variables in a double integral?, (c) How do you change variables in a triple integral?, , True-False Quiz, Determine whether the statement is true or false. If it is true, explain why., If it is false, explain why or give an example that disproves the statement., 2, , 1., , y y, , 2., , yy, , 3., , yy, , 4., , 1, , 0, , 1, , x, , 0, , 0, , 2, , 1, , 1, , x 2 sin共x  y兲 dx dy 苷 y, , 4, , x 2e y dy dx 苷, 1, , 0, , 6, , 0, , sx  y 2 dy dx 苷, , 3, , y y, 1, , 6, , e, , x 2y 2, , y, , 2, , 1, , x, , yy, 0, , 1, , 0, , y, , 2, , 1, , x 2 sin共x  y兲 dy dx, , 0, , 1, , 0, , 1, , 1, , 0, , 2, ,  sy ) sin共x 2 y 2 兲 dx dy, , 7. If D is the disk given by x 2  y 2, , yy s4  x, , 4, then, ,  y 2 dA 苷, , 16, 3, , , , 8. The integral xxxE kr 3 dz dr d represents the moment of, , 4, , 3, , f 共x兲 f 共 y兲 dy dx 苷, , 2, , 9, , D, , x 2 dx y e y dy, , 5. If f is continuous on 关0, 1兴, then, 1, , 4, , y y (x, , sx  y 2 dx dy, , inertia about the z-axis of a solid E with constant density k., , sin y dx dy 苷 0, , yy, , 6., , 9. The integral, 2, , 2, , y yy, , 冋y, , 册, , 2, , 1, , 0, , f 共x兲 dx, , 0, , 0, , 2, , r, , dz dr d, , represents the volume enclosed by the cone z 苷 sx 2  y 2, and the plane z 苷 2., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1066-1078.qk_97817_15_ch15_p1066-1078 11/8/10 3:44 PM Page 1075, , CHAPTER 15, , where E is bounded by the planes y 苷 0, z 苷 0,, x  y 苷 2 and the cylinder y 2  z 2 苷 1 in the first octant, , 26., , xxxE z dV,, , 27., , xxxE yz dV,, , 28., ,  y  z dV, where H is the solid hemisphere that, lies above the xy-plane and has center the origin and radius 1, , CAS, , where E lies above the plane z 苷 0, below the plane, z 苷 y, and inside the cylinder x 2  y 2 苷 4, 3, , xxxH z sx, , 2, , 2, , x 3,  y, find its surface area correct to four decimal places., , 41. Use polar coordinates to evaluate, 3, , y y, , 2, , 2, , xy-plane with vertices 共1, 0兲, 共2, 1兲, and 共4, 0兲, , and y  z 苷 3, , 33. One of the wedges cut from the cylinder x 2  9y 2 苷 a 2 by the, , planes z 苷 0 and z 苷 mx, , 34. Above the paraboloid z 苷 x 2  y 2 and below the half-cone, , z 苷 sx 2  y 2, , 35. Consider a lamina that occupies the region D bounded by, , the parabola x 苷 1  y 2 and the coordinate axes in the first, quadrant with density function 共x, y兲 苷 y., (a) Find the mass of the lamina., (b) Find the center of mass., (c) Find the moments of inertia and radii of gyration about, the x- and y-axes., 36. A lamina occupies the part of the disk x 2  y 2, , a 2 that lies in, , the first quadrant., (a) Find the centroid of the lamina., (b) Find the center of mass of the lamina if the density function, is 共x, y兲 苷 xy 2., , y, , s4x 2y 2, , s4x 2y 2, , 0, , CAS, , 44. Find the center of mass of the solid tetrahedron with vertices, , 共0, 0, 0兲, 共1, 0, 0兲, 共0, 2, 0兲, 共0, 0, 3兲 and density function, 共x, y, z兲 苷 x 2  y 2  z 2., 45. The joint density function for random variables X and Y is, , f 共x, y兲 苷, , 再, , C共x  y兲 if 0 x, 0, otherwise, , and base radius a. (Place the cone so that its base is in the, xy-plane with center the origin and its axis along the positive z-axis.), (b) Find the moment of inertia of the cone about its axis, (the z-axis)., 38. Find the area of the part of the cone z 2 苷 a 2共x 2  y 2 兲 between, , the planes z 苷 1 and z 苷 2., , 39. Find the area of the part of the surface z 苷 x  y that lies, 2, , above the triangle with vertices (0, 0), (1, 0), and (0, 2)., , 3, 0, , y, , 2, , (a) Find the value of the constant C., (b) Find P共X 2, Y 1兲., (c) Find P共X  Y 1兲., 46. A lamp has three bulbs, each of a type with average lifetime, , 800 hours. If we model the probability of failure of the, bulbs by an exponential density function with mean 800,, find the probability that all three bulbs fail within a total of, 1000 hours., 47. Rewrite the integral, 1, , 1, , 1y, , 1, , x2, , 0, , y y y, , f 共x, y, z兲 dz dy dx, , as an iterated integral in the order dx dy dz., 48. Give five other iterated integrals that are equal to, 2, , 37. (a) Find the centroid of a right circular cone with height h, , y 2sx 2  y 2  z 2 dz dx dy, , y 苷 e x, find the approximate value of the integral xxD y 2 dA., (Use a graphing device to estimate the points of intersection, of the curves.), , 30. Under the surface z 苷 x y and above the triangle in the, , 32. Bounded by the cylinder x 2  y 2 苷 4 and the planes z 苷 0, , s4y 2, , 2, ; 43. If D is the region bounded by the curves y 苷 1  x and, , 2, , and 共2, 2, 0兲, , 共x 3  xy 2 兲 dy dx, , 42. Use spherical coordinates to evaluate, , y y, , 31. The solid tetrahedron with vertices 共0, 0, 0兲, 共0, 0, 1兲, 共0, 2, 0兲,, , s9x 2, , s9x 2, , 0, , 29–34 Find the volume of the given solid., , R 苷 关0, 2兴  关1, 4兴, , 1075, , and, , 40. Graph the surface z 苷 x sin y, 3, , 2, , 29. Under the paraboloid z 苷 x 2  4y 2 and above the rectangle, , REVIEW, , y3, , y y y, 0, , 0, , y2, , 0, , f 共x, y, z兲 dz dx dy, , 49. Use the transformation u 苷 x  y, v 苷 x  y to evaluate, , xy, , yy x  y dA, R, , where R is the square with vertices 共0, 2兲, 共1, 1兲, 共2, 2兲,, and 共1, 3兲., 50. Use the transformation x 苷 u 2, y 苷 v 2, z 苷 w 2 to, , find the volume of the region bounded by the surface, sx  sy  sz 苷 1 and the coordinate planes., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1066-1078.qk_97817_15_ch15_p1066-1078 11/8/10 3:44 PM Page 1076, , 1076, , CHAPTER 15, , MULTIPLE INTEGRALS, , 51. Use the change of variables formula and an appropriate trans-, , formation to evaluate xxR xy dA, where R is the square with vertices 共0, 0兲, 共1, 1兲, 共2, 0兲, and 共1, 1兲., , Exercise 52) to show that, lim, rl0, , 52. The Mean Value Theorem for double integrals says that, , if f is a continuous function on a plane region D that is of type, I or II, then there exists a point 共x 0 , y0 兲 in D such that, , yy f 共x, y兲 dA 苷 f 共x , y 兲 A共D兲, 0, , 0, , D, , Use the Extreme Value Theorem (14.7.8) and Property 15.3.11, of integrals to prove this theorem. (Use the proof of the singlevariable version in Section 5.5 as a guide.), 53. Suppose that f is continuous on a disk that contains the, , point 共a, b兲. Let Dr be the closed disk with center 共a, b兲 and, radius r. Use the Mean Value Theorem for double integrals (see, , 54. (a) Evaluate yy, D, , 1, r 2, , yy f 共x, y兲 dA 苷 f 共a, b兲, Dr, , 1, dA, where n is an integer and D is, 共x 2  y 2 兲n兾2, , the region bounded by the circles with center the origin and, radii r and R, 0  r  R., (b) For what values of n does the integral in part (a) have a, limit as r l 0 ?, 1, (c) Find yyy 2, dV, where E is the region, 2, 共x, , y,  z 2 兲n兾2, E, bounded by the spheres with center the origin and radii r, and R, 0  r  R., (d) For what values of n does the integral in part (c) have a, limit as r l 0 ?, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Problems Plus, 1. If 冀x冁 denotes the greatest integer in x, evaluate the integral, , yy 冀x  y冁 dA, R, , where R 苷 兵共x, y兲, , ⱍ1, , x, , 3, 2, , 5其., , y, , 2. Evaluate the integral, 1, , yy, 0, , 1, , 0, , 2, , 2, , e max兵x , y 其 dy dx, , where max 兵x 2, y 2 其 means the larger of the numbers x 2 and y 2., 3. Find the average value of the function f 共x兲 苷, , xx1 cos共t 2 兲 dt on the interval [0, 1]., , 4. If a, b, and c are constant vectors, r is the position vector x i  y j  z k, and E is given by, , the inequalities 0, , aⴢr, , bⴢr, , , 0, , , 0, , yyy 共a ⴢ r兲共b ⴢ r兲共c ⴢ r兲 dV 苷 8, E, , cⴢr, , ⱍ, , , show that, , 共 兲2, a ⴢ 共b  c兲, , ⱍ, , 1, dx dy is an improper integral and could be defined as, 1  xy, the limit of double integrals over the rectangle 关0, t兴  关0, t兴 as t l 1. But if we expand the, integrand as a geometric series, we can express the integral as the sum of an infinite series., Show that, , 5. The double integral y, , 1, , 0, , y, , 1, , 0, , 1, , yy, 0, , 1, , 0, , , 1, 1, dx dy 苷 兺 2, 1  xy, n苷1 n, , 6. Leonhard Euler was able to find the exact sum of the series in Problem 5. In 1736 he proved, , that, , , 兺, , n苷1, , 1, 2, 苷, 2, n, 6, , In this problem we ask you to prove this fact by evaluating the double integral in Problem 5., Start by making the change of variables, uv, s2, , x苷, , y苷, , uv, s2, , This gives a rotation about the origin through the angle 兾4. You will need to sketch the, corresponding region in the u v-plane., [Hint: If, in evaluating the integral, you encounter either of the expressions, 共1  sin  兲兾cos  or 共cos  兲兾共1  sin  兲, you might like to use the identity, cos  苷 sin共共兾2兲   兲 and the corresponding identity for sin .], 7. (a) Show that, 1, , 1, , yyy, 0, , 0, , 1, , 0, , , 1, 1, dx dy dz 苷 兺 3, 1  xyz, n苷1 n, , (Nobody has ever been able to find the exact value of the sum of this series.), (b) Show that, 1, , 1, , 0, , 0, , yyy, , 1, , 0, , , 1, 共1兲 n1, dx dy dz 苷 兺, 1  xyz, n3, n苷1, , Use this equation to evaluate the triple integral correct to two decimal places., , 1077, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_15_ch15_p1066-1078.qk_97817_15_ch15_p1066-1078 11/8/10 3:44 PM Page 1078, , 8. Show that, , y, , , , 0, , arctan  x  arctan x, , dx 苷, ln , x, 2, , by first expressing the integral as an iterated integral., 9. (a) Show that when Laplace’s equation, , 2u, 2u, 2u, ,  2 苷0, 2, 2, x, y, z, is written in cylindrical coordinates, it becomes, 2u, 1 u, 1 2u, 2u, ,  2,  2 苷0, 2, 2, r, r r, r , z, (b) Show that when Laplace’s equation is written in spherical coordinates, it becomes, 2u, 2 u, cot  u, 1 2u, , ,  2, , 2, 2, , , ,  2, , 2, , 1,  2u, 苷0, 2, sin   2, , 10. (a) A lamina has constant density, , and takes the shape of a disk with center the origin and, radius R. Use Newton’s Law of Gravitation (see Section 13.4) to show that the magnitude, of the force of attraction that the lamina exerts on a body with mass m located at the, point 共0, 0, d 兲 on the positive z-axis is, , 冉, , F 苷 2 Gm d, , 1, 1, , d, sR 2  d 2, , 冊, , [Hint: Divide the disk as in Figure 4 in Section 15.4 and first compute the vertical component of the force exerted by the polar subrectangle Rij .], (b) Show that the magnitude of the force of attraction of a lamina with density that occupies an entire plane on an object with mass m located at a distance d from the plane is, F 苷 2 Gm, Notice that this expression does not depend on d., 11. If f is continuous, show that, x, , y, , z, , 0, , 0, , 0, , yyy, n, , 12. Evaluate lim n 2 兺, nl, , n2, , 兺, , i苷1 j苷1, , x, , f 共t兲 dt dz dy 苷 12 y 共x  t兲2 f 共t兲 dt, 0, , 1, ., sn 2  ni  j, , 13. The plane, , x, y, z,   苷1, a, b, c, , a  0,, , b  0,, , c0, , cuts the solid ellipsoid, x2, y2, z2, 2 , 2 , a, b, c2, , 1, , into two pieces. Find the volume of the smaller piece., , 1078, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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16, , Vector Calculus, , Parametric surfaces, which are studied in, Section 16.6, are frequently used by, programmers creating animated films. In, this scene from Antz, Princess Bala is, about to try to rescue Z, who is trapped, in a dewdrop. A parametric surface, represents the dewdrop and a family of, such surfaces depicts its motion. One of, the programmers for this film was heard, to say, “I wish I had paid more attention, in calculus class when we were studying, parametric surfaces. It would sure have, helped me today.”, , © Dreamworks / Photofest, , In this chapter we study the calculus of vector fields. (These are functions that assign vectors to points in, space.) In particular we define line integrals (which can be used to find the work done by a force field in, moving an object along a curve). Then we define surface integrals (which can be used to find the rate, of fluid flow across a surface). The connections between these new types of integrals and the single,, double, and triple integrals that we have already met are given by the higher-dimensional versions of the, Fundamental Theorem of Calculus: Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem., , 1079, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1080, , 16.1, , CHAPTER 16, , VECTOR CALCULUS, , Vector Fields, The vectors in Figure 1 are air velocity vectors that indicate the wind speed and direction, at points 10 m above the surface elevation in the San Francisco Bay area. We see at a, glance from the largest arrows in part (a) that the greatest wind speeds at that time occurred, as the winds entered the bay across the Golden Gate Bridge. Part (b) shows the very different wind pattern 12 hours earlier. Associated with every point in the air we can imagine, a wind velocity vector. This is an example of a velocity vector field., , (a) 6:00 PM, March 1, 2010, , (b) 6:00 AM, March 1, 2010, , FIGURE 1 Velocity vector fields showing San Francisco Bay wind patterns, , Adapted from ONERA photograph, Werle, 1974, , Other examples of velocity vector fields are illustrated in Figure 2: ocean currents and, flow past an airfoil., , Nova Scotia, , (a) Ocean currents off the coast of Nova Scotia, , (b) Airflow past an inclined airfoil, , FIGURE 2 Velocity vector fields, , Another type of vector field, called a force field, associates a force vector with each, point in a region. An example is the gravitational force field that we will look at in, Example 4., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1079-1087.qk_97817_16_ch16_p1079-1087 11/9/10 9:04 AM Page 1082, , 1082, , CHAPTER 16, , VECTOR CALCULUS, , It appears from Figure 5 that each arrow is tangent to a circle with center the origin., To confirm this, we take the dot product of the position vector x 苷 x i ⫹ y j with the, vector F共x兲 苷 F共x, y兲:, x ⴢ F共x兲 苷 共x i ⫹ y j兲 ⴢ 共⫺y i ⫹ x j兲 苷 ⫺xy ⫹ yx 苷 0, This shows that F共x, y兲 is perpendicular to the position vector 具 x, y典 and is therefore, tangent to a circle with center the origin and radius x 苷 sx 2 ⫹ y 2 . Notice also that, , ⱍ ⱍ, , ⱍ F共x, y兲 ⱍ 苷 s共⫺y兲, , 2, , ⱍ ⱍ, , ⫹ x 2 苷 sx 2 ⫹ y 2 苷 x, , so the magnitude of the vector F共x, y兲 is equal to the radius of the circle., Some computer algebra systems are capable of plotting vector fields in two or three, dimensions. They give a better impression of the vector field than is possible by hand, because the computer can plot a large number of representative vectors. Figure 6 shows a, computer plot of the vector field in Example 1; Figures 7 and 8 show two other vector, fields. Notice that the computer scales the lengths of the vectors so they are not too long, and yet are proportional to their true lengths., 5, , _5, , 6, , 5, , _6, , 5, , _5, , _5, , 6, , 5, , _6, , _5, , FIGURE 6, , FIGURE 7, , FIGURE 8, , F(x, y)=k_y, xl, , F(x, y)=ky, sin xl, , F(x, y)=k ln(1+¥), ln(1+≈)l, , v, , EXAMPLE 2 Sketch the vector field on ⺢ 3 given by F共x, y, z兲 苷 z k., , SOLUTION The sketch is shown in Figure 9. Notice that all vectors are vertical and point, , upward above the xy-plane or downward below it. The magnitude increases with the, distance from the xy-plane., z, , 0, y, x, , FIGURE 9, , F(x, y, z)=z k, , We were able to draw the vector field in Example 2 by hand because of its particularly, simple formula. Most three-dimensional vector fields, however, are virtually impossible to, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1079-1087.qk_97817_16_ch16_p1079-1087 11/9/10 9:04 AM Page 1083, , VECTOR FIELDS, , SECTION 16.1, , 1083, , sketch by hand and so we need to resort to a computer algebra system. Examples are, shown in Figures 10, 11, and 12. Notice that the vector fields in Figures 10 and 11 have similar formulas, but all the vectors in Figure 11 point in the general direction of the negative, y-axis because their y-components are all ⫺2. If the vector field in Figure 12 represents a, velocity field, then a particle would be swept upward and would spiral around the z-axis, in the clockwise direction as viewed from above., , 1, z, , 0, , z, , _1, , 1, , 5, , 0, , z3, , _1, 1, _1, , 0, y, , 1, , 1, , 0, , _1, x, , _1, , FIGURE 10, F(x, y, z)=y i+z j+x k, , TEC In Visual 16.1 you can rotate the, vector fields in Figures 10–12 as well as, additional fields., z, , 0, x, , 0, y, , 1, , 1, , 0, , _1, , _1, _1, , x, , FIGURE 11, F(x, y, z)=y i-2 j+x k, , y0, , 0, 1, , 1, , x, , FIGURE 12, y, x, z, F(x, y, z)= i- j+ k, z, z, 4, , EXAMPLE 3 Imagine a fluid flowing steadily along a pipe and let V共x, y, z兲 be the velocity vector at a point 共x, y, z兲. Then V assigns a vector to each point 共x, y, z兲 in a certain, domain E (the interior of the pipe) and so V is a vector field on ⺢ 3 called a velocity field., A possible velocity field is illustrated in Figure 13. The speed at any given point is indicated by the length of the arrow., Velocity fields also occur in other areas of physics. For instance, the vector field in, Example 1 could be used as the velocity field describing the counterclockwise rotation of, a wheel. We have seen other examples of velocity fields in Figures 1 and 2., , y, , FIGURE 13, , EXAMPLE 4 Newton’s Law of Gravitation states that the magnitude of the gravitational, force between two objects with masses m and M is, , Velocity field in fluid flow, , mMG, r2, , ⱍFⱍ 苷, , where r is the distance between the objects and G is the gravitational constant. (This, is an example of an inverse square law.) Let’s assume that the object with mass M is, located at the origin in ⺢ 3. (For instance, M could be the mass of the earth and the origin, would be at its center.) Let the position vector of the object with mass m be x 苷 具x, y, z典., Then r 苷 x , so r 2 苷 x 2. The gravitational force exerted on this second object acts, toward the origin, and the unit vector in this direction is, , ⱍ ⱍ, , ⱍ ⱍ, , ⫺, , x, x, , ⱍ ⱍ, , Therefore the gravitational force acting on the object at x 苷 具x, y, z典 is, 3, , F共x兲 苷 ⫺, , mMG, x, x 3, , ⱍ ⱍ, , [Physicists often use the notation r instead of x for the position vector, so you may see, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1079-1087.qk_97817_16_ch16_p1079-1087 11/9/10 9:04 AM Page 1084, , 1084, , CHAPTER 16, , VECTOR CALCULUS, , Formula 3 written in the form F 苷 ⫺共mMG兾r 3 兲r.] The function given by Equation 3 is, an example of a vector field, called the gravitational field, because it associates a vector, [the force F共x兲] with every point x in space., Formula 3 is a compact way of writing the gravitational field, but we can also write, it in terms of its component functions by using the facts that x 苷 x i ⫹ y j ⫹ z k and, x 苷 sx 2 ⫹ y 2 ⫹ z 2 :, , z, , ⱍ ⱍ, y, , x, , F共x, y, z兲 苷, , ⫺mMGx, ⫺mMGy, ⫺mMGz, i⫹ 2, j⫹ 2, k, 2, 2 3兾2, 2, 2 3兾2, 共x ⫹ y ⫹ z 兲, 共x ⫹ y ⫹ z 兲, 共x ⫹ y 2 ⫹ z 2 兲3兾2, 2, , The gravitational field F is pictured in Figure 14., EXAMPLE 5 Suppose an electric charge Q is located at the origin. According to, Coulomb’s Law, the electric force F共x兲 exerted by this charge on a charge q located at a, point 共x, y, z兲 with position vector x 苷 具x, y, z 典 is, , FIGURE 14, , Gravitational force field, , F共x兲 苷, , 4, , ␧qQ, x, x 3, , ⱍ ⱍ, , where ␧ is a constant (that depends on the units used). For like charges, we have qQ ⬎ 0, and the force is repulsive; for unlike charges, we have qQ ⬍ 0 and the force is attractive., Notice the similarity between Formulas 3 and 4. Both vector fields are examples of force, fields., Instead of considering the electric force F, physicists often consider the force per unit, charge:, E共x兲 苷, , 1, ␧Q, F共x兲 苷, x, q, x 3, , ⱍ ⱍ, , Then E is a vector field on ⺢ 3 called the electric field of Q., , Gradient Fields, If f is a scalar function of two variables, recall from Section 14.6 that its gradient ∇f (or, grad f ) is defined by, ⵜf 共x, y兲 苷 fx 共x, y兲 i ⫹ fy 共x, y兲 j, Therefore ∇f is really a vector field on ⺢ 2 and is called a gradient vector field. Likewise,, if f is a scalar function of three variables, its gradient is a vector field on ⺢ 3 given by, ⵜf 共x, y, z兲 苷 fx 共x, y, z兲 i ⫹ fy 共x, y, z兲 j ⫹ fz 共x, y, z兲 k, , 4, , v, , EXAMPLE 6 Find the gradient vector field of f 共x, y兲 苷 x 2 y ⫺ y 3. Plot the gradient, , vector field together with a contour map of f. How are they related?, _4, , 4, , SOLUTION The gradient vector field is given by, , ⵜf 共x, y兲 苷, _4, , FIGURE 15, , ⭸f, ⭸f, i⫹, j 苷 2xy i ⫹ 共x 2 ⫺ 3y 2 兲 j, ⭸x, ⭸y, , Figure 15 shows a contour map of f with the gradient vector field. Notice that the gradient vectors are perpendicular to the level curves, as we would expect from Section 14.6., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 16.2, , LINE INTEGRALS, , 1087, , Line Integrals, , 16.2, , In this section we define an integral that is similar to a single integral except that instead, of integrating over an interval 关a, b兴, we integrate over a curve C. Such integrals are called, line integrals, although “curve integrals” would be better terminology. They were invented, in the early 19th century to solve problems involving fluid flow, forces, electricity, and, magnetism., We start with a plane curve C given by the parametric equations, x 苷 x共t兲, , 1, , y, , P i*(x *i , y i*), , Pi-1, , Pi, , C, , Pn, , P™, P¡, P¸, , x, , 0, , y 苷 y共t兲, , or, equivalently, by the vector equation r共t兲 苷 x共t兲 i ⫹ y共t兲 j, and we assume that C is a, smooth curve. [This means that r⬘ is continuous and r⬘共t兲 苷 0. See Section 13.3.] If we, divide the parameter interval 关a, b兴 into n subintervals 关ti⫺1, ti 兴 of equal width and we let, x i 苷 x共ti 兲 and yi 苷 y共ti 兲, then the corresponding points Pi 共x i , yi 兲 divide C into n subarcs, with lengths ⌬s1, ⌬s2 , . . . , ⌬sn . (See Figure 1.) We choose any point Pi*共x i*, yi*兲 in the ith, subarc. (This corresponds to a point t*i in 关ti⫺1, ti兴.) Now if f is any function of two variables whose domain includes the curve C, we evaluate f at the point 共x i*, yi*兲, multiply by, the length ⌬si of the subarc, and form the sum, , t *i, a, FIGURE 1, , t i-1, , a艋t艋b, , n, , ti, , 兺 f 共x *, y*兲 ⌬s, , b t, , i, , i, , i, , i苷1, , which is similar to a Riemann sum. Then we take the limit of these sums and make the following definition by analogy with a single integral., 2 Definition If f is defined on a smooth curve C given by Equations 1, then the, line integral of f along C is, n, , y, , C, , 兺 f 共x *, y*兲 ⌬s, , f 共x, y兲 ds 苷 lim, , n l ⬁ i苷1, , i, , i, , i, , if this limit exists., In Section 10.2 we found that the length of C is, L苷, , y, , b, , a, , 冑冉 冊 冉 冊, dx, dt, , 2, , ⫹, , 2, , dy, dt, , dt, , A similar type of argument can be used to show that if f is a continuous function, then the, limit in Definition 2 always exists and the following formula can be used to evaluate the, line integral:, , 3, , y, , C, , b, , 冑冉 冊 冉 冊, , f 共x, y兲 ds 苷 y f ( x共t兲, y共t兲), a, , dx, dt, , 2, , ⫹, , dy, dt, , 2, , dt, , The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1088-1097.qk_97817_16_ch16_p1088-1097 11/9/10 9:05 AM Page 1092, , 1092, , CHAPTER 16, , VECTOR CALCULUS, , In general, a given parametrization x 苷 x共t兲, y 苷 y共t兲, a 艋 t 艋 b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the, parameter t. (See Figure 8, where the initial point A corresponds to the parameter value a, and the terminal point B corresponds to t 苷 b.), If ⫺C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A in Figure 8), then we have, , B, C, A, , a, , b, , t, , y, , B, , A, , _C, , ⫺C, , f 共x, y兲 dx 苷 ⫺y f 共x, y兲 dx, , y, , ⫺C, , C, , f 共x, y兲 dy 苷 ⫺y f 共x, y兲 dy, C, , But if we integrate with respect to arc length, the value of the line integral does not change, when we reverse the orientation of the curve:, , FIGURE 8, , y, , ⫺C, , f 共x, y兲 ds 苷, , y, , C, , f 共x, y兲 ds, , This is because ⌬si is always positive, whereas ⌬ x i and ⌬yi change sign when we reverse, the orientation of C., , Line Integrals in Space, We now suppose that C is a smooth space curve given by the parametric equations, x 苷 x共t兲, , y 苷 y共t兲, , z 苷 z共t兲, , a艋t艋b, , or by a vector equation r共t兲 苷 x共t兲 i ⫹ y共t兲 j ⫹ z共t兲 k. If f is a function of three variables, that is continuous on some region containing C, then we define the line integral of f, along C (with respect to arc length) in a manner similar to that for plane curves:, n, , y, , C, , f 共x, y, z兲 ds 苷 lim, , 兺 f 共x*, y*, z*兲 ⌬s, i, , n l ⬁ i苷1, , i, , i, , i, , We evaluate it using a formula similar to Formula 3:, , 9, , y, , C, , f 共x, y, z兲 ds 苷, , 冑冉 冊 冉 冊 冉 冊, , b, , dx, dt, , y f ( x共t兲, y共t兲, z共t兲), a, , 2, , ⫹, , dy, dt, , 2, , ⫹, , dz, dt, , 2, , dt, , Observe that the integrals in both Formulas 3 and 9 can be written in the more compact, vector notation, , y, , b, , a, , ⱍ, , ⱍ, , f 共r共t兲兲 r⬘共t兲 dt, , For the special case f 共x, y, z兲 苷 1, we get, , y, , C, , ds 苷 y, , b, , a, , ⱍ r⬘共t兲 ⱍ dt 苷 L, , where L is the length of the curve C (see Formula 13.3.3)., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1088-1097.qk_97817_16_ch16_p1088-1097 11/9/10 9:05 AM Page 1094, , 1094, , CHAPTER 16, , VECTOR CALCULUS, , Then dx 苷 0 苷 dy, so, 1, , y dx ⫹ z dy ⫹ x dz 苷 y 3共⫺5兲 dt 苷 ⫺15, , y, , C2, , 0, , Adding the values of these integrals, we obtain, , y, , C, , y dx ⫹ z dy ⫹ x dz 苷 24.5 ⫺ 15 苷 9.5, , Line Integrals of Vector Fields, Recall from Section 5.4 that the work done by a variable force f 共x兲 in moving a particle, from a to b along the x-axis is W 苷 xab f 共x兲 dx. Then in Section 12.3 we found that the, work done by a constant force F in moving an object from a point P to another point Q in, l, space is W 苷 F ⴢ D, where D 苷 PQ is the displacement vector., Now suppose that F 苷 P i ⫹ Q j ⫹ R k is a continuous force field on ⺢ 3, such as the, gravitational field of Example 4 in Section 16.1 or the electric force field of Example 5 in, Section 16.1. (A force field on ⺢ 2 could be regarded as a special case where R 苷 0 and P, and Q depend only on x and y.) We wish to compute the work done by this force in moving a particle along a smooth curve C., We divide C into subarcs Pi⫺1Pi with lengths ⌬si by dividing the parameter interval, 关a, b兴 into subintervals of equal width. (See Figure 1 for the two-dimensional case or, Figure 11 for the three-dimensional case.) Choose a point Pi*共x*i , yi*, zi*兲 on the ith subarc, corresponding to the parameter value t i*. If ⌬si is small, then as the particle moves from, Pi⫺1 to Pi along the curve, it proceeds approximately in the direction of T共t i*兲, the unit tangent vector at Pi*. Thus the work done by the force F in moving the particle from Pi⫺1 to, Pi is approximately, , z, , F(x *i , y*i , z *i ), T(t *i ), Pi-1, , Pi, , 0, , P i*(x *i , y*i , z *i ), , Pn, y, , F共 x*i , yi*, zi*兲 ⴢ 关⌬si T共t i*兲兴 苷 关F共x*i , yi*, zi*兲 ⴢ T共t i*兲兴 ⌬si, , x, , P¸, , and the total work done in moving the particle along C is approximately, FIGURE 11, n, , 兺 关F共x*, y*, z*兲 ⴢ T共x*, y*, z*兲兴 ⌬s, , 11, , i, , i, , i, , i, , i, , i, , i, , i苷1, , where T共x, y, z兲 is the unit tangent vector at the point 共x, y, z兲 on C. Intuitively, we see that, these approximations ought to become better as n becomes larger. Therefore we define the, work W done by the force field F as the limit of the Riemann sums in 11 , namely,, W 苷 y F共x, y, z兲 ⴢ T共x, y, z兲 ds 苷 y F ⴢ T ds, , 12, , C, , C, , Equation 12 says that work is the line integral with respect to arc length of the tangential, component of the force., If the curve C is given by the vector equation r共t兲 苷 x共t兲 i ⫹ y共t兲 j ⫹ z共t兲 k, then, T共t兲 苷 r⬘共t兲兾 r⬘共t兲 , so using Equation 9 we can rewrite Equation 12 in the form, , ⱍ, , ⱍ, , W苷, , y, , b, , a, , 冋, , F共r共t兲兲 ⴢ, , ⱍ, , r⬘共t兲, r⬘共t兲, , ⱍ册ⱍ, , ⱍ, , b, , r⬘共t兲 dt 苷 y F共r共t兲兲 ⴢ r⬘共t兲 dt, a, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1088-1097.qk_97817_16_ch16_p1088-1097 11/9/10 9:06 AM Page 1097, , SECTION 16.2, , 17. Let F be the vector field shown in the figure., , (a) If C1 is the vertical line segment from 共⫺3, ⫺3兲 to 共⫺3, 3兲,, determine whether xC F ⴢ dr is positive, negative, or zero., (b) If C2 is the counterclockwise-oriented circle with radius 3, and center the origin, determine whether xC F ⴢ dr is positive, negative, or zero., 1, , xC F ⴢ dr, where F共x, y, z兲 苷 y sin z i ⫹ z sin x j ⫹ x sin y k, and r共t兲 苷 cos t i ⫹ sin t j ⫹ sin 5t k, 0 艋 t 艋 ␲, , 25., , xC x sin共 y ⫹ z兲 ds, where C has parametric equations x 苷 t 2,, y 苷 t 3, z 苷 t 4, 0 艋 t 艋 5, , 26., , xC ze⫺xy ds, where C has parametric equations x 苷 t, y 苷 t 2,, , z 苷 e⫺t, 0 艋 t 艋 1, , y, 3, CAS, , 2, 1, _2, , 1097, , 24., , 2, , _3, , LINE INTEGRALS, , _1 0, _1, , 2, , 1, , 27–28 Use a graph of the vector field F and the curve C to guess, whether the line integral of F over C is positive, negative, or zero., Then evaluate the line integral., 27. F共x, y兲 苷 共x ⫺ y兲 i ⫹ x y j,, , 3x, , C is the arc of the circle x 2 ⫹ y 2 苷 4 traversed counterclockwise from (2, 0) to 共0, ⫺2兲, , _2, , x, y, i⫹, j,, sx 2 ⫹ y 2, sx 2 ⫹ y 2, 2, C is the parabola y 苷 1 ⫹ x from 共⫺1, 2兲 to (1, 2), , 28. F共x, y兲 苷, , _3, , 18. The figure shows a vector field F and two curves C1 and C2., 29. (a) Evaluate the line integral xC F ⴢ dr, where, , Are the line integrals of F over C1 and C2 positive, negative,, or zero? Explain., y, , ;, , C¡, , 30. (a) Evaluate the line integral xC F ⴢ dr, where, , C™, , ;, , F共x, y, z兲 苷 x i ⫺ z j ⫹ y k and C is given by, r共t兲 苷 2t i ⫹ 3t j ⫺ t 2 k, ⫺1 艋 t 艋 1., (b) Illustrate part (a) by using a computer to graph C and, the vectors from the vector field corresponding to, t 苷 ⫾1 and ⫾ 12 (as in Figure 13)., , CAS, , 31. Find the exact value of xC x 3 y 2 z ds, where C is the curve with, , x, , parametric equations x 苷 e⫺t cos 4 t, y 苷 e⫺t sin 4 t, z 苷 e⫺t,, 0 艋 t 艋 2␲., , 19–22 Evaluate the line integral xC F ⴢ dr, where C is given by the, , 32. (a) Find the work done by the force field F共x, y兲 苷 x 2 i ⫹ x y j, , vector function r共t兲., 19. F共x, y兲 苷 xy i ⫹ 3y 2 j,, , r共t兲 苷 11t 4 i ⫹ t 3 j, 0 艋 t 艋 1, , CAS, , 20. F共x, y, z兲 苷 共x ⫹ y兲 i ⫹ 共 y ⫺ z兲 j ⫹ z k,, 2, , r共t兲 苷 t 2 i ⫹ t 3 j ⫹ t 2 k,, , 0艋t艋1, , 21. F共x, y, z兲 苷 sin x i ⫹ cos y j ⫹ xz k,, , r共t兲 苷 t i ⫺ t j ⫹ t k,, 3, , 2, , 0艋t艋1, , 22. F共x, y, z兲 苷 x i ⫹ y j ⫹ xy k,, , r共t兲 苷 cos t i ⫹ sin t j ⫹ t k,, , 0艋t艋␲, , 23–26 Use a calculator or CAS to evaluate the line integral correct, , to four decimal places., 23., , xC F ⴢ dr, where F共x, y兲 苷 xy i ⫹ sin y j and, r共t兲 苷 e t i ⫹ e⫺t j, 1 艋 t 艋 2, 2, , F共x, y兲 苷 e x⫺1 i ⫹ x y j and C is given by, r共t兲 苷 t 2 i ⫹ t 3 j, 0 艋 t 艋 1., (b) Illustrate part (a) by using a graphing calculator or computer to graph C and the vectors from the vector field, corresponding to t 苷 0, 1兾s2 , and 1 (as in Figure 13)., , on a particle that moves once around the circle, x 2 ⫹ y 2 苷 4 oriented in the counter-clockwise direction., (b) Use a computer algebra system to graph the force field and, circle on the same screen. Use the graph to explain your, answer to part (a)., 33. A thin wire is bent into the shape of a semicircle x 2 ⫹ y 2 苷 4,, , x 艌 0. If the linear density is a constant k, find the mass and, center of mass of the wire., , 34. A thin wire has the shape of the first-quadrant part of the, , circle with center the origin and radius a. If the density, function is ␳ 共x, y兲 苷 kxy, find the mass and center of mass, of the wire., 35. (a) Write the formulas similar to Equations 4 for the center of, , mass 共 x, y, z 兲 of a thin wire in the shape of a space curve C, if the wire has density function ␳ 共x, y, z兲., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1098-1107.qk_97817_16_ch16_p1098-1107 11/9/10 9:06 AM Page 1098, , 1098, , CHAPTER 16, , VECTOR CALCULUS, , (b) Find the center of mass of a wire in the shape of the helix, x 苷 2 sin t, y 苷 2 cos t, z 苷 3t, 0 艋 t 艋 2␲, if the density, is a constant k., 36. Find the mass and center of mass of a wire in the shape of the, , helix x 苷 t, y 苷 cos t, z 苷 sin t, 0 艋 t 艋 2␲, if the density at, any point is equal to the square of the distance from the origin., 37. If a wire with linear density ␳ 共x, y兲 lies along a plane curve C,, , its moments of inertia about the x- and y-axes are defined as, Ix 苷, , y, , C, , y 2␳ 共x, y兲 ds, , I y 苷 y x 2␳ 共x, y兲 ds, C, , Find the moments of inertia for the wire in Example 3., 38. If a wire with linear density ␳ 共x, y, z兲 lies along a space curve, , C, its moments of inertia about the x-, y-, and z-axes are, defined as, I x 苷 y 共 y 2 ⫹ z 2 兲␳ 共x, y, z兲 ds, C, , I y 苷 y 共x 2 ⫹ z 2 兲␳ 共x, y, z兲 ds, C, , I z 苷 y 共x 2 ⫹ y 2 兲␳ 共x, y, z兲 ds, C, , 45. A 160-lb man carries a 25-lb can of paint up a helical staircase, , that encircles a silo with a radius of 20 ft. If the silo is 90 ft, high and the man makes exactly three complete revolutions, climbing to the top, how much work is done by the man, against gravity?, 46. Suppose there is a hole in the can of paint in Exercise 45 and, , 9 lb of paint leaks steadily out of the can during the man’s, ascent. How much work is done?, 47. (a) Show that a constant force field does zero work on a, , particle that moves once uniformly around the circle, x 2 ⫹ y 2 苷 1., (b) Is this also true for a force field F共x兲 苷 k x, where k is a, constant and x 苷 具 x, y 典 ?, 48. The base of a circular fence with radius 10 m is given by, , x 苷 10 cos t, y 苷 10 sin t. The height of the fence at position, 共x, y兲 is given by the function h共x, y兲 苷 4 ⫹ 0.01共x 2 ⫺ y 2 兲, so, the height varies from 3 m to 5 m. Suppose that 1 L of paint, covers 100 m2. Sketch the fence and determine how much paint, you will need if you paint both sides of the fence., 49. If C is a smooth curve given by a vector function r共t兲,, , a 艋 t 艋 b, and v is a constant vector, show that, , y, , C, , Find the moments of inertia for the wire in Exercise 35., , 50. If C is a smooth curve given by a vector function r共t兲,, , a 艋 t 艋 b, show that, , 39. Find the work done by the force field F共x, y兲 苷 x i ⫹ 共 y ⫹ 2兲 j, , in moving an object along an arch of the cycloid, r共t兲 苷 共t ⫺ sin t兲 i ⫹ 共1 ⫺ cos t兲 j, 0 艋 t 艋 2␲., 40. Find the work done by the force field F共x, y兲 苷 x 2 i ⫹ ye x j on, , a particle that moves along the parabola x 苷 y 2 ⫹ 1 from 共1, 0兲, to 共2, 1兲., , 41. Find the work done by the force field, , F共x, y, z兲 苷 具x ⫺ y 2, y ⫺ z 2, z ⫺ x 2 典 on a particle that moves, along the line segment from 共0, 0, 1兲 to 共2, 1, 0兲., , v ⴢ d r 苷 v ⴢ 关r共b兲 ⫺ r共a兲兴, , C, , [ⱍ, , ⱍ, , ⱍ, , ⱍ], , r ⴢ dr 苷 2 r共b兲 2 ⫺ r共a兲 2, 1, , y, , 51. An object moves along the curve C shown in the figure from, , (1, 2) to (9, 8). The lengths of the vectors in the force field F, are measured in newtons by the scales on the axes. Estimate, the work done by F on the object., y, (meters), , C, , 42. The force exerted by an electric charge at the origin on a, , charged particle at a point 共x, y, z兲 with position vector, r 苷 具x, y, z典 is F共r兲 苷 Kr兾 r 3 where K is a constant. (See, Example 5 in Section 16.1.) Find the work done as the particle, moves along a straight line from 共2, 0, 0兲 to 共2, 1, 5兲., , ⱍ ⱍ, , 43. The position of an object with mass m at time t is, , r共t兲 苷 at 2 i ⫹ bt 3 j, 0 艋 t 艋 1., (a) What is the force acting on the object at time t ?, (b) What is the work done by the force during the time interval, 0 艋 t 艋 1?, 44. An object with mass m moves with position function, , r共t兲 苷 a sin t i ⫹ b cos t j ⫹ ct k, 0 艋 t 艋 ␲兾2. Find the work, done on the object during this time period., , C, , 1, 0, , 1, , x, (meters), , 52. Experiments show that a steady current I in a long wire pro-, , duces a magnetic field B that is tangent to any circle that lies in, the plane perpendicular to the wire and whose center is the axis, of the wire (as in the figure). Ampère’s Law relates the electric, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 16.3, , THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS, , current to its magnetic effects and states that, , y, , C, , 1099, , I, , B ⴢ dr 苷 ␮ 0 I, , where I is the net current that passes through any surface, bounded by a closed curve C, and ␮ 0 is a constant called the, permeability of free space. By taking C to be a circle with, radius r, show that the magnitude B 苷 B of the magnetic, field at a distance r from the center of the wire is, , ⱍ ⱍ, , B, , ␮0 I, B苷, 2␲ r, , The Fundamental Theorem for Line Integrals, , 16.3, , Recall from Section 4.3 that Part 2 of the Fundamental Theorem of Calculus can be written as, , y, , 1, , b, , a, , F⬘共x兲 dx 苷 F共b兲 ⫺ F共a兲, , where F⬘ is continuous on 关a, b兴. We also called Equation 1 the Net Change Theorem: The, integral of a rate of change is the net change., If we think of the gradient vector ∇ f of a function f of two or three variables as a sort, of derivative of f , then the following theorem can be regarded as a version of the Fundamental Theorem for line integrals., 2 Theorem Let C be a smooth curve given by the vector function r共t兲, a 艋 t 艋 b., Let f be a differentiable function of two or three variables whose gradient vector, ∇ f is continuous on C. Then, , y, , B(x™, y™), , A(x¡, y¡), , y, , C, , 0, , C, , x, , NOTE Theorem 2 says that we can evaluate the line integral of a conservative vector, field (the gradient vector field of the potential function f ) simply by knowing the value of, f at the endpoints of C. In fact, Theorem 2 says that the line integral of ∇f is the net, change in f. If f is a function of two variables and C is a plane curve with initial point, A共x 1, y1 兲 and terminal point B共x 2 , y2 兲, as in Figure 1, then Theorem 2 becomes, , (a), z, , C, A(x¡, y¡, z¡), , y, , C, , B(x™, y™, z™), , 0, , ⵜf ⴢ dr 苷 f 共x 2 , y2 兲 ⫺ f 共x 1, y1 兲, , If f is a function of three variables and C is a space curve joining the point A共x 1, y1, z1 兲, to the point B共x 2 , y2 , z2 兲, then we have, , y, , x, , y, (b), FIGURE 1, , ⵜf ⴢ dr 苷 f 共r共b兲兲 ⫺ f 共r共a兲兲, , C, , ⵜ f ⴢ dr 苷 f 共x 2 , y2 , z2 兲 ⫺ f 共x 1, y1, z1 兲, , Let’s prove Theorem 2 for this case., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1098-1107.qk_97817_16_ch16_p1098-1107 11/11/10 9:40 AM Page 1101, , SECTION 16.3, , THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS, , 1101, , A curve is called closed if its terminal point coincides with its initial point, that is,, r共b兲 苷 r共a兲. (See Figure 2.) If xC F ⴢ dr is independent of path in D and C is any closed, path in D, we can choose any two points A and B on C and regard C as being composed, of the path C1 from A to B followed by the path C2 from B to A. (See Figure 3.) Then, C, , y, , C, , FIGURE 2, , A closed curve, , F ⴢ dr 苷 y F ⴢ dr ⫹, C1, , y, , C2, , F ⴢ dr 苷 y F ⴢ dr ⫺ y, , ⫺C2, , C1, , F ⴢ dr 苷 0, , since C1 and ⫺C2 have the same initial and terminal points., Conversely, if it is true that xC F ⴢ dr 苷 0 whenever C is a closed path in D, then we, demonstrate independence of path as follows. Take any two paths C1 and C2 from A to B, in D and define C to be the curve consisting of C1 followed by ⫺C2. Then, , C™, B, , 0 苷 y F ⴢ dr 苷 y F ⴢ dr ⫹ y, , A, , C, , C¡, , F ⴢ dr 苷, , ⫺C2, , C1, , y, , C1, , F ⴢ dr ⫺ y F ⴢ dr, C2, , and so xC F ⴢ dr 苷 xC F ⴢ dr. Thus we have proved the following theorem., , FIGURE 3, , 1, , 2, , 3 Theorem xC F ⴢ dr is independent of path in D if and only if xC F ⴢ dr 苷 0 for, every closed path C in D., , Since we know that the line integral of any conservative vector field F is independent, of path, it follows that xC F ⴢ dr 苷 0 for any closed path. The physical interpretation is that, the work done by a conservative force field (such as the gravitational or electric field in, Section 16.1) as it moves an object around a closed path is 0., The following theorem says that the only vector fields that are independent of path are, conservative. It is stated and proved for plane curves, but there is a similar version for, space curves. We assume that D is open, which means that for every point P in D there is, a disk with center P that lies entirely in D. (So D doesn’t contain any of its boundary, points.) In addition, we assume that D is connected: This means that any two points in D, can be joined by a path that lies in D., 4 Theorem Suppose F is a vector field that is continuous on an open connected, region D. If xC F ⴢ dr is independent of path in D, then F is a conservative vector, field on D ; that is, there exists a function f such that ∇ f 苷 F., PROOF Let A共a, b兲 be a fixed point in D. We construct the desired potential function f by, , defining, f 共x, y兲 苷 y, , 共x, y兲, , 共a, b兲, , for any point 共x, y兲 in D. Since xC F ⴢ dr is independent of path, it does not matter, which path C from 共a, b兲 to 共x, y兲 is used to evaluate f 共x, y兲. Since D is open, there exists, a disk contained in D with center 共x, y兲. Choose any point 共x 1, y兲 in the disk with x 1 ⬍ x, and let C consist of any path C1 from 共a, b兲 to 共x 1, y兲 followed by the horizontal line segment C2 from 共x 1, y兲 to 共x, y兲. (See Figure 4.) Then, , y, (x¡, y), , C¡, , C™, (x, y), , f 共x, y兲 苷, , D, (a, b), 0, , FIGURE 4, , F ⴢ dr, , x, , y, , C1, , F ⴢ dr ⫹ y F ⴢ dr 苷 y, C2, , 共x1, y兲, , 共a, b兲, , F ⴢ dr ⫹ y F ⴢ dr, C2, , Notice that the first of these integrals does not depend on x, so, ⭸, ⭸, f 共x, y兲 苷 0 ⫹, ⭸x, ⭸x, , y, , C2, , F ⴢ dr, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1098-1107.qk_97817_16_ch16_p1098-1107 11/9/10 9:06 AM Page 1102, , 1102, , CHAPTER 16, , VECTOR CALCULUS, , If we write F 苷 P i ⫹ Q j, then, , y, , C2, , F ⴢ dr 苷 y P dx ⫹ Q dy, C2, , On C2 , y is constant, so dy 苷 0. Using t as the parameter, where x 1 艋 t 艋 x, we have, ⭸, ⭸, f 共x, y兲 苷, ⭸x, ⭸x, , P dx ⫹ Q dy 苷, , y, , C2, , ⭸, ⭸x, , y, , x, , x1, , P共t, y兲 dt 苷 P共x, y兲, , y, (x, y), , by Part 1 of the Fundamental Theorem of Calculus (see Section 4.3). A similar argument,, using a vertical line segment (see Figure 5), shows that, , C™, C¡, , (x, y¡), , ⭸, ⭸, f 共x, y兲 苷, ⭸y, ⭸y, , D, (a, b), x, , 0, , Thus, , y, , C2, , P dx ⫹ Q dy 苷, , F苷Pi⫹Qj苷, , ⭸, ⭸y, , y, , y, , y1, , Q共x, t兲 dt 苷 Q共x, y兲, , ⭸f, ⭸f, i⫹, j 苷 ∇f, ⭸x, ⭸y, , which says that F is conservative., , FIGURE 5, , The question remains: How is it possible to determine whether or not a vector field, F is conservative? Suppose it is known that F 苷 P i ⫹ Q j is conservative, where P and, Q have continuous first-order partial derivatives. Then there is a function f such that, F 苷 ∇f , that is,, ⭸f, ⭸f, P苷, and, Q苷, ⭸x, ⭸y, simple,, not closed, , not simple,, not closed, , Therefore, by Clairaut’s Theorem,, ⭸P, ⭸2 f, ⭸2 f, ⭸Q, 苷, 苷, 苷, ⭸y, ⭸y ⭸x, ⭸x ⭸y, ⭸x, , simple,, closed, , not simple,, closed, , FIGURE 6, , Types of curves, , simply-connected region, , regions that are not simply-connected, FIGURE 7, , 5 Theorem If F共x, y兲 苷 P共x, y兲 i ⫹ Q共x, y兲 j is a conservative vector field,, where P and Q have continuous first-order partial derivatives on a domain D, then, throughout D we have, , ⭸P, ⭸Q, 苷, ⭸y, ⭸x, The converse of Theorem 5 is true only for a special type of region. To explain this, we, first need the concept of a simple curve, which is a curve that doesn’t intersect itself anywhere between its endpoints. [See Figure 6; r共a兲 苷 r共b兲 for a simple closed curve, but, r共t1 兲 苷 r共t2 兲 when a ⬍ t1 ⬍ t2 ⬍ b.], In Theorem 4 we needed an open connected region. For the next theorem we need a, stronger condition. A simply-connected region in the plane is a connected region D such, that every simple closed curve in D encloses only points that are in D. Notice from Figure, 7 that, intuitively speaking, a simply-connected region contains no hole and can’t consist, of two separate pieces., In terms of simply-connected regions, we can now state a partial converse to Theorem 5, that gives a convenient method for verifying that a vector field on ⺢ 2 is conservative. The, proof will be sketched in the next section as a consequence of Green’s Theorem., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1098-1107.qk_97817_16_ch16_p1098-1107 11/9/10 9:06 AM Page 1103, , SECTION 16.3, , THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS, , 1103, , 6 Theorem Let F 苷 P i ⫹ Q j be a vector field on an open simply-connected, region D. Suppose that P and Q have continuous first-order derivatives and, , ⭸P, ⭸Q, 苷, ⭸y, ⭸x, , throughout D, , Then F is conservative., , v, , 10, , EXAMPLE 2 Determine whether or not the vector field, , F共x, y兲 苷 共x ⫺ y兲 i ⫹ 共x ⫺ 2兲 j, _10, , 10, , is conservative., SOLUTION Let P共x, y兲 苷 x ⫺ y and Q共x, y兲 苷 x ⫺ 2. Then, , C, , ⭸P, 苷 ⫺1, ⭸y, , _10, , FIGURE 8, Figures 8 and 9 show the vector fields in, Examples 2 and 3, respectively. The vectors in, Figure 8 that start on the closed curve C all, appear to point in roughly the same direction as, C. So it looks as if xC F ⴢ dr ⬎ 0 and therefore, F is not conservative. The calculation in Example, 2 confirms this impression. Some of the vectors, near the curves C1 and C2 in Figure 9 point in, approximately the same direction as the curves,, whereas others point in the opposite direction., So it appears plausible that line integrals around, all closed paths are 0. Example 3 shows that F, is indeed conservative., , C™, , _2, , Since ⭸P兾⭸y 苷 ⭸Q兾⭸x, F is not conservative by Theorem 5., , v, , EXAMPLE 3 Determine whether or not the vector field, , F共x, y兲 苷 共3 ⫹ 2 xy兲 i ⫹ 共x 2 ⫺ 3y 2 兲 j, is conservative., SOLUTION Let P共x, y兲 苷 3 ⫹ 2xy and Q共x, y兲 苷 x 2 ⫺ 3y 2. Then, , ⭸P, ⭸Q, 苷 2x 苷, ⭸y, ⭸x, Also, the domain of F is the entire plane 共D 苷 ⺢ 2 兲, which is open and simplyconnected. Therefore we can apply Theorem 6 and conclude that F is conservative., , 2, , C¡, , ⭸Q, 苷1, ⭸x, , 2, , In Example 3, Theorem 6 told us that F is conservative, but it did not tell us how to find, the (potential) function f such that F 苷 ∇f . The proof of Theorem 4 gives us a clue as to, how to find f . We use “partial integration” as in the following example., EXAMPLE 4, , _2, , FIGURE 9, , (a) If F共x, y兲 苷 共3 ⫹ 2xy兲 i ⫹ 共x 2 ⫺ 3y 2 兲 j, find a function f such that F 苷 ∇f ., (b) Evaluate the line integral xC F ⴢ dr, where C is the curve given by, r共t兲 苷 e t sin t i ⫹ e t cos t j, , 0艋t艋␲, , SOLUTION, , (a) From Example 3 we know that F is conservative and so there exists a function f, with ∇ f 苷 F, that is,, 7, , fx 共x, y兲 苷 3 ⫹ 2xy, , 8, , fy 共x, y兲 苷 x 2 ⫺ 3y 2, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1108, , 16.4, , CHAPTER 16, , VECTOR CALCULUS, , Green’s Theorem, , y, , D, C, 0, , x, , Green’s Theorem gives the relationship between a line integral around a simple closed, curve C and a double integral over the plane region D bounded by C. (See Figure 1. We, assume that D consists of all points inside C as well as all points on C.) In stating Green’s, Theorem we use the convention that the positive orientation of a simple closed curve C, refers to a single counterclockwise traversal of C. Thus if C is given by the vector function r共t兲, a  t  b, then the region D is always on the left as the point r共t兲 traverses C., (See Figure 2.), y, , y, , FIGURE 1, , C, D, , D, C, 0, , FIGURE 2, , x, , 0, , (a) Positive orientation, , x, , (b) Negative orientation, , Green’s Theorem Let C be a positively oriented, piecewise-smooth, simple closed, , curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then, Recall that the left side of this equation, is another way of writing xC F ⴢ dr, where, F 苷 P i  Q j., , y, , C, , P dx  Q dy 苷, , yy, D, , 冉, , Q, P, , x, y, , 冊, , dA, , NOTE The notation, , y, , 䊊, C, , P dx  Q dy, , gC P dx  Q dy, , or, , is sometimes used to indicate that the line integral is calculated using the positive orientation of the closed curve C. Another notation for the positively oriented boundary curve of, D is D, so the equation in Green’s Theorem can be written as, , 1, , yy, D, , 冉, , Q, P, , x, y, , 冊, , dA 苷, , y, , D, , P dx  Q dy, , Green’s Theorem should be regarded as the counterpart of the Fundamental Theorem of, Calculus for double integrals. Compare Equation 1 with the statement of the Fundamental, Theorem of Calculus, Part 2, in the following equation:, , y, , b, , a, , F共x兲 dx 苷 F共b兲  F共a兲, , In both cases there is an integral involving derivatives (F, Q兾x, and P兾y) on the left, side of the equation. And in both cases the right side involves the values of the original, functions (F , Q, and P ) only on the boundary of the domain. (In the one-dimensional case,, the domain is an interval 关a, b兴 whose boundary consists of just two points, a and b.), , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:15 AM Page 1109, , SECTION 16.4, , GREEN’S THEOREM, , 1109, , Green’s Theorem is not easy to prove in general, but we can give a proof for the special case where the region is both type I and type II (see Section 15.3). Let’s call such, regions simple regions., PROOF OF GREEN’S THEOREM FOR THE CASE IN WHICH D IS A SIMPLE REGION Notice that, , George Green, Green’s Theorem is named after the selftaught English scientist George Green, (1793–1841). He worked full-time in his father’s, bakery from the age of nine and taught himself, mathematics from library books. In 1828 he, published privately An Essay on the Application, of Mathematical Analysis to the Theories of, Electricity and Magnetism, but only 100 copies, were printed and most of those went to his, friends. This pamphlet contained a theorem, that is equivalent to what we know as Green’s, Theorem, but it didn’t become widely known, at that time. Finally, at age 40, Green entered, Cambridge University as an undergraduate, but died four years after graduation. In 1846, William Thomson (Lord Kelvin) located a copy, of Green’s essay, realized its significance, and, had it reprinted. Green was the first person to, try to formulate a mathematical theory of electricity and magnetism. His work was the basis, for the subsequent electromagnetic theories of, Thomson, Stokes, Rayleigh, and Maxwell., , Green’s Theorem will be proved if we can show that, , y, , 2, , C, , D, , y, , 3, , C, , Q dy 苷 yy, D, , Q, dA, x, , We prove Equation 2 by expressing D as a type I region:, , ⱍ, , D 苷 兵共x, y兲 a  x  b, t1共x兲  y  t 2共x兲其, where t1 and t 2 are continuous functions. This enables us to compute the double integral, on the right side of Equation 2 as follows:, , yy, , 4, , b t 共x兲 P, b, P, dA 苷 y y, 共x, y兲 dy dx 苷 y 关P共x, t 2共x兲兲  P共x, t1共x兲兲兴 dx, a t 共x兲 y, a, y, 2, , 1, , where the last step follows from the Fundamental Theorem of Calculus., Now we compute the left side of Equation 2 by breaking up C as the union of the four, curves C1 , C2 , C3 , and C4 shown in Figure 3. On C1 we take x as the parameter and write, the parametric equations as x 苷 x, y 苷 t1共x兲, a  x  b. Thus, , y=g™(x), C£, C¢, , P, dA, y, , and, , D, , y, , P dx 苷 yy, , D, , C™, , y, , C¡, , C1, , P共x, y兲 dx 苷, , y, , b, , a, , P共x, t1共x兲兲 dx, , y=g¡(x), 0, , FIGURE 3, , a, , b, , x, , Observe that C3 goes from right to left but C3 goes from left to right, so we can write, the parametric equations of C3 as x 苷 x, y 苷 t 2共x兲, a  x  b. Therefore, , y, , C3, , P共x, y兲 dx 苷 y, , b, , C3, , P共x, y兲 dx 苷 y P共x, t 2共x兲兲 dx, a, , On C2 or C4 (either of which might reduce to just a single point), x is constant, so dx 苷 0, and, , y, , C2, , P共x, y兲 dx 苷 0 苷 y P共x, y兲 dx, C4, , Hence, , y, , C, , P共x, y兲 dx 苷 y P共x, y兲 dx  y P共x, y兲 dx  y P共x, y兲 dx  y P共x, y兲 dx, C1, , C2, , C3, , b, , b, , a, , a, , C4, , 苷 y P共x, t1共x兲兲 dx  y P共x, t 2共x兲兲 dx, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:15 AM Page 1110, , 1110, , CHAPTER 16, , VECTOR CALCULUS, , Comparing this expression with the one in Equation 4, we see that, P共x, y兲 dx 苷 yy, , y, , C, , D, , P, dA, y, , Equation 3 can be proved in much the same way by expressing D as a type II region (see, Exercise 30). Then, by adding Equations 2 and 3, we obtain Green’s Theorem., EXAMPLE 1 Evaluate xC x 4 dx  xy dy, where C is the triangular curve consisting of the, , line segments from 共0, 0兲 to 共1, 0兲, from 共1, 0兲 to 共0, 1兲, and from 共0, 1兲 to 共0, 0兲., y, , SOLUTION Although the given line integral could be evaluated as usual by the methods of, , Section 16.2, that would involve setting up three separate integrals along the three sides, of the triangle, so let’s use Green’s Theorem instead. Notice that the region D enclosed by, C is simple and C has positive orientation (see Figure 4). If we let P共x, y兲 苷 x 4 and, Q共x, y兲 苷 xy, then we have, , y=1-x, , (0, 1), , C, D, (0, 0), , x, , (1, 0), , y, , C, , x 4 dx  xy dy 苷, , yy, D, , FIGURE 4, , 苷y, , 1, , 0, , 冉, , Q, P, , x, y, , [y], , 2 y苷1x, y苷0, , 1, 2, , 苷  16 共1  x兲3, , v, , ], , dA 苷 y, , 1, , 0, , y, , 1x, , 0, , 共y  0兲 dy dx, , 1, , dx 苷 2 y 共1  x兲2 dx, , 1, 0, , 1, , 0, , 苷 16, , (, , ), , xC 共3y  e sin x 兲 dx  7x  sy 4  1 dy, where C is the circle, EXAMPLE 2 Evaluate 䊊, , x  y 2 苷 9., 2, , 冊, , SOLUTION The region D bounded by C is the disk x 2  y 2  9, so let’s change to polar, , coordinates after applying Green’s Theorem:, , y, , 䊊, C, , 共3y  e sin x 兲 dx  (7x  sy 4  1 ) dy, , Instead of using polar coordinates, we could, simply use the fact that D is a disk of radius 3, and write, , yy 4 dA 苷 4 ⴢ  共3兲, , 2, , 苷, , yy, D, , 苷 36, , 苷, , D, , 2, , 冋, , y y, 0, , 册, , , , (7x  sy 4  1 )  y, 共3y  e sin x兲 dA, x, 3, , 0, , 2, , 共7  3兲 r dr d 苷 4 y d, 0, , y, , 3, , 0, , r dr 苷 36, , In Examples 1 and 2 we found that the double integral was easier to evaluate than the, line integral. (Try setting up the line integral in Example 2 and you’ll soon be convinced!), But sometimes it’s easier to evaluate the line integral, and Green’s Theorem is used in the, reverse direction. For instance, if it is known that P共x, y兲 苷 Q共x, y兲 苷 0 on the curve C,, then Green’s Theorem gives, , yy, D, , 冉, , Q, P, , x, y, , 冊, , dA 苷 y P dx  Q dy 苷 0, C, , no matter what values P and Q assume in the region D., Another application of the reverse direction of Green’s Theorem is in computing areas., Since the area of D is xxD 1 dA, we wish to choose P and Q so that, Q, P, , 苷1, x, y, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:15 AM Page 1111, , SECTION 16.4, , GREEN’S THEOREM, , 1111, , There are several possibilities:, P共x, y兲 苷 0, , P共x, y兲 苷 y, , P共x, y兲 苷  12 y, , Q共x, y兲 苷 x, , Q共x, y兲 苷 0, , Q共x, y兲 苷 2 x, 1, , Then Green’s Theorem gives the following formulas for the area of D :, 5, , 䊊 x dy 苷 y, 䊊 y dx 苷 2 y, 䊊 x dy  y dx, A苷y, 1, , C, , C, , C, , EXAMPLE 3 Find the area enclosed by the ellipse, , x2, y2, , 苷 1., a2, b2, , SOLUTION The ellipse has parametric equations x 苷 a cos t and y 苷 b sin t, where, , 0  t  2. Using the third formula in Equation 5, we have, A 苷 12 y x dy  y dx, C, , 2, , 苷 12 y 共a cos t兲共b cos t兲 dt  共b sin t兲共a sin t兲 dt, 0, , ab, 苷, 2, Wheel, Pole arm, 0, , 10, , 5, , Pivot, , 7, , 8, , 4, , 3, , 9, , 6, , 0, 5, , 7, , 2, , Pole, , 4, , Tracer arm, , Tracer, FIGURE 5, , A Keuffel and Esser polar planimeter, , y, , 2, , 0, , dt 苷  ab, , Formula 5 can be used to explain how planimeters work. A planimeter is a mechanical instrument used for measuring the area of a region by tracing its boundary curve. These, devices are useful in all the sciences: in biology for measuring the area of leaves or wings,, in medicine for measuring the size of cross-sections of organs or tumors, in forestry for, estimating the size of forested regions from photographs., Figure 5 shows the operation of a polar planimeter: The pole is fixed and, as the tracer, is moved along the boundary curve of the region, the wheel partly slides and partly rolls, perpendicular to the tracer arm. The planimeter measures the distance that the wheel rolls, and this is proportional to the area of the enclosed region. The explanation as a consequence of Formula 5 can be found in the following articles:, ■, , R. W. Gatterman, “The planimeter as an example of Green’s Theorem” Amer. Math., Monthly, Vol. 88 (1981), pp. 701–4., , ■, , Tanya Leise, “As the planimeter wheel turns” College Math. Journal, Vol. 38, (2007), pp. 24 –31., , Extended Versions of Green’s Theorem, Although we have proved Green’s Theorem only for the case where D is simple, we can, now extend it to the case where D is a finite union of simple regions. For example, if D is, the region shown in Figure 6, then we can write D 苷 D1 傼 D2 , where D1 and D2 are both, simple. The boundary of D1 is C1 傼 C3 and the boundary of D2 is C2 傼 共C3兲 so, applying Green’s Theorem to D1 and D2 separately, we get, C¡, D¡, , D™, C£, , y, , C™, , C1傼C3, , _C£, , 冉, yy 冉, yy, D1, , y, FIGURE 6, , P dx  Q dy 苷, , C2傼共C3 兲, , P dx  Q dy 苷, , D2, , Q, P, , x, y, Q, P, , x, y, , 冊, 冊, , dA, , dA, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:15 AM Page 1112, , 1112, , CHAPTER 16, , VECTOR CALCULUS, , If we add these two equations, the line integrals along C3 and C3 cancel, so we get, , C, , y, , C1傼C2, , P dx  Q dy 苷, , 冉, , yy, D, , Q, P, , x, y, , 冊, , dA, , which is Green’s Theorem for D 苷 D1 傼 D2 , since its boundary is C 苷 C1 傼 C2 ., The same sort of argument allows us to establish Green’s Theorem for any finite union, of nonoverlapping simple regions (see Figure 7)., FIGURE 7, , v, , EXAMPLE 4 Evaluate 䊊, xC y 2 dx  3xy dy , where C is the boundary of the semiannular, , region D in the upper half-plane between the circles x 2  y 2 苷 1 and x 2  y 2 苷 4 ., , SOLUTION Notice that although D is not simple, the y -axis divides it into two simple, , y, , regions (see Figure 8). In polar coordinates we can write, , ≈+¥=4, , D 苷 兵共r,  兲 1  r  2, 0     其, , ⱍ, , C, , D, , Therefore Green’s Theorem gives, 0, , ≈+¥=1, , x, , y, , 䊊, C, , y 2 dx  3xy dy 苷, , yy, D, , FIGURE 8, , 冋, , 册, , , , 共3xy兲 , 共y 2 兲 dA, x, y, , 苷 yy y dA 苷, , , , y y, 0, , 2, , 1, , 共r sin  兲 r dr d, , D, , , , 苷 y sin  d, 0, , C™, D, C¡, , y, , 2, , 1, , [, ,  1, 0 3, , ][r], , r 2 dr 苷 cos , , 3 2, 1, , 苷, , 14, 3, , Green’s Theorem can be extended to apply to regions with holes, that is, regions that, are not simply-connected. Observe that the boundary C of the region D in Figure 9 consists of two simple closed curves C1 and C2 . We assume that these boundary curves are, oriented so that the region D is always on the left as the curve C is traversed. Thus the, positive direction is counterclockwise for the outer curve C1 but clockwise for the inner, curve C2 . If we divide D into two regions D and D by means of the lines shown in, Figure 10 and then apply Green’s Theorem to each of D and D , we get, , FIGURE 9, , yy, Dª, , D, , 冉, , Q, P, , x, y, , 冊, , dA 苷, , yy, D, , 苷y, , D, , Dªª, , 冉, , Q, P, , x, y, , 冊, , dA , , yy, D, , P dx  Q dy  y, , D, , 冉, , Q, P, , x, y, , 冊, , dA, , P dx  Q dy, , Since the line integrals along the common boundary lines are in opposite directions, they, cancel and we get, FIGURE 10, , yy, D, , 冉, , Q, P, , x, y, , 冊, , dA 苷, , y, , C1, , P dx  Q dy  y P dx  Q dy 苷 y P dx  Q dy, C2, , C, , which is Green’s Theorem for the region D., , v, , EXAMPLE 5 If F共x, y兲 苷 共y i  x j兲兾共x 2  y 2 兲, show that xC F ⴢ dr 苷 2 for every, , positively oriented simple closed path that encloses the origin., SOLUTION Since C is an arbitrary closed path that encloses the origin, it’s difficult to, , compute the given integral directly. So let’s consider a counterclockwise-oriented circle C, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:15 AM Page 1113, , GREEN’S THEOREM, , SECTION 16.4, , with center the origin and radius a, where a is chosen to be small enough that C lies, inside C. (See Figure 11.) Let D be the region bounded by C and C. Then its positively, oriented boundary is C 傼 共C兲 and so the general version of Green’s Theorem gives, , y, , C, Cª, D, , 1113, , x, , y, , C, , P dx  Q dy  y, , C, , P dx  Q dy 苷, , 冉, yy 冋, , yy, D, , 苷, , FIGURE 11, , D, , y, , Therefore, , C, , Q, P, , x, y, , 冊, , dA, , y2  x2, y2  x2, 2, 2 2 , 共x  y 兲, 共x 2  y 2 兲 2, , 册, , dA 苷 0, , P dx  Q dy 苷 y P dx  Q dy, C, , y, , that is,, , C, , F ⴢ dr 苷 y F ⴢ dr, C, , We now easily compute this last integral using the parametrization given by, r共t兲 苷 a cos t i  a sin t j , 0  t  2. Thus, , y, , C, , F ⴢ dr 苷 y F ⴢ dr 苷 y, C, , 苷y, , 2, , 0, , 2, , 0, , F共r共t兲兲 ⴢ r共t兲 dt, , 共a sin t兲共a sin t兲  共a cos t兲共a cos t兲, dt 苷, a 2 cos 2 t  a 2 sin 2 t, , y, , 2, , 0, , dt 苷 2, , We end this section by using Green’s Theorem to discuss a result that was stated in the, preceding section., SKETCH OF PROOF OF THEOREM 16.3.6 We’re assuming that F 苷 P i  Q j is a vector field, , on an open simply-connected region D, that P and Q have continuous first-order partial, derivatives, and that, P, Q, 苷, throughout D, y, x, If C is any simple closed path in D and R is the region that C encloses, then Green’s Theorem gives, , y, , 䊊, C, , 䊊 P dx  Q dy 苷, F ⴢ dr 苷 y, C, , yy, R, , 冉, , Q, P, , x, y, , 冊, , dA 苷, , yy 0 dA 苷 0, R, , A curve that is not simple crosses itself at one or more points and can be broken up, into a number of simple curves. We have shown that the line integrals of F around these, simple curves are all 0 and, adding these integrals, we see that xC F ⴢ dr 苷 0 for any, closed curve C. Therefore xC F ⴢ dr is independent of path in D by Theorem 16.3.3. It follows that F is a conservative vector field., , 16.4, , Exercises, , 1– 4 Evaluate the line integral by two methods: (a) directly and, (b) using Green’s Theorem., 1., , x 共x  y兲 dx  共x  y兲 dy,, C is the circle with center the origin and radius 2, , 䊊, C, , ;, , Graphing calculator or computer required, , xy dx  x 2 dy,, C is the rectangle with vertices 共0, 0兲, 共3, 0兲, 共3, 1兲, and 共0, 1兲, , x, , 2., , 䊊, C, , 3., , x y dx  x 2 y 3 dy,, C is the triangle with vertices 共0, 0兲, (1, 0), and (1, 2), , x, , 䊊, C, , CAS Computer algebra system required, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 16.5, , by proving Equation 3., 31. Use Green’s Theorem to prove the change of variables, , formula for a double integral (Formula 15.10.9) for the case, where f 共x, y兲 苷 1:, 共x, y兲, yy dx dy 苷 yy 共u, v兲 du dv, R, S, , 16.5, , 1115, , Here R is the region in the xy-plane that corresponds to the, region S in the uv-plane under the transformation given by, x 苷 t共u, v兲, y 苷 h共u, v兲., [Hint: Note that the left side is A共R兲 and apply the first, part of Equation 5. Convert the line integral over R to a, line integral over S and apply Green’s Theorem in the, uv-plane.], , 30. Complete the proof of the special case of Green’s Theorem, , 冟, , CURL AND DIVERGENCE, , 冟, , Curl and Divergence, In this section we define two operations that can be performed on vector fields and that, play a basic role in the applications of vector calculus to fluid flow and electricity and magnetism. Each operation resembles differentiation, but one produces a vector field whereas, the other produces a scalar field., , Curl, If F 苷 P i  Q j  R k is a vector field on ⺢ 3 and the partial derivatives of P, Q, and R, all exist, then the curl of F is the vector field on ⺢ 3 defined by, , 1, , curl F 苷, , 冉, , R, Q, , y, z, , 冊 冉, i, , P, R, , z, x, , 冊 冉, j, , Q, P, , x, y, , 冊, , k, , As an aid to our memory, let’s rewrite Equation 1 using operator notation. We introduce the vector differential operator ∇ (“del”) as, , , , j, k, x, y, z, , ∇ 苷i, , It has meaning when it operates on a scalar function to produce the gradient of f :, ∇f 苷 i, , f, f, f, f, f, f, j, k, 苷, i, j, k, x, y, z, x, y, z, , If we think of ∇ as a vector with components 兾x, 兾y, and 兾z, we can also consider, the formal cross product of ∇ with the vector field F as follows:, , ⱍ ⱍ, , i, , F苷, x, P, 苷, , 冉, , j, , y, Q, , R, Q, , y, z, , k, , z, R, , 冊 冉, i, , P, R, , z, x, , 冊 冉, j, , Q, P, , x, y, , 冊, , k, , 苷 curl F, So the easiest way to remember Definition 1 is by means of the symbolic expression, 2, , curl F 苷 ∇  F, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:16 AM Page 1116, , 1116, , CHAPTER 16, , VECTOR CALCULUS, , EXAMPLE 1 If F共x, y, z兲 苷 xz i  xyz j  y 2 k, find curl F., SOLUTION Using Equation 2, we have, , curl F 苷, , 苷, , CAS Most computer algebra systems have com-, , mands that compute the curl and divergence of, vector fields. If you have access to a CAS, use, these commands to check the answers to the, examples and exercises in this section., , ⱍ, , i, , F苷, x, xz, , 冋, , j, k, , , y z, xyz y 2, , ⱍ, , 册 冋, 册, , , , 共y 2 兲 , 共xyz兲 i , y, z, , , 冋, , 册, , , , 共y 2 兲 , 共xz兲 j, x, z, , , , 共xyz兲 , 共xz兲 k, x, y, , 苷 共2y  xy兲 i  共0  x兲 j  共yz  0兲 k, 苷 y共2  x兲 i  x j  yz k, Recall that the gradient of a function f of three variables is a vector field on ⺢ 3 and so, we can compute its curl. The following theorem says that the curl of a gradient vector field, is 0., , 3 Theorem If f is a function of three variables that has continuous second-order, partial derivatives, then, , curl共 f 兲 苷 0, , PROOF We have, , Notice the similarity to what we know, from Section 12.4: a  a 苷 0 for every, three-dimensional vector a., , curl共 f 兲 苷, , 苷, , ⱍ ⱍ, , i, ,  共 f 兲 苷 x, f, x, , 冉, , 2 f, 2 f, , y z, z y, , j, , y, f, y, , 冊 冉, i, , k, , z, f, z, , 2 f, 2 f, , z x, x z, , 冊 冉, j, , 2 f, 2 f, , x y, y x, , 冊, , k, , 苷0i0j0k苷0, by Clairaut’s Theorem., Since a conservative vector field is one for which F 苷 ∇f , Theorem 3 can be rephrased, as follows:, Compare this with Exercise 29 in, Section 16.3., , If F is conservative, then curl F 苷 0., This gives us a way of verifying that a vector field is not conservative., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:16 AM Page 1117, , SECTION 16.5, , v, , CURL AND DIVERGENCE, , 1117, , EXAMPLE 2 Show that the vector field F共x, y, z兲 苷 xz i  xyz j  y 2 k is not, , conservative., SOLUTION In Example 1 we showed that, , curl F 苷 y共2  x兲 i  x j  yz k, This shows that curl F 苷 0 and so, by Theorem 3, F is not conservative., The converse of Theorem 3 is not true in general, but the following theorem says the, converse is true if F is defined everywhere. (More generally it is true if the domain is, simply-connected, that is, “has no hole.”) Theorem 4 is the three-dimensional version, of Theorem 16.3.6. Its proof requires Stokes’ Theorem and is sketched at the end of, Section 16.8., 4, , Theorem If F is a vector field defined on all of ⺢ 3 whose component func-, , tions have continuous partial derivatives and curl F 苷 0, then F is a conservative, vector field., , v, , EXAMPLE 3, , (a) Show that, F共x, y, z兲 苷 y 2 z 3 i  2xyz 3 j  3xy 2 z 2 k, is a conservative vector field., (b) Find a function f such that F 苷, , f., , SOLUTION, , (a) We compute the curl of F :, , curl F 苷, , ⱍ, , i, j, k, , , , F苷, x, y, z, y 2 z 3 2xyz 3 3xy 2 z 2, , ⱍ, , 苷 共6xyz 2  6xyz 2 兲 i  共3y 2 z 2  3y 2 z 2 兲 j  共2yz 3  2yz 3 兲 k, 苷0, Since curl F 苷 0 and the domain of F is ⺢ 3, F is a conservative vector field by, Theorem 4., (b) The technique for finding f was given in Section 16.3. We have, 5, , fx 共x, y, z兲 苷 y 2 z 3, , 6, , fy 共x, y, z兲 苷 2xyz 3, , 7, , fz 共x, y, z兲 苷 3xy 2 z 2, , Integrating 5 with respect to x, we obtain, 8, , f 共x, y, z兲 苷 xy 2 z 3  t共 y, z兲, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:16 AM Page 1118, , 1118, , CHAPTER 16, , VECTOR CALCULUS, , Differentiating 8 with respect to y, we get fy 共x, y, z兲 苷 2xyz 3  ty 共y, z兲, so comparison, with 6 gives ty 共 y, z兲 苷 0. Thus t共y, z兲 苷 h共z兲 and, fz 共x, y, z兲 苷 3xy 2 z 2  h共z兲, Then 7 gives h共z兲 苷 0. Therefore, f 共x, y, z兲 苷 xy 2 z 3  K, , curl F(x, y, z), , (x, y, z), , FIGURE 1, , The reason for the name curl is that the curl vector is associated with rotations. One, connection is explained in Exercise 37. Another occurs when F represents the velocity, field in fluid flow (see Example 3 in Section 16.1). Particles near (x, y, z) in the fluid tend, to rotate about the axis that points in the direction of curl F共x, y, z兲, and the length of this, curl vector is a measure of how quickly the particles move around the axis (see Figure 1)., If curl F 苷 0 at a point P, then the fluid is free from rotations at P and F is called irrotational at P. In other words, there is no whirlpool or eddy at P. If curl F 苷 0, then a, tiny paddle wheel moves with the fluid but doesn’t rotate about its axis. If curl F 苷 0, the, paddle wheel rotates about its axis. We give a more detailed explanation in Section 16.8 as, a consequence of Stokes’ Theorem., , Divergence, If F 苷 P i  Q j  R k is a vector field on ⺢ 3 and P兾x, Q兾y, and R兾z exist, then, the divergence of F is the function of three variables defined by, , div F 苷, , 9, , P, Q, R, , , x, y, z, , Observe that curl F is a vector field but div F is a scalar field. In terms of the gradient operator  苷 共兾x兲 i  共兾y兲 j  共兾z兲 k, the divergence of F can be written symbolically, as the dot product of  and F :, div F 苷  ⴢ F, , 10, , EXAMPLE 4 If F共x, y, z兲 苷 xz i  xyz j  y 2 k, find div F., SOLUTION By the definition of divergence (Equation 9 or 10) we have, , div F 苷  ⴢ F 苷, , , , , 共xz兲 , 共xyz兲 , 共y 2 兲 苷 z  xz, x, y, z, , If F is a vector field on ⺢ 3, then curl F is also a vector field on ⺢ 3. As such, we can, compute its divergence. The next theorem shows that the result is 0., 11 Theorem If F 苷 P i  Q j  R k is a vector field on ⺢ 3 and P, Q, and R have, , continuous second-order partial derivatives, then, div curl F 苷 0, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:16 AM Page 1119, , CURL AND DIVERGENCE, , SECTION 16.5, , 1119, , PROOF Using the definitions of divergence and curl, we have, , div curl F 苷  ⴢ 共  F兲, , Note the analogy with the scalar triple, product: a ⴢ 共a  b兲 苷 0., , 冉, , R, Q, , y, z, , 冊 冉, , , x, , 苷, , 2R, 2Q, 2P, 2R, 2Q, 2P, , , , , , x y, x z, y z, y x, z x, z y, , , , , y, , P, R, , z, x, , 冊 冉, , 苷, , , , , z, , Q, P, , x, y, , 冊, , 苷0, because the terms cancel in pairs by Clairaut’s Theorem., , v, , EXAMPLE 5 Show that the vector field F共x, y, z兲 苷 xz i  xyz j  y 2 k can’t be, , written as the curl of another vector field, that is, F 苷 curl G., SOLUTION In Example 4 we showed that, , div F 苷 z  xz, and therefore div F 苷 0. If it were true that F 苷 curl G, then Theorem 11 would give, div F 苷 div curl G 苷 0, which contradicts div F 苷 0. Therefore F is not the curl of another vector field., The reason for this interpretation of div F will, be explained at the end of Section 16.9 as a, consequence of the Divergence Theorem., , Again, the reason for the name divergence can be understood in the context of fluid, flow. If F共x, y, z兲 is the velocity of a fluid (or gas), then div F共x, y, z兲 represents the net rate, of change (with respect to time) of the mass of fluid (or gas) flowing from the point 共x, y, z兲, per unit volume. In other words, div F共x, y, z兲 measures the tendency of the fluid to diverge, from the point 共x, y, z兲. If div F 苷 0, then F is said to be incompressible., Another differential operator occurs when we compute the divergence of a gradient, vector field f . If f is a function of three variables, we have, div共f 兲 苷  ⴢ 共f 兲 苷, , 2 f, 2 f, 2 f,  2  2, 2, x, y, z, , and this expression occurs so often that we abbreviate it as  2 f . The operator, 2 苷  ⴢ , is called the Laplace operator because of its relation to Laplace’s equation, 2 f 苷, , 2 f, 2 f, 2 f, , , 苷0, x 2, y 2, z 2, , We can also apply the Laplace operator  2 to a vector field, F苷PiQjRk, in terms of its components:,  2 F 苷  2P i   2Q j   2R k, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:16 AM Page 1122, , 1122, , CHAPTER 16, , VECTOR CALCULUS, , 23–29 Prove the identity, assuming that the appropriate partial, derivatives exist and are continuous. If f is a scalar field and F, G, are vector fields, then f F, F ⴢ G, and F  G are defined by, , 共 f F兲共x, y, z兲 苷 f 共x, y, z兲 F共x, y, z兲, 共F ⴢ G兲共x, y, z兲 苷 F共x, y, z兲 ⴢ G共x, y, z兲, 共F  G兲共x, y, z兲 苷 F共x, y, z兲  G共x, y, z兲, 23. div共F  G兲 苷 div F  div G, , Exercise 33) to show that if t is harmonic on D, then, , x䊊C Dn t ds 苷 0. Here Dn t is the normal derivative of t defined, in Exercise 33., 36. Use Green’s first identity to show that if f is harmonic, , on D, and if f 共x, y兲 苷 0 on the boundary curve C, then, xx D f 2 dA 苷 0. (Assume the same hypotheses as in, Exercise 33.), , ⱍ ⱍ, , 37. This exercise demonstrates a connection between the curl, , vector and rotations. Let B be a rigid body rotating about the, z-axis. The rotation can be described by the vector w 苷 k,, where is the angular speed of B, that is, the tangential speed, of any point P in B divided by the distance d from the axis of, rotation. Let r 苷 具x, y, z 典 be the position vector of P., (a) By considering the angle in the figure, show that the, velocity field of B is given by v 苷 w  r., (b) Show that v 苷  y i  x j., (c) Show that curl v 苷 2w., , 24. curl共F  G兲 苷 curl F  curl G, 25. div共 f F兲 苷 f div F  F ⴢ  f, 26. curl共 f F兲 苷 f curl F  共 f 兲  F, 27. div共F  G兲 苷 G ⴢ curl F  F ⴢ curl G, 28. div共 f   t兲 苷 0, 29. curl共curl F兲 苷 grad共div F兲   2 F, , z, , ⱍ ⱍ, , 30–32 Let r 苷 x i  y j  z k and r 苷 r ., , w, , 30. Verify each identity., , (a)  ⴢ r 苷 3, (c)  2 r 3 苷 12r, , (b)  ⴢ 共r r兲 苷 4r, B, , 31. Verify each identity., , (a) r 苷 r兾r, (c) 共1兾r兲 苷 r兾r 3, , d, , (b)   r 苷 0, (d)  ln r 苷 r兾r 2, , v, P, , 32. If F 苷 r兾r p, find div F. Is there a value of p for which, , ¨, , div F 苷 0 ?, , 0, , 33. Use Green’s Theorem in the form of Equation 13 to prove, , y, , Green’s first identity:, , yy f  t dA 苷 y, 2, , 䊊, C, , D, , D, , where D and C satisfy the hypotheses of Green’s Theorem, and the appropriate partial derivatives of f and t exist and are, continuous. (The quantity t ⴢ n 苷 Dn t occurs in the line integral. This is the directional derivative in the direction of the, normal vector n and is called the normal derivative of t.), 34. Use Green’s first identity (Exercise 33) to prove Green’s, , second identity:, , yy 共 f  t  t f 兲 dA 苷 y, 2, , 2, , x, , f 共t兲 ⴢ n ds  yy  f ⴢ t dA, , 䊊, C, , 共 f t  t f 兲 ⴢ n ds, , 38. Maxwell’s equations relating the electric field E and magnetic, , field H as they vary with time in a region containing no charge, and no current can be stated as follows:, div E 苷 0, curl E 苷 , , div H 苷 0, 1 H, c t, , (a)   共  E兲 苷 , , 1 2 E, c 2 t 2, , (b)   共  H兲 苷 , , 1 2 H, c 2 t 2, , (c)  2 E 苷, , 1 2 E, c 2 t 2, , (d)  2 H 苷, , 1 2 H, c 2 t 2, , 35. Recall from Section 14.3 that a function t is called harmonic, , on D if it satisfies Laplace’s equation, that is,  2t 苷 0 on D., Use Green’s first identity (with the same hypotheses as in, , 1 E, c t, , where c is the speed of light. Use these equations to prove the, following:, , D, , where D and C satisfy the hypotheses of Green’s Theorem, and the appropriate partial derivatives of f and t exist and are, continuous., , curl H 苷, , [Hint: Use Exercise 29.], , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SECTION 16.6, , 1123, , form f 苷 div G must satisfy? Show that the answer to, this question is “No” by proving that every continuous, function f on ⺢ 3 is the divergence of some vector field., [Hint: Let G共x, y, z兲 苷 具t共x, y, z兲, 0, 0典 ,where, t共x, y, z兲 苷 x0x f 共t, y, z兲 dt.], , 39. We have seen that all vector fields of the form F 苷 t, , satisfy the equation curl F 苷 0 and that all vector fields of the, form F 苷 curl G satisfy the equation div F 苷 0 (assuming, continuity of the appropriate partial derivatives). This suggests, the question: Are there any equations that all functions of the, , 16.6, , PARAMETRIC SURFACES AND THEIR AREAS, , Parametric Surfaces and Their Areas, So far we have considered special types of surfaces: cylinders, quadric surfaces, graphs of, functions of two variables, and level surfaces of functions of three variables. Here we use, vector functions to describe more general surfaces, called parametric surfaces, and compute their areas. Then we take the general surface area formula and see how it applies to, special surfaces., , Parametric Surfaces, In much the same way that we describe a space curve by a vector function r共t兲 of a single, parameter t, we can describe a surface by a vector function r共u, v兲 of two parameters u, and v. We suppose that, r共u, v兲 苷 x共u, v兲 i  y共u, v兲 j  z共u, v兲 k, , 1, , is a vector-valued function defined on a region D in the uv-plane. So x, y, and z, the component functions of r, are functions of the two variables u and v with domain D. The set of, all points 共x, y, z兲 in ⺢ 3 such that, x 苷 x共u, v兲, , 2, , y 苷 y共u, v兲, , z 苷 z共u, v兲, , and 共u, v兲 varies throughout D, is called a parametric surface S and Equations 2 are called, parametric equations of S. Each choice of u and v gives a point on S; by making all, choices, we get all of S. In other words, the surface S is traced out by the tip of the position, vector r共u, v兲 as 共u, v兲 moves throughout the region D. (See Figure 1.), √, , z, , S, D, , r, (u, √), , r(u, √), 0, , u, , 0, , FIGURE 1, x, , A parametric surface, , y, , EXAMPLE 1 Identify and sketch the surface with vector equation, , r共u, v兲 苷 2 cos u i  v j  2 sin u k, SOLUTION The parametric equations for this surface are, , x 苷 2 cos u, , y苷v, , z 苷 2 sin u, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:16 AM Page 1124, , 1124, , CHAPTER 16, , VECTOR CALCULUS, , So for any point 共x, y, z兲 on the surface, we have, , z, (0, 0, 2), , x 2  z 2 苷 4 cos 2u  4 sin 2u 苷 4, This means that vertical cross-sections parallel to the xz-plane (that is, with y constant), are all circles with radius 2. Since y 苷 v and no restriction is placed on v, the surface is a, circular cylinder with radius 2 whose axis is the y-axis (see Figure 2)., , 0, x, , y, , In Example 1 we placed no restrictions on the parameters u and v and so we obtained the, entire cylinder. If, for instance, we restrict u and v by writing the parameter domain as, , (2, 0, 0), , 0  u  兾2, , FIGURE 2, z, (0, 3, 2), 0, x, , y, , FIGURE 3, , 0v3, , then x 0, z 0, 0  y  3, and we get the quarter-cylinder with length 3 illustrated in, Figure 3., If a parametric surface S is given by a vector function r共u, v兲, then there are two useful, families of curves that lie on S, one family with u constant and the other with v constant., These families correspond to vertical and horizontal lines in the uv-plane. If we keep u constant by putting u 苷 u 0 , then r共u 0 , v兲 becomes a vector function of the single parameter v, and defines a curve C1 lying on S. (See Figure 4.), z, , √, , (u¸, √¸), √=√ ¸, , TEC Visual 16.6 shows animated versions, , of Figures 4 and 5, with moving grid curves, for, several parametric surfaces., , D, , r, , C¡, C™, , u=u¸, , 0, , 0, , u, , y, , x, , FIGURE 4, , Similarly, if we keep v constant by putting v 苷 v0 , we get a curve C2 given by r共u, v0 兲, that lies on S. We call these curves grid curves. (In Example 1, for instance, the grid curves, obtained by letting u be constant are horizontal lines whereas the grid curves with v constant, are circles.) In fact, when a computer graphs a parametric surface, it usually depicts the surface by plotting these grid curves, as we see in the following example., √ constant, , EXAMPLE 2 Use a computer algebra system to graph the surface, u constant, , r共u, v兲 苷 具共2  sin v兲 cos u, 共2  sin v兲 sin u, u  cos v 典, Which grid curves have u constant? Which have v constant?, SOLUTION We graph the portion of the surface with parameter domain 0  u  4 ,, , 0  v  2 in Figure 5. It has the appearance of a spiral tube. To identify the grid, curves, we write the corresponding parametric equations:, x 苷 共2  sin v兲 cos u, , x, , FIGURE 5, , y 苷 共2  sin v兲 sin u, , z 苷 u  cos v, , y, , If v is constant, then sin v and cos v are constant, so the parametric equations resemble, those of the helix in Example 4 in Section 13.1. Thus the grid curves with v constant are, the spiral curves in Figure 5. We deduce that the grid curves with u constant must be, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:16 AM Page 1125, , SECTION 16.6, , PARAMETRIC SURFACES AND THEIR AREAS, , 1125, , curves that look like circles in the figure. Further evidence for this assertion is that if u is, kept constant, u 苷 u 0 , then the equation z 苷 u 0  cos v shows that the z-values vary, from u 0  1 to u 0  1., In Examples 1 and 2 we were given a vector equation and asked to graph the corresponding parametric surface. In the following examples, however, we are given the more, challenging problem of finding a vector function to represent a given surface. In the rest of, this chapter we will often need to do exactly that., EXAMPLE 3 Find a vector function that represents the plane that passes through the point, P0 with position vector r0 and that contains two nonparallel vectors a and b., P, , distance in the direction of a and another distance in the direction of b. So there are, scalars u and v such that P, A, 0 P 苷 ua  v b. (Figure 6 illustrates how this works, by, means of the Parallelogram Law, for the case where u and v are positive. See also, Exercise 46 in Section 12.2.) If r is the position vector of P, then, , √b, b, a, , P¸, , SOLUTION If P is any point in the plane, we can get from P0 to P by moving a certain, , ua, , r 苷 OP, A0  P, A, 0 P 苷 r0  u a  vb, , FIGURE 6, , So the vector equation of the plane can be written as, r共u, v兲 苷 r0  ua  v b, where u and v are real numbers., If we write r 苷 具x, y, z 典 , r0 苷 具x0 , y0 , z0 典 , a 苷 具a1 , a2 , a3 典 , and b 苷 具b1 , b2 , b3 典 ,, then we can write the parametric equations of the plane through the point 共x0 , y0 , z0 兲 as, follows:, x 苷 x0  ua1  v b1, , ¨, 2π, , v, , y 苷 y0  ua2  v b2, , z 苷 z0  ua3  v b3, , EXAMPLE 4 Find a parametric representation of the sphere, , D, , x 2  y 2  z2 苷 a 2, , ˙=c, , 苷 a in spherical coordinates, so let’s, choose the angles  and in spherical coordinates as the parameters (see Section 15.9)., Then, putting 苷 a in the equations for conversion from spherical to rectangular coordinates (Equations 15.9.1), we obtain, , SOLUTION The sphere has a simple representation, ¨=k, , k, 0, , c, , ˙, , π, , x 苷 a sin  cos, , r, , z 苷 a cos , , as the parametric equations of the sphere. The corresponding vector equation is, r共, 兲 苷 a sin  cos, , z, , ˙=c, , 0, , y, x, ¨=k, , FIGURE 7, , y 苷 a sin  sin, , i  a sin  sin, , j  a cos  k, , We have 0    and 0   2 , so the parameter domain is the rectangle, D 苷 关0, 兴  关0, 2 兴. The grid curves with  constant are the circles of constant latitude (including the equator). The grid curves with constant are the meridians (semicircles), which connect the north and south poles (see Figure 7)., NOTE We saw in Example 4 that the grid curves for a sphere are curves of constant latitude and longitude. For a general parametric surface we are really making a map and the, grid curves are similar to lines of latitude and longitude. Describing a point on a parametric surface (like the one in Figure 5) by giving specific values of u and v is like giving, the latitude and longitude of a point., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:17 AM Page 1126, , 1126, , CHAPTER 16, , VECTOR CALCULUS, , One of the uses of parametric surfaces is in, computer graphics. Figure 8 shows the result of, trying to graph the sphere x 2  y 2  z 2 苷 1, by solving the equation for z and graphing the, top and bottom hemispheres separately. Part, of the sphere appears to be missing because, of the rectangular grid system used by the, computer. The much better picture in Figure 9, was produced by a computer using the, parametric equations found in Example 4., , FIGURE 8, , FIGURE 9, , EXAMPLE 5 Find a parametric representation for the cylinder, , x2  y2 苷 4, , 0z1, , SOLUTION The cylinder has a simple representation r 苷 2 in cylindrical coordinates, so, , we choose as parameters, tions of the cylinder are, , and z in cylindrical coordinates. Then the parametric equax 苷 2 cos, , y 苷 2 sin, , z苷z, , where 0   2 and 0  z  1., , v, , EXAMPLE 6 Find a vector function that represents the elliptic paraboloid z 苷 x 2  2y 2., , SOLUTION If we regard x and y as parameters, then the parametric equations are simply, , x苷x, , y苷y, , z 苷 x 2  2y 2, , and the vector equation is, r共x, y兲 苷 x i  y j  共x 2  2y 2 兲 k, TEC In Module 16.6 you can investigate, several families of parametric surfaces., , In general, a surface given as the graph of a function of x and y, that is, with an equation, of the form z 苷 f 共x, y兲, can always be regarded as a parametric surface by taking x and y, as parameters and writing the parametric equations as, x苷x, , y苷y, , z 苷 f 共x, y兲, , Parametric representations (also called parametrizations) of surfaces are not unique. The, next example shows two ways to parametrize a cone., EXAMPLE 7 Find a parametric representation for the surface z 苷 2sx 2  y 2 , that is, the, , top half of the cone z 2 苷 4x 2  4y 2., , SOLUTION 1 One possible representation is obtained by choosing x and y as parameters:, , x苷x, , y苷y, , z 苷 2sx 2  y 2, , So the vector equation is, r共x, y兲 苷 x i  y j  2sx 2  y 2 k, SOLUTION 2 Another representation results from choosing as parameters the polar, , coordinates r and . A point 共x, y, z兲 on the cone satisfies x 苷 r cos , y 苷 r sin , and, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:17 AM Page 1127, , SECTION 16.6, For some purposes the parametric representations in Solutions 1 and 2 are equally good,, but Solution 2 might be preferable in certain, situations. If we are interested only in the part, of the cone that lies below the plane z 苷 1,, for instance, all we have to do in Solution 2 is, change the parameter domain to, 0  r  12, , 0, , r共r, 兲 苷 r cos i  r sin j  2r k, where r, , 0 and 0   2 ., , Surfaces of Revolution, , 2, , Surfaces of revolution can be represented parametrically and thus graphed using a computer. For instance, let’s consider the surface S obtained by rotating the curve y 苷 f 共x兲,, a  x  b, about the x-axis, where f 共x兲 0. Let be the angle of rotation as shown in Figure 10. If 共x, y, z兲 is a point on S, then, , 0, , x苷x, , 3, y, , x, x, , ¨, , y 苷 f 共x兲 cos, , z 苷 f 共x兲 sin, , Therefore we take x and as parameters and regard Equations 3 as parametric equations of, S. The parameter domain is given by a  x  b, 0   2 ., , y=ƒ, z, , 1127, , z 苷 2sx 2  y 2 苷 2r. So a vector equation for the cone is, , z, , ƒ, , PARAMETRIC SURFACES AND THEIR AREAS, , (x, y, z), , EXAMPLE 8 Find parametric equations for the surface generated by rotating the curve, y 苷 sin x, 0  x  2 , about the x-axis. Use these equations to graph the surface of revolution., , ƒ, , SOLUTION From Equations 3, the parametric equations are, , x苷x, , FIGURE 10, z, , y 苷 sin x cos, , z 苷 sin x sin, , and the parameter domain is 0  x  2 , 0   2 . Using a computer to plot these, equations and rotate the image, we obtain the graph in Figure 11., , y, , x, , We can adapt Equations 3 to represent a surface obtained through revolution about the, y- or z-axis (see Exercise 30)., , Tangent Planes, , FIGURE 11, , We now find the tangent plane to a parametric surface S traced out by a vector function, r共u, v兲 苷 x共u, v兲 i  y共u, v兲 j  z共u, v兲 k, at a point P0 with position vector r共u0 , v0 兲. If we keep u constant by putting u 苷 u0 , then, r共u0 , v兲 becomes a vector function of the single parameter v and defines a grid curve C1, lying on S. (See Figure 12.) The tangent vector to C1 at P0 is obtained by taking the partial, derivative of r with respect to v :, rv 苷, , 4, , x, y, z, 共u0 , v0 兲 i , 共u0 , v0 兲 j , 共u0 , v0 兲 k, v, v, v, z, , √, , P¸, (u ¸, √¸), √=√¸, D, 0, , FIGURE 12, , ru, , r√, C¡, , r, , u=u ¸, , 0, , u, x, , C™, y, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1128-1129.qk_97817_16_ch16_p1128-1129 11/9/10 9:22 AM Page 1128, , 1128, , CHAPTER 16, , VECTOR CALCULUS, , Similarly, if we keep v constant by putting v 苷 v0 , we get a grid curve C2 given by, r共u, v0 兲 that lies on S, and its tangent vector at P0 is, ru 苷, , 5, , x, y, z, 共u0 , v0 兲 i , 共u0 , v0 兲 j , 共u0 , v0 兲 k, u, u, u, , If ru  rv is not 0, then the surface S is called smooth (it has no “corners”). For a smooth, surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the, vector ru  rv is a normal vector to the tangent plane., Figure 13 shows the self-intersecting, surface in Example 9 and its tangent plane, at 共1, 1, 3兲., , EXAMPLE 9 Find the tangent plane to the surface with parametric equations x 苷 u 2,, y 苷 v 2, z 苷 u  2v at the point 共1, 1, 3兲., , v, , SOLUTION We first compute the tangent vectors:, , z, , (1, 1, 3), , ru 苷, , x, y, z, i, j, k 苷 2u i  k, u, u, u, , rv 苷, , x, y, z, i, j, k 苷 2v j  2 k, v, v, v, , y, x, , Thus a normal vector to the tangent plane is, , ⱍ ⱍ, , i, ru  rv 苷 2u, 0, , FIGURE 13, , j, 0, 2v, , k, 1 苷 2v i  4u j  4uv k, 2, , Notice that the point 共1, 1, 3兲 corresponds to the parameter values u 苷 1 and v 苷 1, so, the normal vector there is, 2 i  4 j  4 k, Therefore an equation of the tangent plane at 共1, 1, 3兲 is, 2共x  1兲  4共y  1兲  4共z  3兲 苷 0, x  2y  2z  3 苷 0, , or, , Surface Area, Now we define the surface area of a general parametric surface given by Equation 1. For, simplicity we start by considering a surface whose parameter domain D is a rectangle, and, we divide it into subrectangles Rij . Let’s choose 共u i*, vj*兲 to be the lower left corner of Rij., (See Figure 14.), √, , z, , R ij, , r, , Î√, , Pij, , Sij, , Îu, , (u *i , √ *j ), , FIGURE 14, , The image of the, subrectangle Rij is the patch Sij ., , 0, , 0, , u, x, , y, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1128-1129.qk_97817_16_ch16_p1128-1129 11/9/10 9:22 AM Page 1129, , PARAMETRIC SURFACES AND THEIR AREAS, , SECTION 16.6, , 1129, , The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij, with position vector r共u i*, vj*兲 as one of its corners. Let, ru* 苷 ru共u i*, vj*兲, , Sij, , Pij, (a), , rv* 苷 rv共u i*, vj*兲, , and, , be the tangent vectors at Pij as given by Equations 5 and 4., Figure 15(a) shows how the two edges of the patch that meet at Pij can be approximated, by vectors. These vectors, in turn, can be approximated by the vectors u r*u and v r*v, because partial derivatives can be approximated by difference quotients. So we approximate Sij by the parallelogram determined by the vectors u ru* and v r*v . This parallelogram, is shown in Figure 15(b) and lies in the tangent plane to S at Pij . The area of this parallelogram is, 共u ru*兲  共v r*兲, 苷 ru*  r*v u v, v, , ⱍ, , ⱍ ⱍ, , ⱍ, , and so an approximation to the area of S is, m, , n, , 兺 兺 ⱍ r*  r* ⱍ u v, u, , v, , i苷1 j苷1, , Î√ r √*, Îu r u*, , (b), , Our intuition tells us that this approximation gets better as we increase the number of subrectangles, and we recognize the double sum as a Riemann sum for the double integral, xxD ru  rv du dv. This motivates the following definition., , ⱍ, , ⱍ, , 6, , Definition If a smooth parametric surface S is given by the equation, , r共u, v兲 苷 x共u, v兲 i  y共u, v兲 j  z共u, v兲 k, , FIGURE 15, , Approximating a patch, by a parallelogram, , 共u, v兲 僆 D, , and S is covered just once as 共u, v兲 ranges throughout the parameter domain D,, then the surface area of S is, , ⱍ, , ⱍ, , A共S兲 苷 yy ru  rv dA, D, , where, , ru 苷, , x, y, z, i, j, k, u, u, u, , rv 苷, , x, y, z, i, j, k, v, v, v, , EXAMPLE 10 Find the surface area of a sphere of radius a., SOLUTION In Example 4 we found the parametric representation, , x 苷 a sin  cos , , y 苷 a sin  sin , , z 苷 a cos , , where the parameter domain is, D 苷 兵共, 兲, , ⱍ0, , , , ,0, , , , 2 其, , We first compute the cross product of the tangent vectors:, , ⱍ, , i, x, r  r 苷 , x, , , j, y, , y, , , ⱍⱍ, , k, z, i,  苷 a cos  cos , z, a sin  sin , , , j, a cos  sin , a sin  cos , , k, a sin , 0, , ⱍ, , 苷 a 2 sin 2 cos  i  a 2 sin2 sin  j  a 2 sin  cos  k, Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1133, , SECTION 16.6, , 26. The part of the plane z 苷 x  3 that lies inside the cylinder, , 38. r共u, v兲 苷 共1, , x y 苷1, 2, , PARAMETRIC SURFACES AND THEIR AREAS, , 2, , u2, , v 2兲 i, , vj, , u k; 共 1,, , 1,, , 1133, 1兲, , 39–50 Find the area of the surface., CAS, , 27–28 Use a computer algebra system to produce a graph that, , 39. The part of the plane 3x  2y  z 苷 6 that lies in the, , looks like the given one., , first octant, , 27., , 28., , 40. The part of the plane with vector equation, r共u, v兲 苷 具u  v, 2 3u, 1  u v 典 that is given by, 0  u  2, 1  v  1, , 3, , z, , z 0, , _3, _3, , y, , 0 5, , x, , 41. The part of the plane x  2y  3z 苷 1 that lies inside the, , 0, , cylinder x 2  y 2 苷 3, , 42. The part of the cone z 苷 sx 2  y 2 that lies between the, , _1, _1, , 0, , y, , 0, , 1 1, , 0, , plane y 苷 x and the cylinder y 苷 x 2, , _1, x, , 43. The surface z 苷 3 共x 3兾2  y 3兾2 兲, 0  x  1, 0  y  1, 2, , 44. The part of the surface z 苷 1  3x  2y 2 that lies above the, , triangle with vertices 共0, 0兲, 共0, 1兲, and 共2, 1兲, , ; 29. Find parametric equations for the surface obtained by, rotating the curve y 苷 e , 0  x  3, about the x-axis and, use them to graph the surface., x, , 45. The part of the surface z 苷 xy that lies within the, , cylinder x 2  y 2 苷 1, , ; 30. Find parametric equations for the surface obtained by, , 46. The part of the paraboloid x 苷 y 2  z 2 that lies inside the, , rotating the curve x 苷 4y 2 y 4, 2  y  2, about the, y-axis and use them to graph the surface., , cylinder y 2  z 2 苷 9, , 47. The part of the surface y 苷 4x  z 2 that lies between the, , ; 31. (a) What happens to the spiral tube in Example 2 (see Fig-, , planes x 苷 0, x 苷 1, z 苷 0, and z 苷 1, , ure 5) if we replace cos u by sin u and sin u by cos u ?, (b) What happens if we replace cos u by cos 2u and sin u, by sin 2u?, , 48. The helicoid (or spiral ramp) with vector equation, r共u, v兲 苷 u cos v i  u sin v j  v k, 0  u  1, 0  v  , 49. The surface with parametric equations x 苷 u 2 , y 苷 u v,, 1, z 苷 2 v 2 , 0  u  1, 0  v  2, , ; 32. The surface with parametric equations, x 苷 2 cos   r cos共兾2兲, , 50. The part of the sphere x 2  y 2  z 2 苷 b 2 that lies inside the, , y 苷 2 sin   r cos共兾2兲, , cylinder x 2  y 2 苷 a 2, where 0, , z 苷 r sin共兾2兲, , ⱍ ⱍ, , 52–53 Find the area of the surface correct to four decimal places, , by expressing the area in terms of a single integral and using, your calculator to estimate the integral., , parametric surface at the specified point., , 34. x 苷 u 2  1,, , z苷u, , y 苷 v 3  1,, , v;, , 共2, 3, 0兲, , cylinder x 2  y 2 苷 1, , 37–38 Find an equation of the tangent plane to the given, , parametric surface at the specified point. Graph the surface and, the tangent plane., 37. r共u, v兲 苷 u 2 i  2u sin v j  u cos v k;, , 53. The part of the surface z 苷 e, , u 苷 1, v 苷 兾3, , 36. r共u, v兲 苷 sin u i  cos u sin v j  sin v k;, u 苷 兾6 , v 苷 兾6, , CAS, , 52. The part of the surface z 苷 cos共x 2  y 2 兲 that lies inside the, , z 苷 u  v ; 共5, 2, 3兲, , 35. r共u, v兲 苷 u cos v i  u sin v j  v k;, , u 苷 1, v 苷 0, , ⱍ ⱍ, , x 2  y 2  R 2, and you know that fx  1 and fy  1,, what can you say about A共S兲?, , 33–36 Find an equation of the tangent plane to the given, , y 苷 3u 2,, , b, , 51. If the equation of a surface S is z 苷 f 共x, y兲, where, , where 12  r  12 and 0    2, is called a Möbius, strip. Graph this surface with several viewpoints. What is, unusual about it?, , 33. x 苷 u  v,, , a, , x2 y2, , that lies above the, , disk x  y  4, 2, , CAS, , 2, , 54. Find, to four decimal places, the area of the part of the sur-, , face z 苷 共1  x 2 兲兾共1  y 2 兲 that lies above the square, x  y  1. Illustrate by graphing this part of the, surface., , ⱍ ⱍ ⱍ ⱍ, , 55. (a) Use the Midpoint Rule for double integrals (see Sec-, , tion 15.1) with six squares to estimate the area of the, surface z 苷 1兾共1  x 2  y 2 兲, 0  x  6, 0  y  4., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1134, CAS, , CAS, , CHAPTER 16, , VECTOR CALCULUS, , 61. Find the area of the part of the sphere x 2  y 2  z 2 苷 4z, , (b) Use a computer algebra system to approximate the surface area in part (a) to four decimal places. Compare, with the answer to part (a)., , that lies inside the paraboloid z 苷 x 2  y 2., , 62. The figure shows the surface created when the cylinder, , y 2  z 2 苷 1 intersects the cylinder x 2  z 2 苷 1. Find the, area of this surface., , 56. Find the area of the surface with vector equation, r共u, v兲 苷 具 cos 3u cos 3v, sin 3u cos 3v, sin 3v 典 , 0  u  ,, 0  v  2. State your answer correct to four decimal, , z, , places., CAS, , 57. Find the exact area of the surface z 苷 1  2x  3y  4y 2,, , 1  x  4, 0  y  1., , x, , 58. (a) Set up, but do not evaluate, a double integral for the area, of the surface with parametric equations x 苷 au cos v,, y 苷 bu sin v, z 苷 u 2, 0  u  2, 0  v  2., , (b) Eliminate the parameters to show that the surface is an, elliptic paraboloid and set up another double integral for, the surface area., (c) Use the parametric equations in part (a) with a 苷 2 and, b 苷 3 to graph the surface., (d) For the case a 苷 2, b 苷 3, use a computer algebra system, to find the surface area correct to four decimal places., , ;, CAS, , 59. (a) Show that the parametric equations x 苷 a sin u cos v,, y 苷 b sin u sin v, z 苷 c cos u, 0  u  , 0  v  2,, , 63. Find the area of the part of the sphere x 2  y 2  z 2 苷 a 2, , that lies inside the cylinder x 2  y 2 苷 ax., , 64. (a) Find a parametric representation for the torus obtained, , ;, , represent an ellipsoid., (b) Use the parametric equations in part (a) to graph the, ellipsoid for the case a 苷 1, b 苷 2, c 苷 3., (c) Set up, but do not evaluate, a double integral for the surface area of the ellipsoid in part (b)., , ;, , y, , by rotating about the z-axis the circle in the xz-plane, with center 共b, 0, 0兲 and radius a b. [Hint: Take as, parameters the angles  and shown in the figure.], (b) Use the parametric equations found in part (a) to graph, the torus for several values of a and b., (c) Use the parametric representation from part (a) to find, the surface area of the torus., z, (x, y, z), , 60. (a) Show that the parametric equations x 苷 a cosh u cos v,, y 苷 b cosh u sin v, z 苷 c sinh u, represent a hyperboloid, , ;, , 16.7, , of one sheet., (b) Use the parametric equations in part (a) to graph the, hyperboloid for the case a 苷 1, b 苷 2, c 苷 3., (c) Set up, but do not evaluate, a double integral for the surface area of the part of the hyperboloid in part (b) that, lies between the planes z 苷 3 and z 苷 3., , 0, , å, , ¨, , x, , y, , (b, 0, 0), , Surface Integrals, The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. Suppose f is a function of three variables, whose domain includes a surface S. We will define the surface integral of f over S in such, a way that, in the case where f 共x, y, z兲 苷 1, the value of the surface integral is equal to the, surface area of S. We start with parametric surfaces and then deal with the special case, where S is the graph of a function of two variables., , Parametric Surfaces, Suppose that a surface S has a vector equation, r共u, v兲 苷 x共u, v兲 i  y共u, v兲 j  z共u, v兲 k, , 共u, v兲 僆 D, , We first assume that the parameter domain D is a rectangle and we divide it into subrect-, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1135, , SECTION 16.7, √, , 1135, , angles Rij with dimensions u and v. Then the surface S is divided into corresponding, patches Sij as in Figure 1. We evaluate f at a point Pij* in each patch, multiply by the area, Sij of the patch, and form the Riemann sum, , R ij, Î√, , D, , SURFACE INTEGRALS, , Îu, , m, , n, , 兺 兺 f 共P*兲 S, ij, , 0, , ij, , i苷1 j苷1, , u, , Then we take the limit as the number of patches increases and define the surface integral, of f over the surface S as, , r, , m, , z, , S, , P *ij, , S, , Sij, , 0, , x, , yy f 共x, y, z兲 dS 苷, , 1, , y, , lim, , n, , 兺 兺 f 共P*兲 S, ij, , m, n l  i苷1 j苷1, , ij, , Notice the analogy with the definition of a line integral (16.2.2) and also the analogy with, the definition of a double integral (15.1.5)., To evaluate the surface integral in Equation 1 we approximate the patch area Sij by the, area of an approximating parallelogram in the tangent plane. In our discussion of surface, area in Section 16.6 we made the approximation, , ⱍ, , ⱍ, , Sij ⬇ ru  rv u v, FIGURE 1, , where, , ru 苷, , x, y, z, i, j, k, u, u, u, , x, , rv 苷, , v, , y, , i, , v, , j, , z, v, , k, , are the tangent vectors at a corner of Sij . If the components are continuous and ru and rv, are nonzero and nonparallel in the interior of D, it can be shown from Definition 1, even, when D is not a rectangle, that, We assume that the surface is covered only, once as 共u, v兲 ranges throughout D. The value, of the surface integral does not depend on the, parametrization that is used., , 2, , yy f 共x, y, z兲 dS 苷 yy f 共r共u, v兲兲 ⱍ r, , u, , S, , ⱍ, ,  rv dA, , D, , This should be compared with the formula for a line integral:, , y, , C, , f 共x, y, z兲 ds 苷, , y, , b, , a, , ⱍ, , ⱍ, , f 共r共t兲兲 r 共t兲 dt, , Observe also that, , yy 1 dS 苷 yy ⱍ r, , u, , S, , D, , ⱍ, ,  rv dA 苷 A共S兲, , Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. When using this formula, remember that f 共r共u, v兲兲 is, evaluated by writing x 苷 x共u, v兲, y 苷 y共u, v兲, and z 苷 z共u, v兲 in the formula for f 共x, y, z兲., EXAMPLE 1 Compute the surface integral xxS x 2 dS, where S is the unit sphere, , x 2  y 2  z 2 苷 1., , SOLUTION As in Example 4 in Section 16.6, we use the parametric representation, , x 苷 sin  cos , , y 苷 sin  sin , , z 苷 cos , , 0 , , 0    2, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1139, , SECTION 16.7, , SURFACE INTEGRALS, , 1139, , Oriented Surfaces, , P, , To define surface integrals of vector fields, we need to rule out nonorientable surfaces such, as the Möbius strip shown in Figure 4. [It is named after the German geometer August, Möbius (1790–1868).] You can construct one for yourself by taking a long rectangular, strip of paper, giving it a half-twist, and taping the short edges together as in Figure 5., If an ant were to crawl along the Möbius strip starting at a point P, it would end up on, the “other side” of the strip (that is, with its upper side pointing in the opposite direction)., Then, if the ant continued to crawl in the same direction, it would end up back at the, same point P without ever having crossed an edge. (If you have constructed a Möbius strip,, try drawing a pencil line down the middle.) Therefore a Möbius strip really has only, one side. You can graph the Möbius strip using the parametric equations in Exercise 32 in, Section 16.6., , FIGURE 4, , A Möbius strip, , TEC Visual 16.7 shows a Möbius strip, , with a normal vector that can be moved along, the surface., , B, , C, , A, , D, , B, , D, , A, , C, , FIGURE 5, , Constructing a Möbius strip, z, , From now on we consider only orientable (two-sided) surfaces. We start with a surface, S that has a tangent plane at every point 共x, y, z兲 on S (except at any boundary point). There, are two unit normal vectors n1 and n 2 苷 n1 at 共x, y, z兲. (See Figure 6.), If it is possible to choose a unit normal vector n at every such point 共x, y, z兲 so that n, varies continuously over S, then S is called an oriented surface and the given choice of n, provides S with an orientation. There are two possible orientations for any orientable surface (see Figure 7)., , n¡, , n™, 0, x, , FIGURE 6, , n, , n, , n, , n, , n, , y, , n, , FIGURE 7, , n, , n, n, , n, , The two orientations, of an orientable surface, , For a surface z 苷 t共x, y兲 given as the graph of t, we use Equation 3 to associate with, the surface a natural orientation given by the unit normal vector, , 5, , n苷, , t, t, i, jk, x, y, , 冑 冉 冊 冉 冊, t, x, , 1, , 2, , , , t, y, , 2, , Since the k-component is positive, this gives the upward orientation of the surface., If S is a smooth orientable surface given in parametric form by a vector function, r共u, v兲, then it is automatically supplied with the orientation of the unit normal vector, 6, , n苷, , ⱍ, , ru  rv, ru  rv, , ⱍ, , and the opposite orientation is given by n. For instance, in Example 4 in Section 16.6 we, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1140, , 1140, , VECTOR CALCULUS, , CHAPTER 16, , found the parametric representation, r共,  兲 苷 a sin  cos  i  a sin  sin  j  a cos  k, for the sphere x 2  y 2  z 2 苷 a 2. Then in Example 10 in Section 16.6 we found that, r  r 苷 a 2 sin 2 cos  i  a 2 sin 2 sin  j  a 2 sin  cos  k, , ⱍr, , and, , , , ⱍ, ,  r 苷 a 2 sin , , So the orientation induced by r共,  兲 is defined by the unit normal vector, n苷, , ⱍ, , r  r, 1, 苷 sin  cos  i  sin  sin  j  cos  k 苷 r共,  兲, r  r, a, , ⱍ, , Observe that n points in the same direction as the position vector, that is, outward from the, sphere (see Figure 8). The opposite (inward) orientation would have been obtained (see, Figure 9) if we had reversed the order of the parameters because r  r 苷 r  r ., z, , z, , 0, y, , y, , x, , x, , FIGURE 8, , FIGURE 9, , Positive orientation, , Negative orientation, , For a closed surface, that is, a surface that is the boundary of a solid region E , the, convention is that the positive orientation is the one for which the normal vectors point, outward from E, and inward-pointing normals give the negative orientation (see Figures 8, and 9)., , Surface Integrals of Vector Fields, z, , F=∏v, , n, Sij, S, 0, x, , FIGURE 10, , y, , Suppose that S is an oriented surface with unit normal vector n, and imagine a fluid with, density 共x, y, z兲 and velocity field v共x, y, z兲 flowing through S. (Think of S as an imaginary surface that doesn’t impede the fluid flow, like a fishing net across a stream.) Then the, rate of flow (mass per unit time) per unit area is v. If we divide S into small patches Sij ,, as in Figure 10 (compare with Figure 1), then Sij is nearly planar and so we can approximate the mass of fluid per unit time crossing Sij in the direction of the normal n by the, quantity, 共 v ⴢ n兲A共Sij 兲, where , v, and n are evaluated at some point on Sij . (Recall that the component of the vector v in the direction of the unit vector n is v ⴢ n.) By summing these quantities and taking the limit we get, according to Definition 1, the surface integral of the function v ⴢ n, over S :, 7, , yy, S, , v ⴢ n dS 苷 yy 共x, y, z兲v共x, y, z兲 ⴢ n共x, y, z兲 dS, S, , and this is interpreted physically as the rate of flow through S., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1141, , SURFACE INTEGRALS, , SECTION 16.7, , 1141, , If we write F 苷 v, then F is also a vector field on ⺢ 3 and the integral in Equation 7, becomes, , yy F ⴢ n dS, S, , A surface integral of this form occurs frequently in physics, even when F is not v, and is, called the surface integral (or flux integral) of F over S., , 8 Definition If F is a continuous vector field defined on an oriented surface S, with unit normal vector n, then the surface integral of F over S is, , yy F ⴢ dS 苷 yy F ⴢ n dS, S, , S, , This integral is also called the flux of F across S., , In words, Definition 8 says that the surface integral of a vector field over S is equal to, the surface integral of its normal component over S (as previously defined)., If S is given by a vector function r共u, v兲, then n is given by Equation 6, and from Definition 8 and Equation 2 we have, , yy F ⴢ dS 苷 yy F ⴢ, S, , S, , 苷, , yy, D, , 冋, , ⱍ, , ru  rv, dS, ru  rv, , ⱍ, , F共r共u, v兲兲 ⴢ, , ⱍ, , ru  rv, ru  rv, , 册, , r, ⱍ ⱍ, , u, , ⱍ, ,  rv dA, , where D is the parameter domain. Thus we have, Compare Equation 9 to the similar expression, for evaluating line integrals of vector fields in, Definition 16.2.13:, , y, , C, , F ⴢ dr 苷, , y, , b, , a, , 9, , z, , u, , S, , F共r共t兲兲 ⴢ r 共t兲 dt, , Figure 11 shows the vector field F in Example 4, at points on the unit sphere., , yy F ⴢ dS 苷 yy F ⴢ 共r, ,  rv 兲 dA, , D, , EXAMPLE 4 Find the flux of the vector field F共x, y, z兲 苷 z i  y j  x k across the unit, sphere x 2  y 2  z 2 苷 1., SOLUTION As in Example 1, we use the parametric representation, , r共,  兲 苷 sin  cos  i  sin  sin  j  cos  k, , 0, , , , , , 0, , , , 2, , F共r共,  兲兲 苷 cos  i  sin  sin  j  sin  cos  k, , Then, , and, from Example 10 in Section 16.6,, y, x, , FIGURE 11, , r  r 苷 sin 2 cos  i  sin 2 sin  j  sin  cos  k, Therefore, F共r共,  兲兲 ⴢ 共r  r 兲 苷 cos  sin 2 cos   sin 3 sin 2  sin 2 cos  cos , , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1142, , 1142, , CHAPTER 16, , VECTOR CALCULUS, , and, by Formula 9, the flux is, , yy F ⴢ dS 苷 yy F ⴢ 共r  r 兲 dA, S, , D, , 苷y, , 2, , y, , 0, , , , 共2 sin 2 cos  cos   sin 3 sin 2 兲 d d, , 0, , , , 苷 2 y sin2 cos  d, , y, , 0, , 苷0, 苷, , y, , , , 0, , sin 3 d, , y, , 2, , 0, , 2, , 0, , , , cos  d  y sin3 d, 0, , 冉, , sin 2 d, , y, , 2, , sin2 d, , 0, , 2, , since y cos  d 苷 0, 0, , 冊, , 4, 3, , by the same calculation as in Example 1., If, for instance, the vector field in Example 4 is a velocity field describing the flow of a, fluid with density 1, then the answer, 4兾3, represents the rate of flow through the unit, sphere in units of mass per unit time., In the case of a surface S given by a graph z 苷 t共x, y兲, we can think of x and y as parameters and use Equation 3 to write, , 冉, , F ⴢ 共rx  ry兲 苷 共P i  Q j  R k兲 ⴢ , , 冊, , t, t, i, jk, x, y, , Thus Formula 9 becomes, , yy F ⴢ dS 苷 yy, , 10, , S, , D, , 冉, , P, , 冊, , t, t, Q,  R dA, x, y, , This formula assumes the upward orientation of S; for a downward orientation we multiply by 1. Similar formulas can be worked out if S is given by y 苷 h共x, z兲 or x 苷 k共y, z兲., (See Exercises 37 and 38.), , v EXAMPLE 5 Evaluate xxS F ⴢ dS, where F共x, y, z兲 苷 y i  x j  z k and S is the, boundary of the solid region E enclosed by the paraboloid z 苷 1  x 2  y 2 and the, plane z 苷 0., z, , SOLUTION S consists of a parabolic top surface S1 and a circular bottom surface S2. (See, Figure 12.) Since S is a closed surface, we use the convention of positive (outward), orientation. This means that S1 is oriented upward and we can use Equation 10 with, D being the projection of S1 onto the xy-plane, namely, the disk x 2  y 2 1. Since, , S¡, S™, y, , P共x, y, z兲 苷 y, , x, , FIGURE 12, , on S1 and, , Q共x, y, z兲 苷 x, t, 苷 2x, x, , R共x, y, z兲 苷 z 苷 1  x 2  y 2, t, 苷 2y, y, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1144, , 1144, , CHAPTER 16, , VECTOR CALCULUS, , where K is an experimentally determined constant called the conductivity of the substance. The rate of heat flow across the surface S in the body is then given by the surface, integral, , yy F ⴢ dS 苷 K yy ∇u ⴢ dS, S, , S, , v EXAMPLE 6 The temperature u in a metal ball is proportional to the square of the, distance from the center of the ball. Find the rate of heat flow across a sphere S of, radius a with center at the center of the ball., SOLUTION Taking the center of the ball to be at the origin, we have, , u共x, y, z兲 苷 C共x 2  y 2  z 2 兲, where C is the proportionality constant. Then the heat flow is, F共x, y, z兲 苷 K u 苷 KC共2x i  2y j  2z k兲, where K is the conductivity of the metal. Instead of using the usual parametrization of, the sphere as in Example 4, we observe that the outward unit normal to the sphere, x 2  y 2  z 2 苷 a 2 at the point 共x, y, z兲 is, n苷, , 1, 共x i  y j  z k兲, a, , Fⴢn苷, , and so, , 2KC 2, 共x  y 2  z 2 兲, a, , But on S we have x 2  y 2  z 2 苷 a 2, so F ⴢ n 苷 2aKC. Therefore the rate of heat, flow across S is, , yy F ⴢ dS 苷 yy F ⴢ n dS 苷 2aKC yy dS, S, , S, , S, , 苷 2aKCA共S兲 苷 2aKC共4 a 2 兲 苷 8KC a 3, , 16.7, , Exercises, , 1. Let S be the boundary surface of the box enclosed by the, , planes x 苷 0, x 苷 2, y 苷 0, y 苷 4, z 苷 0, and z 苷 6. Approximate xxS e0.1共xyz兲 dS by using a Riemann sum as in Definition 1, taking the patches Sij to be the rectangles that are the, faces of the box S and the points Pij* to be the centers of the, rectangles., 2. A surface S consists of the cylinder x 2  y 2 苷 1, 1, , z 1,, together with its top and bottom disks. Suppose you know that, f is a continuous function with, f 共 1, 0, 0兲 苷 2, , f 共0,, , 1, 0兲 苷 3, , f 共0, 0,, , 1兲 苷 4, , Estimate the value of xxS f 共x, y, z兲 dS by using a Riemann sum,, taking the patches Sij to be four quarter-cylinders and the top, and bottom disks., CAS Computer algebra system required, , 3. Let H be the hemisphere x 2  y 2  z 2 苷 50, z  0, and, , suppose f is a continuous function with f 共3, 4, 5兲 苷 7,, f 共3, 4, 5兲 苷 8, f 共3, 4, 5兲 苷 9, and f 共3, 4, 5兲 苷 12., By dividing H into four patches, estimate the value of, xxH f 共x, y, z兲 dS., , 4. Suppose that f 共x, y, z兲 苷 t (sx 2  y 2  z 2 ), where t is a, , function of one variable such that t共2兲 苷 5. Evaluate, xxS f 共x, y, z兲 dS, where S is the sphere x 2  y 2  z 2 苷 4., , 5–20 Evaluate the surface integral., 5., , xxS 共x  y  z兲 dS,, S is the parallelogram with parametric equations x 苷 u  v,, y 苷 u  v, z 苷 1  2u  v, 0 u 2, 0 v 1, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1146, , VECTOR CALCULUS, , CHAPTER 16, , 38. Find a formula for xxS F ⴢ dS similar to Formula 10 for the case, , 44. Seawater has density 1025 kg兾m3 and flows in a velocity field, , 39. Find the center of mass of the hemisphere x 2  y 2  z 2 苷 a 2,, , 45. Use Gauss’s Law to find the charge contained in the solid, , where S is given by x 苷 k共 y, z兲 and n is the unit normal that, points forward (that is, toward the viewer when the axes are, drawn in the usual way)., , v 苷 y i  x j, where x, y, and z are measured in meters and the, components of v in meters per second. Find the rate of flow, outward through the hemisphere x 2  y 2  z 2 苷 9 , z  0 ., , z  0, if it has constant density., , hemisphere x 2  y 2  z 2, , E共x, y, z兲 苷 x i  y j  2z k, , 40. Find the mass of a thin funnel in the shape of a cone, , z 苷 sx 2  y 2 , 1 z, 共x, y, z兲 苷 10  z., , 4, if its density function is, , 46. Use Gauss’s Law to find the charge enclosed by the cube, , with vertices 共 1,, , 41. (a) Give an integral expression for the moment of inertia I z, , ductivity K 苷 6.5 is u共x, y, z兲 苷 2y 2  2z 2. Find the rate of, heat flow inward across the cylindrical surface y 2  z 2 苷 6,, 0 x 4., , above the plane z 苷 4. If S has constant density k, find, (a) the center of mass and (b) the moment of inertia about, the z-axis., , 48. The temperature at a point in a ball with conductivity K is, , inversely proportional to the distance from the center of the, ball. Find the rate of heat flow across a sphere S of radius a, with center at the center of the ball., , 43. A fluid has density 870 kg兾m3 and flows with velocity, , ⱍ ⱍ, , 49. Let F be an inverse square field, that is, F共r兲 苷 cr兾 r, , v 苷 z i  y 2 j  x 2 k , where x, y, and z are measured in, meters and the components of v in meters per second. Find the, rate of flow outward through the cylinder x 2  y 2 苷 4 ,, 0 z 1., , 3, for, some constant c, where r 苷 x i  y j  z k. Show that the flux, of F across a sphere S with center the origin is independent of, the radius of S., , Stokes’ Theorem, , z, , n, n, , S, , C, 0, , FIGURE 1, , 1兲 if the electric field is, , 47. The temperature at the point 共x, y, z兲 in a substance with con-, , 42. Let S be the part of the sphere x 2  y 2  z 2 苷 25 that lies, , x, , 1,, , E共x, y, z兲 苷 x i  y j  z k, , about the z-axis of a thin sheet in the shape of a surface S if, the density function is ., (b) Find the moment of inertia about the z-axis of the funnel in, Exercise 40., , 16.8, , a 2, z  0, if the electric field is, , y, , Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem., Whereas Green’s Theorem relates a double integral over a plane region D to a line integral, around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface, S to a line integral around the boundary curve of S (which is a space curve). Figure 1 shows, an oriented surface with unit normal vector n. The orientation of S induces the positive, orientation of the boundary curve C shown in the figure. This means that if you walk in, the positive direction around C with your head pointing in the direction of n, then the surface will always be on your left., Stokes’ Theorem Let S be an oriented piecewise-smooth surface that is bounded, by a simple, closed, piecewise-smooth boundary curve C with positive orientation., Let F be a vector field whose components have continuous partial derivatives on, an open region in ⺢ 3 that contains S. Then, , y, , C, , F ⴢ dr 苷 yy curl F ⴢ dS, S, , Since, , y, , C, , F ⴢ dr 苷, , y, , C, , F ⴢ T ds, , and, , yy curl F ⴢ dS 苷 yy curl F ⴢ n dS, S, , S, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1138-1147.qk_97817_16_ch16_p1138-1147 11/9/10 9:25 AM Page 1147, , STOKES’ THEOREM, , SECTION 16.8, George Stokes, Stokes’ Theorem is named after the Irish mathematical physicist Sir George Stokes (1819–1903)., Stokes was a professor at Cambridge University, (in fact he held the same position as Newton,, Lucasian Professor of Mathematics) and was, especially noted for his studies of fluid flow, and light. What we call Stokes’ Theorem was, actually discovered by the Scottish physicist, Sir William Thomson (1824–1907, known as, Lord Kelvin). Stokes learned of this theorem, in a letter from Thomson in 1850 and asked, students to prove it on an examination at, Cambridge University in 1854. We don’t know, if any of those students was able to do so., , 1147, , Stokes’ Theorem says that the line integral around the boundary curve of S of the tangential component of F is equal to the surface integral over S of the normal component of the, curl of F., The positively oriented boundary curve of the oriented surface S is often written as, S, so Stokes’ Theorem can be expressed as, , yy curl F ⴢ dS 苷 y, , 1, , S, , S, , F ⴢ dr, , There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental, Theorem of Calculus. As before, there is an integral involving derivatives on the left side, of Equation 1 (recall that curl F is a sort of derivative of F ) and the right side involves the, values of F only on the boundary of S., In fact, in the special case where the surface S is flat and lies in the xy-plane with, upward orientation, the unit normal is k, the surface integral becomes a double integral,, and Stokes’ Theorem becomes, , y, , C, , F ⴢ dr 苷 yy curl F ⴢ dS 苷, S, , yy 共curl F兲 ⴢ k dA, S, , This is precisely the vector form of Green’s Theorem given in Equation 16.5.12. Thus we, see that Green’s Theorem is really a special case of Stokes’ Theorem., Although Stokes’ Theorem is too difficult for us to prove in its full generality, we can, give a proof when S is a graph and F, S, and C are well behaved., z, , PROOF OF A SPECIAL CASE OF STOKES’ THEOREM We assume that the equation of S is, , z 苷 t共x, y兲, 共x, y兲 僆 D, where t has continuous second-order partial derivatives and D, is a simple plane region whose boundary curve C1 corresponds to C. If the orientation of, S is upward, then the positive orientation of C corresponds to the positive orientation of, C1. (See Figure 2.) We are also given that F 苷 P i  Q j  R k, where the partial derivatives of P, Q, and R are continuous., Since S is a graph of a function, we can apply Formula 16.7.10 with F replaced by, curl F. The result is, , n, z=g(x, y), S, 0, x, , C, , D, C¡, , FIGURE 2, , y, , 2, , yy curl F ⴢ dS, S, , 苷, , yy, D, , 冋冉, , , R, Q, , y, z, , 冊 冉, z, , x, , P, R, , z, x, , 冊 冉, z, , y, , Q, P, , x, y, , 冊册, , dA, , where the partial derivatives of P, Q, and R are evaluated at 共x, y, t共x, y兲兲. If, x 苷 x共t兲, , y 苷 y共t兲, , a, , t, , b, , is a parametric representation of C1, then a parametric representation of C is, x 苷 x共t兲, , y 苷 y共t兲, , z 苷 t ( x共t兲, y共t兲), , a, , t, , b, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:25 AM Page 1148, , 1148, , CHAPTER 16, , VECTOR CALCULUS, , This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows:, , y, , C, , F ⴢ dr 苷, , 冉, 冊, 冉, 冊册, y冋, 冊 冉 冊 册, y 冋冉, 冊 冉 冊, y冉, 冊 冉 冊册, yy 冋 冉, , y, , P, , dx, dy, dz, ⫹Q, ⫹R, dt, dt, dt, , P, , dx, dy, ⭸z dx, ⭸z dy, ⫹Q, ⫹R, ⫹, dt, dt, ⭸x dt, ⭸y dt, , b, , a, , 苷, , b, , a, , 苷, , b, , P⫹R, , a, , 苷, , P⫹R, , C1, , 苷, , D, , ⭸, ⭸x, , ⭸z, ⭸x, , ⭸z, ⭸x, , dt, , dx, ⭸z, ⫹ Q⫹R, dt, ⭸y, , dx ⫹ Q ⫹ R, , Q⫹R, , ⭸z, ⭸y, , ⫺, , ⭸, ⭸y, , ⭸z, ⭸y, , dy, dt, , dt, , dt, , dy, , P⫹R, , ⭸z, ⭸x, , dA, , where we have used Green’s Theorem in the last step. Then, using the Chain Rule again, and remembering that P, Q, and R are functions of x, y, and z and that z is itself a function, of x and y, we get, , y, , C, , F ⴢ dr 苷, , yy, D, , 冋冉, , ⭸Q, ⭸Q ⭸z, ⭸R ⭸z, ⭸R ⭸z ⭸z, ⭸2z, ⫹, ⫹, ⫹, ⫹R, ⭸x, ⭸z ⭸x, ⭸x ⭸y, ⭸z ⭸x ⭸y, ⭸x ⭸y, ⫺, , 冉, , 冊, , ⭸P, ⭸P ⭸z, ⭸R ⭸z, ⭸R ⭸z ⭸z, ⭸2z, ⫹, ⫹, ⫹, ⫹R, ⭸y, ⭸z ⭸y, ⭸y ⭸x, ⭸z ⭸y ⭸x, ⭸y ⭸x, , 冊册, , dA, , Four of the terms in this double integral cancel and the remaining six terms can be, arranged to coincide with the right side of Equation 2. Therefore, , y, , C, , F ⴢ dr 苷 yy curl F ⴢ dS, S, , v, , curve of intersection of the plane y ⫹ z 苷 2 and the cylinder x 2 ⫹ y 2 苷 1. (Orient C to, be counterclockwise when viewed from above.), , z, , S, , ⱍ, , i, ⭸, curl F 苷, ⭸x, ⫺y 2, , D 0, y, , FIGURE 3, , xC F ⴢ dr could be, evaluated directly, it’s easier to use Stokes’ Theorem. We first compute, , SOLUTION The curve C (an ellipse) is shown in Figure 3. Although, C, y+z=2, , x, , EXAMPLE 1 Evaluate xC F ⴢ dr, where F共x, y, z兲 苷 ⫺y 2 i ⫹ x j ⫹ z 2 k and C is the, , j, ⭸, ⭸y, x, , ⱍ, , k, ⭸, 苷 共1 ⫹ 2y兲 k, ⭸z, z2, , Although there are many surfaces with boundary C, the most convenient choice is the, elliptical region S in the plane y ⫹ z 苷 2 that is bounded by C. If we orient S upward,, then C has the induced positive orientation. The projection D of S onto the xy-plane is, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1149, , SECTION 16.8, , STOKES’ THEOREM, , 1149, , the disk x 2 ⫹ y 2 艋 1 and so using Equation 16.7.10 with z 苷 t共x, y兲 苷 2 ⫺ y, we have, , y, , C, , F ⴢ dr 苷 yy curl F ⴢ dS 苷 yy 共1 ⫹ 2y兲 dA, S, , 苷y, , D, 2␲, , 0, , 苷, , y, , 2␲, , 0, , y, , 1, , 0, , 冋, , 共1 ⫹ 2r sin ␪ 兲 r dr d␪, r2, r3, ⫹2, sin ␪, 2, 3, , 册, , 1, , d␪ 苷 y, , 2␲, , 0, , 0, , ( 12 ⫹ 23 sin ␪) d␪, , 苷 12 共2␲兲 ⫹ 0 苷 ␲, z, , ≈+¥+z@=4, S, C, , v EXAMPLE 2 Use Stokes’ Theorem to compute the integral xxS curl F ⴢ dS, where, F共x, y, z兲 苷 xz i ⫹ yz j ⫹ xy k and S is the part of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 4 that, lies inside the cylinder x 2 ⫹ y 2 苷 1 and above the xy-plane. (See Figure 4.), SOLUTION To find the boundary curve C we solve the equations x 2 ⫹ y 2 ⫹ z 2 苷 4 and, , x 2 ⫹ y 2 苷 1. Subtracting, we get z 2 苷 3 and so z 苷 s3 (since z ⬎ 0). Thus C is the, circle given by the equations x 2 ⫹ y 2 苷 1, z 苷 s3 . A vector equation of C is, , 0, , y, x, , FIGURE 4, , 0 艋 t 艋 2␲, , r共t兲 苷 cos t i ⫹ sin t j ⫹ s3 k, , ≈+¥=1, , r⬘共t兲 苷 ⫺sin t i ⫹ cos t j, , so, Also, we have, , F共r共t兲兲 苷 s3 cos t i ⫹ s3 sin t j ⫹ cos t sin t k, Therefore, by Stokes’ Theorem,, , yy curl F ⴢ dS 苷 y, , C, , F ⴢ dr 苷 y, , 2␲, , 0, , F共r共t兲兲 ⴢ r⬘共t兲 dt, , S, , 苷y, , 2␲, , 0, , 苷 s3, , (⫺s3 cos t sin t ⫹ s3 sin t cos t) dt, , y, , 2␲, , 0, , 0 dt 苷 0, , Note that in Example 2 we computed a surface integral simply by knowing the values, of F on the boundary curve C. This means that if we have another oriented surface with, the same boundary curve C, then we get exactly the same value for the surface integral!, In general, if S1 and S2 are oriented surfaces with the same oriented boundary curve C, and both satisfy the hypotheses of Stokes’ Theorem, then, 3, , yy curl F ⴢ dS 苷 y, , C, , F ⴢ dr 苷 yy curl F ⴢ dS, , S1, , S2, , This fact is useful when it is difficult to integrate over one surface but easy to integrate, over the other., We now use Stokes’ Theorem to throw some light on the meaning of the curl vector., Suppose that C is an oriented closed curve and v represents the velocity field in fluid flow., Consider the line integral, , y, , C, , v ⴢ dr 苷 y v ⴢ T ds, C, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1150, , 1150, , CHAPTER 16, , VECTOR CALCULUS, , and recall that v ⴢ T is the component of v in the direction of the unit tangent vector T., This means that the closer the direction of v is to the direction of T, the larger the value of, v ⴢ T. Thus xC v ⴢ dr is a measure of the tendency of the fluid to move around C and is, called the circulation of v around C. (See Figure 5.), , T, , C, , T, v, , C, , v, , (a) jC v ⭈ dr>0, positive circulation, , FIGURE 5, , (b) jC v ⭈ dr<0, negative circulation, , Now let P0共x 0 , y0 , z0 兲 be a point in the fluid and let Sa be a small disk with radius a and, center P0. Then (curl F兲共P兲 ⬇ 共curl F兲共P0兲 for all points P on Sa because curl F is continuous. Thus, by Stokes’ Theorem, we get the following approximation to the circulation, around the boundary circle Ca :, , y, , Ca, , v ⴢ dr 苷 yy curl v ⴢ dS 苷 yy curl v ⴢ n dS, Sa, , Sa, , ⬇ yy curl v共P0 兲 ⴢ n共P0 兲 dS 苷 curl v共P0 兲 ⴢ n共P0 兲␲ a 2, Sa, , Imagine a tiny paddle wheel placed in the, fluid at a point P, as in Figure 6; the paddle, wheel rotates fastest when its axis is parallel, to curl v., , curl v, , FIGURE 6, , This approximation becomes better as a l 0 and we have, curl v共P0 兲 ⴢ n共P0 兲 苷 lim, , 4, , al0, , 1, ␲a 2, , y, , Ca, , v ⴢ dr, , Equation 4 gives the relationship between the curl and the circulation. It shows that, curl v ⭈ n is a measure of the rotating effect of the fluid about the axis n. The curling effect, is greatest about the axis parallel to curl v., Finally, we mention that Stokes’ Theorem can be used to prove Theorem 16.5.4 (which, states that if curl F 苷 0 on all of ⺢ 3, then F is conservative). From our previous work, (Theorems 16.3.3 and 16.3.4), we know that F is conservative if xC F ⴢ dr 苷 0 for every, closed path C. Given C, suppose we can find an orientable surface S whose boundary is, C. (This can be done, but the proof requires advanced techniques.) Then Stokes’ Theorem, gives, , y, , C, , F ⴢ dr 苷, , yy curl F ⴢ dS 苷 yy 0 ⴢ dS 苷 0, S, , S, , A curve that is not simple can be broken into a number of simple curves, and the integrals, around these simple curves are all 0. Adding these integrals, we obtain xC F ⴢ dr 苷 0 for, any closed curve C., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1151, , SECTION 16.8, , 16.8, , STOKES’ THEOREM, , 1151, , Exercises, 10. F共x, y, z兲 苷 xy i ⫹ 2z j ⫹ 3y k,, , 1. A hemisphere H and a portion P of a paraboloid are shown., , C is the curve of intersection of the plane x ⫹ z 苷 5 and the cylinder x 2 ⫹ y 2 苷 9, , Suppose F is a vector field on ⺢3 whose components have continuous partial derivatives. Explain why, , 11. (a) Use Stokes’ Theorem to evaluate xC F ⴢ dr, where, , yy curl F ⴢ dS 苷 yy curl F ⴢ dS, H, , P, , F共x, y, z兲 苷 x 2 z i ⫹ x y 2 j ⫹ z 2 k, z, , z, , 4, , 4, , ;, , P, H, , ;, , and C is the curve of intersection of the plane, x ⫹ y ⫹ z 苷 1 and the cylinder x 2 ⫹ y 2 苷 9 oriented, counterclockwise as viewed from above., (b) Graph both the plane and the cylinder with domains, chosen so that you can see the curve C and the surface, that you used in part (a)., (c) Find parametric equations for C and use them to graph C., 12. (a) Use Stokes’ Theorem to evaluate xC F ⴢ dr, where, , x, , 2, , 2, , x, , y, , 2, , 2, , y, , 2–6 Use Stokes’ Theorem to evaluate xxS curl F ⴢ dS., 2. F共x, y, z兲 苷 2y cos z i ⫹ e x sin z j ⫹ xe y k,, , S is the hemisphere x ⫹ y ⫹ z 苷 9, z 艌 0, oriented, upward, 2, , 2, , 3. F共x, y, z兲 苷 x z i ⫹ y z j ⫹ xyz k,, 2 2, , 4. F共x, y, z兲 苷 tan⫺1共x 2 yz 2 兲 i ⫹ x 2 y j ⫹ x 2 z 2 k,, , S is the cone x 苷 sy 2 ⫹ z 2 , 0 艋 x 艋 2, oriented in the direction of the positive x-axis, , 5. F共x, y, z兲 苷 x yz i ⫹ x y j ⫹ x yz k,, 2, , S consists of the top and the four sides (but not the bottom), of the cube with vertices 共⫾1, ⫾1, ⫾1兲, oriented outward, 6. F共x, y, z兲 苷 e, , ;, , 2 2, , S is the part of the paraboloid z 苷 x 2 ⫹ y 2 that lies inside the, cylinder x 2 ⫹ y 2 苷 4, oriented upward, , i ⫹ e j ⫹ x z k,, S is the half of the ellipsoid 4x 2 ⫹ y 2 ⫹ 4z 2 苷 4 that lies to, the right of the xz-plane, oriented in the direction of the, positive y-axis, xy, , ;, , 2, , F共x, y, z兲 苷 x 2 y i ⫹ 13 x 3 j ⫹ x y k and C is the curve of, intersection of the hyperbolic paraboloid z 苷 y 2 ⫺ x 2 and, the cylinder x 2 ⫹ y 2 苷 1 oriented counterclockwise as, viewed from above., (b) Graph both the hyperbolic paraboloid and the cylinder, with domains chosen so that you can see the curve C and, the surface that you used in part (a)., (c) Find parametric equations for C and use them to graph C., , xz, , 2, , 13–15 Verify that Stokes’ Theorem is true for the given vector, field F and surface S., 13. F共x, y, z兲 苷 ⫺y i ⫹ x j ⫺ 2 k,, , S is the cone z 2 苷 x 2 ⫹ y 2, 0 艋 z 艋 4, oriented downward, , 14. F共x, y, z兲 苷 ⫺2yz i ⫹ y j ⫹ 3x k,, , S is the part of the paraboloid z 苷 5 ⫺ x 2 ⫺ y 2 that lies, above the plane z 苷 1, oriented upward, , 15. F共x, y, z兲 苷 y i ⫹ z j ⫹ x k,, , S is the hemisphere x 2 ⫹ y 2 ⫹ z 2 苷 1, y 艌 0, oriented in the, direction of the positive y-axis, , 16. Let C be a simple closed smooth curve that lies in the plane, , x ⫹ y ⫹ z 苷 1. Show that the line integral, , xC z dx ⫺ 2x dy ⫹ 3y dz, , 7–10 Use Stokes’ Theorem to evaluate xC F ⴢ dr. In each case C is, , oriented counterclockwise as viewed from above., 7. F共x, y, z兲 苷 共x ⫹ y 2 兲 i ⫹ 共 y ⫹ z 2 兲 j ⫹ 共z ⫹ x 2 兲 k,, , depends only on the area of the region enclosed by C and not, on the shape of C or its location in the plane., , C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), 8. F共x, y, z兲 苷 i ⫹ 共x ⫹ yz兲 j ⫹ ( xy ⫺ sz ) k,, , 17. A particle moves along line segments from the origin to the, , C is the boundary of the part of the plane 3x ⫹ 2y ⫹ z 苷 1, in the first octant, , 9. F共x, y, z兲 苷 yz i ⫹ 2 xz j ⫹ e xy k,, , C is the circle x 2 ⫹ y 2 苷 16, z 苷 5, , ;, , Graphing calculator or computer required, , points 共1, 0, 0兲, 共1, 2, 1兲, 共0, 2, 1兲, and back to the origin, under the influence of the force field, F共x, y, z兲 苷 z 2 i ⫹ 2xy j ⫹ 4y 2 k, Find the work done., , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1152, , CHAPTER 16, , VECTOR CALCULUS, , 20. Suppose S and C satisfy the hypotheses of Stokes’ Theorem, , 18. Evaluate, , xC 共 y ⫹ sin x兲 dx ⫹ 共z, , 2, , and f , t have continuous second-order partial derivatives. Use, Exercises 24 and 26 in Section 16.5 to show the following., (a) xC 共 f ⵜt兲 ⴢ dr 苷 xxS 共ⵜ f ⫻ ⵜt兲 ⴢ dS, , ⫹ cos y兲 dy ⫹ x dz, 3, , where C is the curve r共t兲 苷 具sin t, cos t, sin 2t典 , 0 艋 t 艋 2␲., [Hint: Observe that C lies on the surface z 苷 2 x y.], 19. If S is a sphere and F satisfies the hypotheses of Stokes’, , Theorem, show that xxS curl F ⴢ dS 苷 0., , WRITING PROJECT, , The photograph shows a stained-glass, window at Cambridge University in honor of, George Green., , Courtesy of the Masters and Fellows of Gonville and, Caius College, Cambridge University, England, , (b), , xC 共 f ⵜ f 兲 ⴢ dr 苷 0, , (c), , xC 共 f ⵜt ⫹ t ⵜ f 兲 ⴢ dr 苷 0, , THREE MEN AND TWO THEOREMS, Although two of the most important theorems in vector calculus are named after George Green, and George Stokes, a third man, William Thomson (also known as Lord Kelvin), played a large, role in the formulation, dissemination, and application of both of these results. All three men, were interested in how the two theorems could help to explain and predict physical phenomena, in electricity and magnetism and fluid flow. The basic facts of the story are given in the margin, notes on pages 1109 and 1147., Write a report on the historical origins of Green’s Theorem and Stokes’ Theorem. Explain the, similarities and relationship between the theorems. Discuss the roles that Green, Thomson, and, Stokes played in discovering these theorems and making them widely known. Show how both, theorems arose from the investigation of electricity and magnetism and were later used to study a, variety of physical problems., The dictionary edited by Gillispie [2] is a good source for both biographical and scientific, information. The book by Hutchinson [5] gives an account of Stokes’ life and the book by, Thompson [8] is a biography of Lord Kelvin. The articles by Grattan-Guinness [3] and Gray [4], and the book by Cannell [1] give background on the extraordinary life and works of Green., Additional historical and mathematical information is found in the books by Katz [6] and, Kline [7]., 1. D. M. Cannell, George Green, Mathematician and Physicist 1793–1841: The Background to, , His Life and Work (Philadelphia: Society for Industrial and Applied Mathematics, 2001)., 2. C. C. Gillispie, ed., Dictionary of Scientific Biography (New York: Scribner’s, 1974). See the, , article on Green by P. J. Wallis in Volume XV and the articles on Thomson by Jed Buchwald, and on Stokes by E. M. Parkinson in Volume XIII., 3. I. Grattan-Guinness, “Why did George Green write his essay of 1828 on electricity and, , magnetism?” Amer. Math. Monthly, Vol. 102 (1995), pp. 387–96., 4. J. Gray, “There was a jolly miller.” The New Scientist, Vol. 139 (1993), pp. 24–27., 5. G. E. Hutchinson, The Enchanted Voyage and Other Studies (Westport, CT : Greenwood, , Press, 1978)., 6. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993),, , pp. 678–80., 7. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford, , University Press, 1972), pp. 683–85., 8. Sylvanus P. Thompson, The Life of Lord Kelvin (New York: Chelsea, 1976)., , 16.9, , The Divergence Theorem, In Section 16.5 we rewrote Green’s Theorem in a vector version as, , y, , C, , F ⴢ n ds 苷, , yy div F共x, y兲 dA, D, , where C is the positively oriented boundary curve of the plane region D. If we were seek-, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1153, , SECTION 16.9, , THE DIVERGENCE THEOREM, , 1153, , ing to extend this theorem to vector fields on ⺢ 3, we might make the guess that, , yy F ⴢ n dS 苷 yyy div F共x, y, z兲 dV, , 1, , S, , E, , where S is the boundary surface of the solid region E. It turns out that Equation 1 is true,, under appropriate hypotheses, and is called the Divergence Theorem. Notice its similarity, to Green’s Theorem and Stokes’ Theorem in that it relates the integral of a derivative of a, function (div F in this case) over a region to the integral of the original function F over the, boundary of the region., At this stage you may wish to review the various types of regions over which we were, able to evaluate triple integrals in Section 15.7. We state and prove the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3 and we call such regions, simple solid regions. (For instance, regions bounded by ellipsoids or rectangular boxes, are simple solid regions.) The boundary of E is a closed surface, and we use the convention, introduced in Section 16.7, that the positive orientation is outward; that is, the unit, normal vector n is directed outward from E., , The Divergence Theorem is sometimes called, Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855), who, discovered this theorem during his investigation, of electrostatics. In Eastern Europe the Divergence Theorem is known as Ostrogradsky’s, Theorem after the Russian mathematician, Mikhail Ostrogradsky (1801–1862), who published this result in 1826., , The Divergence Theorem Let E be a simple solid region and let S be the boundary, , surface of E, given with positive (outward) orientation. Let F be a vector field, whose component functions have continuous partial derivatives on an open region, that contains E. Then, , yy F ⴢ dS 苷 yyy div F dV, S, , E, , Thus the Divergence Theorem states that, under the given conditions, the flux of F, across the boundary surface of E is equal to the triple integral of the divergence of F, over E., PROOF Let F 苷 P i ⫹ Q j ⫹ R k. Then, , div F 苷, , yyy div F dV 苷 yyy, , so, , E, , E, , ⭸P, ⭸Q, ⭸R, ⫹, ⫹, ⭸x, ⭸y, ⭸z, , ⭸P, dV ⫹, ⭸x, , yyy, E, , ⭸Q, ⭸R, dV ⫹ yyy, dV, ⭸y, ⭸z, E, , If n is the unit outward normal of S, then the surface integral on the left side of the Divergence Theorem is, , yy F ⴢ dS 苷 yy F ⴢ n dS 苷 yy 共P i ⫹ Q j ⫹ R k兲 ⴢ n dS, S, , S, , S, , 苷 yy P i ⴢ n dS ⫹ yy Q j ⴢ n dS ⫹ yy R k ⴢ n dS, S, , S, , S, , Therefore, to prove the Divergence Theorem, it suffices to prove the following three, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1155, , THE DIVERGENCE THEOREM, , SECTION 16.9, , 1155, , Comparison with Equation 5 shows that, , yy R k ⴢ n dS 苷 yyy, S, , Notice that the method of proof of the, Divergence Theorem is very similar to that, of Green’s Theorem., , E, , ⭸R, dV, ⭸z, , Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2, or type 3 region, respectively., , v EXAMPLE 1 Find the flux of the vector field F共x, y, z兲 苷 z i ⫹ y j ⫹ x k over the unit, sphere x 2 ⫹ y 2 ⫹ z 2 苷 1., SOLUTION First we compute the divergence of F :, , div F 苷, , ⭸, ⭸, ⭸, 共z兲 ⫹, 共 y兲 ⫹, 共x兲 苷 1, ⭸x, ⭸y, ⭸z, , The unit sphere S is the boundary of the unit ball B given by x 2 ⫹ y 2 ⫹ z 2 艋 1. Thus the, Divergence Theorem gives the flux as, The solution in Example 1 should be compared, with the solution in Example 4 in Section 16.7., , S, , v, , z, (0, 0, 1), , yy F ⴢ dS 苷 yyy div F dV 苷 yyy 1 dV 苷 V共B兲 苷, B, , 4, 3, , ␲ 共1兲3 苷, , B, , 4␲, 3, , EXAMPLE 2 Evaluate xxS F ⴢ dS, where, , F共x, y, z兲 苷 xy i ⫹ ( y 2 ⫹ e xz ) j ⫹ sin共xy兲 k, 2, , y=2-z, , and S is the surface of the region E bounded by the parabolic cylinder z 苷 1 ⫺ x 2 and, the planes z 苷 0, y 苷 0, and y ⫹ z 苷 2. (See Figure 2.), 0, , SOLUTION It would be extremely difficult to evaluate the given surface integral directly., , (1, 0, 0), x, , (0, 2, 0) y, , (We would have to evaluate four surface integrals corresponding to the four pieces of S.), Furthermore, the divergence of F is much less complicated than F itself:, , z=1-≈, , div F 苷, , FIGURE 2, , ⭸, ⭸, ( y 2 ⫹ e xz 2 ) ⫹ ⭸ 共sin xy兲 苷 y ⫹ 2y 苷 3y, 共xy兲 ⫹, ⭸x, ⭸y, ⭸z, , Therefore we use the Divergence Theorem to transform the given surface integral into a, triple integral. The easiest way to evaluate the triple integral is to express E as a type 3, region:, E 苷 兵 共x, y, z兲, , ⱍ ⫺1 艋 x 艋 1,, , 0 艋 z 艋 1 ⫺ x 2, 0 艋 y 艋 2 ⫺ z 其, , Then we have, , yy F ⴢ dS 苷 yyy div F dV 苷 yyy 3y dV, S, , E, , E, , 苷3y, , 1, , ⫺1, , 苷, , 3, 2, , y, , y, , 1, , ⫺1, 1, , 1⫺x, , 2, , 0, , 冋, , y, , 2⫺z, , 0, , ⫺, , y dy dz dx 苷 3 y, , 共2 ⫺ z兲3, 3, , 1, , ⫺1, , 册, , 0, , 2, , 共2 ⫺ z兲2, dz dx, 2, , 1⫺x 2, 1, , dx 苷 ⫺ 2 y 关共x 2 ⫹ 1兲3 ⫺ 8兴 dx, 1, , 0, , 苷 ⫺y 共x 6 ⫹ 3x 4 ⫹ 3x 2 ⫺ 7兲 dx 苷, 0, , y, , 1⫺x, , ⫺1, , 184, 35, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1157, , SECTION 16.9, , THE DIVERGENCE THEOREM, , 1157, , Another application of the Divergence Theorem occurs in fluid flow. Let v共x, y, z兲 be, the velocity field of a fluid with constant density ␳. Then F 苷 ␳ v is the rate of flow per, unit area. If P0共x 0 , y0 , z0 兲 is a point in the fluid and Ba is a ball with center P0 and very small, radius a, then div F共P兲 ⬇ div F共P0 兲 for all points in Ba since div F is continuous. We approximate the flux over the boundary sphere Sa as follows:, , yy F ⴢ dS 苷 yyy div F dV ⬇ yyy div F共P 兲 dV 苷 div F共P 兲V共B 兲, 0, , Sa, , Ba, , 0, , a, , Ba, , This approximation becomes better as a l 0 and suggests that, y, , div F共P0 兲 苷 lim, , 8, , al0, , P¡, , yy F ⴢ dS, Sa, , Equation 8 says that div F共P0 兲 is the net rate of outward flux per unit volume at P0. (This, is the reason for the name divergence.) If div F共P兲 ⬎ 0, the net flow is outward near P and, P is called a source. If div F共P兲 ⬍ 0, the net flow is inward near P and P is called a sink., For the vector field in Figure 4, it appears that the vectors that end near P1 are shorter, than the vectors that start near P1. Thus the net flow is outward near P1, so div F共P1兲 ⬎ 0, and P1 is a source. Near P2 , on the other hand, the incoming arrows are longer than the, outgoing arrows. Here the net flow is inward, so div F共P2 兲 ⬍ 0 and P2 is a sink. We, can use the formula for F to confirm this impression. Since F 苷 x 2 i ⫹ y 2 j, we have, div F 苷 2x ⫹ 2y, which is positive when y ⬎ ⫺x. So the points above the line y 苷 ⫺x, are sources and those below are sinks., , x, , P™, , FIGURE 4, , The vector field F=≈ i+¥ j, , 16.9, , 1, V共Ba 兲, , Exercises, , 1– 4 Verify that the Divergence Theorem is true for the vector field, F on the region E., 1. F共x, y, z兲 苷 3x i ⫹ x y j ⫹ 2 xz k,, , E is the cube bounded by the planes x 苷 0, x 苷 1, y 苷 0,, y 苷 1, z 苷 0, and z 苷 1, , 2. F共x, y, z兲 苷 x i ⫹ x y j ⫹ z k,, , E is the solid bounded by the paraboloid z 苷 4 ⫺ x ⫺ y, and the xy-plane, , S is the surface of the solid bounded by the cylinder, y 2 ⫹ z 2 苷 1 and the planes x 苷 ⫺1 and x 苷 2, 8. F共x, y, z兲 苷 共x 3 ⫹ y 3 兲 i ⫹ 共 y 3 ⫹ z 3 兲 j ⫹ 共z 3 ⫹ x 3 兲 k,, , S is the sphere with center the origin and radius 2, 9. F共x, y, z兲 苷 x 2 sin y i ⫹ x cos y j ⫺ xz sin y k,, , 2, , 2, , 7. F共x, y, z兲 苷 3x y 2 i ⫹ xe z j ⫹ z 3 k,, , 2, , 3. F共x, y, z兲 苷 具 z, y, x典 ,, , E is the solid ball x 2 ⫹ y 2 ⫹ z 2 艋 16, , 4. F共x, y, z兲 苷 具x 2, ⫺y, z典 ,, , E is the solid cylinder y 2 ⫹ z 2 艋 9, 0 艋 x 艋 2, , S is the “fat sphere” x 8 ⫹ y 8 ⫹ z 8 苷 8, , 10. F共x, y, z兲 苷 z i ⫹ y j ⫹ zx k,, , S is the surface of the tetrahedron enclosed by the coordinate, planes and the plane, x, y, z, ⫹ ⫹ 苷1, a, b, c, where a, b, and c are positive numbers, , 5–15 Use the Divergence Theorem to calculate the surface integral, , xxS F ⴢ dS; that is, calculate the flux of F across S., 5. F共x, y, z兲 苷 xye z i ⫹ xy 2z 3 j ⫺ ye z k,, , S is the surface of the box bounded by the coordinate planes, and the planes x 苷 3, y 苷 2, and z 苷 1, 6. F共x, y, z兲 苷 x yz i ⫹ x y z j ⫹ xyz k,, 2, , 2, , 2, , S is the surface of the box enclosed by the planes x 苷 0,, x 苷 a, y 苷 0, y 苷 b, z 苷 0, and z 苷 c, where a, b, and c are, positive numbers, , CAS Computer algebra system required, , 11. F共x, y, z兲 苷 共cos z ⫹ x y 2 兲 i ⫹ xe⫺z j ⫹ 共sin y ⫹ x 2 z兲 k,, , S is the surface of the solid bounded by the paraboloid, z 苷 x 2 ⫹ y 2 and the plane z 苷 4, 12. F共x, y, z兲 苷 x 4 i ⫺ x 3z 2 j ⫹ 4 x y 2z k,, , S is the surface of the solid bounded by the cylinder, x 2 ⫹ y 2 苷 1 and the planes z 苷 x ⫹ 2 and z 苷 0, , ⱍ ⱍ, , 13. F 苷 r r, where r 苷 x i ⫹ y j ⫹ z k,, , S consists of the hemisphere z 苷 s1 ⫺ x 2 ⫺ y 2 and the disk, x 2 ⫹ y 2 艋 1 in the xy-plane, , 1. Homework Hints available at stewartcalculus.com, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1158-1164.qk_97817_16_ch16_p1158-1164 11/9/10 9:26 AM Page 1158, , 1158, , CHAPTER 16, , VECTOR CALCULUS, , ⱍ ⱍ, , 14. F 苷 r 2 r, where r 苷 x i ⫹ y j ⫹ z k,, , S is the sphere with radius R and center the origin, CAS, , 15. F共x, y, z兲 苷 e y tan z i ⫹ y s3 ⫺ x 2 j ⫹ x sin y k,, , 23. Verify that div E 苷 0 for the electric field E共x兲 苷, , ␧Q, , ⱍxⱍ, , 3, , x., , 24. Use the Divergence Theorem to evaluate, , S is the surface of the solid that lies above the xy-plane, and below the surface z 苷 2 ⫺ x 4 ⫺ y 4, ⫺1 艋 x 艋 1,, ⫺1 艋 y 艋 1, , yy 共2x ⫹ 2y ⫹ z, , 2, , 兲 dS, , S, , where S is the sphere x 2 ⫹ y 2 ⫹ z 2 苷 1., CAS, , 16. Use a computer algebra system to plot the vector field, , F共x, y, z兲 苷 sin x cos 2 y i ⫹ sin 3 y cos 4z j ⫹ sin 5z cos 6x k, in the cube cut from the first octant by the planes x 苷 ␲兾2,, y 苷 ␲兾2, and z 苷 ␲兾2. Then compute the flux across the, surface of the cube., 17. Use the Divergence Theorem to evaluate xxS F ⴢ dS, where, , F共x, y, z兲 苷 z 2 x i ⫹ ( 13 y 3 ⫹ tan z) j ⫹ 共x 2z ⫹ y 2 兲 k, and S is the top half of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 1., [Hint: Note that S is not a closed surface. First compute, integrals over S1 and S2, where S1 is the disk x 2 ⫹ y 2 艋 1,, oriented downward, and S2 苷 S 傼 S1.], ⫺1, , 25–30 Prove each identity, assuming that S and E satisfy the, conditions of the Divergence Theorem and the scalar functions, and components of the vector fields have continuous secondorder partial derivatives., 25., , 3, , 26. V共E 兲 苷, , 19. A vector field F is shown. Use the interpretation of diver-, , gence derived in this section to determine whether div F, is positive or negative at P1 and at P2., 2, , 1, 3, , yy F ⴢ dS,, , where F共x, y, z兲 苷 x i ⫹ y j ⫹ z k, , S, , 27., , 2, , Find the flux of F across the part of the paraboloid, x 2 ⫹ y 2 ⫹ z 苷 2 that lies above the plane z 苷 1 and is, oriented upward., , where a is a constant vector, , S, , 18. Let F共x, y, z兲 苷 z tan 共 y 兲 i ⫹ z ln共x ⫹ 1兲 j ⫹ z k., 2, , yy a ⴢ n dS 苷 0,, , yy curl F ⴢ dS 苷 0, , 28., , S, , 29., , n, , f dS 苷 yyy ⵜ 2 f dV, , S, , E, , yy 共 f ⵜt兲 ⴢ n dS 苷 yyy 共 f ⵜ t ⫹ ⵜ f ⴢ ⵜt兲 dV, 2, , S, , 30., , yy D, , E, , yy 共 f ⵜt ⫺ t ⵜ f 兲 ⴢ n dS 苷 yyy 共 f ⵜ t ⫺ t ⵜ, 2, , S, , 2, , f 兲 dV, , E, , 31. Suppose S and E satisfy the conditions of the Divergence, , Theorem and f is a scalar function with continuous partial, derivatives. Prove that, , P¡, _2, , 2, P™, , S, , _2, , 20. (a) Are the points P1 and P2 sources or sinks for the vector, , field F shown in the figure? Give an explanation based, solely on the picture., (b) Given that F共x, y兲 苷 具x, y 2 典 , use the definition of divergence to verify your answer to part (a)., 2, P¡, _2, , yy f n dS 苷 yyy ⵜ f dV, E, , These surface and triple integrals of vector functions are, vectors defined by integrating each component function., [Hint: Start by applying the Divergence Theorem to F 苷 f c,, where c is an arbitrary constant vector.], 32. A solid occupies a region E with surface S and is immersed, , in a liquid with constant density ␳. We set up a coordinate, system so that the xy-plane coincides with the surface of the, liquid, and positive values of z are measured downward into, the liquid. Then the pressure at depth z is p 苷 ␳ tz, where t, is the acceleration due to gravity (see Section 8.3). The total, buoyant force on the solid due to the pressure distribution is, given by the surface integral, , 2, , F 苷 ⫺yy pn dS, , P™, , S, , _2, CAS, , 21–22 Plot the vector field and guess where div F ⬎ 0 and, , where div F ⬍ 0 . Then calculate div F to check your guess., 21. F共x, y兲 苷 具xy, x ⫹ y 2 典, , 22. F共x, y兲 苷 具x 2, y 2 典, , where n is the outer unit normal. Use the result of Exercise 31 to show that F 苷 ⫺W k, where W is the weight of, the liquid displaced by the solid. (Note that F is directed, upward because z is directed downward.) The result is, Archimedes’ Principle: The buoyant force on an object, equals the weight of the displaced liquid., , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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SUMMARY, , SECTION 16.10, , 1159, , 16.10 Summary, The main results of this chapter are all higher-dimensional versions of the Fundamental, Theorem of Calculus. To help you remember them, we collect them together here (without hypotheses) so that you can see more easily their essential similarity. Notice that in, each case we have an integral of a “derivative” over a region on the left side, and the right, side involves the values of the original function only on the boundary of the region., , Fundamental Theorem of Calculus, , y, , b, , a, , F⬘共x兲 dx 苷 F共b兲 ⫺ F共a兲, , a, , b, , r(b), , Fundamental Theorem for Line Integrals, , y, , C, , ⵜf ⴢ dr 苷 f 共r共b兲兲 ⫺ f 共r共a兲兲, , C, , r(a), , Green’s Theorem, , yy, D, , 冉, , ⭸Q, ⭸P, ⫺, ⭸x, ⭸y, , 冊, , C, , dA 苷 y P dx ⫹ Q dy, , D, , C, , n, , Stokes’ Theorem, , yy curl F ⴢ dS 苷 y, , C, , F ⴢ dr, , S, , S, , C, , n, S, , Divergence Theorem, , yyy div F dV 苷 yy F ⴢ dS, E, , S, , E, , n, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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1160, , 16, , CHAPTER 16, , VECTOR CALCULUS, , Review, , Concept Check, 1. What is a vector field? Give three examples that have physical, , meaning., , 10. If F 苷 P i ⫹ Q j, how do you test to determine whether F is, , 2. (a) What is a conservative vector field?, , conservative? What if F is a vector field on ⺢3 ?, , (b) What is a potential function?, 3. (a) Write the definition of the line integral of a scalar function, , (b), (c), , (d), (e), , (c) If F is a velocity field in fluid flow, what are the physical, interpretations of curl F and div F ?, , f along a smooth curve C with respect to arc length., How do you evaluate such a line integral?, Write expressions for the mass and center of mass of a thin, wire shaped like a curve C if the wire has linear density, function ␳ 共x, y兲., Write the definitions of the line integrals along C of a scalar, function f with respect to x, y, and z., How do you evaluate these line integrals?, , 4. (a) Define the line integral of a vector field F along a smooth, , curve C given by a vector function r共t兲., (b) If F is a force field, what does this line integral represent?, (c) If F 苷 具P, Q, R 典 , what is the connection between the line, integral of F and the line integrals of the component functions P, Q, and R?, 5. State the Fundamental Theorem for Line Integrals., 6. (a) What does it mean to say that xC F ⴢ dr is independent, , of path?, (b) If you know that xC F ⴢ dr is independent of path, what can, you say about F ?, , 11. (a) What is a parametric surface? What are its grid curves?, , (b) Write an expression for the area of a parametric surface., (c) What is the area of a surface given by an equation, z 苷 t共x, y兲?, 12. (a) Write the definition of the surface integral of a scalar func-, , tion f over a surface S., (b) How do you evaluate such an integral if S is a parametric, surface given by a vector function r共u, v兲?, (c) What if S is given by an equation z 苷 t共x, y兲?, (d) If a thin sheet has the shape of a surface S, and the density, at 共x, y, z兲 is ␳ 共x, y, z兲, write expressions for the mass and, center of mass of the sheet., 13. (a) What is an oriented surface? Give an example of a non-, , orientable surface., (b) Define the surface integral (or flux) of a vector field F over, an oriented surface S with unit normal vector n., (c) How do you evaluate such an integral if S is a parametric, surface given by a vector function r共u, v兲?, (d) What if S is given by an equation z 苷 t共x, y兲?, , 7. State Green’s Theorem., , 14. State Stokes’ Theorem., , 8. Write expressions for the area enclosed by a curve C in terms, , 15. State the Divergence Theorem., , of line integrals around C., , 16. In what ways are the Fundamental Theorem for Line Integrals,, , 9. Suppose F is a vector field on ⺢3., , (a) Define curl F., , (b) Define div F., , Green’s Theorem, Stokes’ Theorem, and the Divergence, Theorem similar?, , True-False Quiz, Determine whether the statement is true or false. If it is true, explain why., If it is false, explain why or give an example that disproves the statement., 1. If F is a vector field, then div F is a vector field., 2. If F is a vector field, then curl F is a vector field., 3. If f has continuous partial derivatives of all orders on ⺢ 3, then, , div共curl ⵜ f 兲 苷 0., , 8. The work done by a conservative force field in moving a par-, , ticle around a closed path is zero., 9. If F and G are vector fields, then, , curl共F ⫹ G兲 苷 curl F ⫹ curl G, 10. If F and G are vector fields, then, , 4. If f has continuous partial derivatives on ⺢ and C is any, , curl共F ⴢ G兲 苷 curl F ⴢ curl G, , 3, , circle, then xC ⵜ f ⴢ dr 苷 0., , 5. If F 苷 P i ⫹ Q j and Py 苷 Q x in an open region D, then F is, , conservative., 6., , x⫺C, , f 共x, y兲 ds 苷 ⫺xC f 共x, y兲 ds, , 7. If F and G are vector fields and div F 苷 div G, then F 苷 G., , 11. If S is a sphere and F is a constant vector field, then, , xxS F ⴢ dS 苷 0., 12. There is a vector field F such that, , curl F 苷 x i ⫹ y j ⫹ z k, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1158-1164.qk_97817_16_ch16_p1158-1164 11/9/10 9:26 AM Page 1162, , 1162, , CHAPTER 16, , VECTOR CALCULUS, , 25. Find the area of the part of the surface z 苷 x 2 ⫹ 2y that lies, , above the triangle with vertices 共0, 0兲, 共1, 0兲, and 共1, 2兲., , 26. (a) Find an equation of the tangent plane at the point, , 共4, ⫺2, 1兲 to the parametric surface S given by, r共u, v兲 苷 v 2 i ⫺ u v j ⫹ u 2 k, , 0 艋 u 艋 3, ⫺3 艋 v 艋 3, , 37. Let, , F共x, y, z兲 苷 共3x 2 yz ⫺ 3y兲 i ⫹ 共x 3 z ⫺ 3x兲 j ⫹ 共x 3 y ⫹ 2z兲 k, Evaluate xC F ⴢ dr, where C is the curve with initial point, 共0, 0, 2兲 and terminal point 共0, 3, 0兲 shown in the figure., z, , (b) Use a computer to graph the surface S and the tangent, plane found in part (a)., (c) Set up, but do not evaluate, an integral for the surface, area of S., (d) If, z2, x2, y2, F共x, y, z兲 苷, k, 2 i ⫹, 2 j ⫹, 1⫹x, 1⫹y, 1 ⫹ z2, , ;, , CAS, , (0, 0, 2), , 0, , (0, 3, 0), (1, 1, 0), , y, , (3, 0, 0), , find xxS F ⴢ dS correct to four decimal places., , x, , 27–30 Evaluate the surface integral., , 38. Let, , 27., , where S is the part of the paraboloid z 苷 x 2 ⫹ y 2, that lies under the plane z 苷 4, , F共x, y兲 苷, , 28., , xxS 共x 2 z ⫹ y 2 z兲 dS,, , Evaluate x䊊C F ⴢ dr, where C is shown in the figure., , 29., , xxS F ⴢ dS,, , 30., , xxS z dS,, , where S is the part of the plane, z 苷 4 ⫹ x ⫹ y that lies inside the cylinder x 2 ⫹ y 2 苷 4, , 共2 x 3 ⫹ 2 x y 2 ⫺ 2y兲 i ⫹ 共2y 3 ⫹ 2 x 2 y ⫹ 2 x兲 j, x2 ⫹ y2, , y, , where F共x, y, z兲 苷 x z i ⫺ 2y j ⫹ 3x k and S is, the sphere x 2 ⫹ y 2 ⫹ z 2 苷 4 with outward orientation, , C, , xxS F ⴢ dS, where F共x, y, z兲 苷 x 2 i ⫹ x y j ⫹ z k and S is the, part of the paraboloid z 苷 x 2 ⫹ y 2 below the plane z 苷 1, with upward orientation, , x, , 0, , 31. Verify that Stokes’ Theorem is true for the vector field, , F共x, y, z兲 苷 x 2 i ⫹ y 2 j ⫹ z 2 k, where S is the part of the, paraboloid z 苷 1 ⫺ x 2 ⫺ y 2 that lies above the xy-plane and, S has upward orientation., 32. Use Stokes’ Theorem to evaluate xxS curl F ⴢ dS, where, , 39. Find xxS F ⴢ n dS, where F共x, y, z兲 苷 x i ⫹ y j ⫹ z k and S is, , the outwardly oriented surface shown in the figure (the boundary surface of a cube with a unit corner cube removed)., , F共x, y, z兲 苷 x 2 yz i ⫹ yz 2 j ⫹ z 3e xy k, S is the part of the, sphere x 2 ⫹ y 2 ⫹ z 2 苷 5 that lies above the plane z 苷 1,, and S is oriented upward., , z, , (0, 2, 2), , 33. Use Stokes’ Theorem to evaluate xC F ⴢ dr, where, , F共x, y, z兲 苷 x y i ⫹ yz j ⫹ z x k, and C is the triangle with, vertices 共1, 0, 0兲, 共0, 1, 0兲, and 共0, 0, 1兲, oriented counterclockwise as viewed from above., , (2, 0, 2), , 1, , 34. Use the Divergence Theorem to calculate the surface, , integral xxS F ⴢ dS, where F共x, y, z兲 苷 x i ⫹ y j ⫹ z k and, S is the surface of the solid bounded by the cylinder, x 2 ⫹ y 2 苷 1 and the planes z 苷 0 and z 苷 2., 3, , 3, , 35. Verify that the Divergence Theorem is true for the vector, , field F共x, y, z兲 苷 x i ⫹ y j ⫹ z k, where E is the unit ball, x 2 ⫹ y 2 ⫹ z 2 艋 1., 36. Compute the outward flux of, , xi⫹yj⫹zk, F共x, y, z兲 苷 2, 共x ⫹ y 2 ⫹ z 2 兲 3兾2, through the ellipsoid 4 x 2 ⫹ 9y 2 ⫹ 6z 2 苷 36., , 1, , 3, , 1, y, , S, , x, , (2, 2, 0), , 40. If the components of F have continuous second partial, , derivatives and S is the boundary surface of a simple solid, region, show that xxS curl F ⴢ dS 苷 0., 41. If a is a constant vector, r 苷 x i ⫹ y j ⫹ z k, and S is an, , oriented, smooth surface with a simple, closed, smooth, positively oriented boundary curve C, show that, , yy 2a ⴢ dS 苷 y, , C, , 共a ⫻ r兲 ⴢ dr, , S, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Problems Plus, 1. Let S be a smooth parametric surface and let P be a point such that each line that starts, , at P intersects S at most once. The solid angle ⍀共S 兲 subtended by S at P is the set of lines, starting at P and passing through S. Let S共a兲 be the intersection of ⍀共S 兲 with the surface of, the sphere with center P and radius a. Then the measure of the solid angle (in steradians) is, defined to be, area of S共a兲, ⍀共S 兲 苷, a2, , ⱍ, , ⱍ, , Apply the Divergence Theorem to the part of ⍀共S 兲 between S共a兲 and S to show that, , ⱍ ⍀共S 兲 ⱍ 苷 yy, S, , rⴢn, dS, r3, , ⱍ ⱍ, , where r is the radius vector from P to any point on S, r 苷 r , and the unit normal vector n, is directed away from P., This shows that the definition of the measure of a solid angle is independent of the radius a, of the sphere. Thus the measure of the solid angle is equal to the area subtended on a unit, sphere. (Note the analogy with the definition of radian measure.) The total solid angle subtended by a sphere at its center is thus 4␲ steradians., S, S(a), , P, , a, , 2. Find the positively oriented simple closed curve C for which the value of the line integral, , y, , C, , 共 y 3 ⫺ y兲 dx ⫺ 2x 3 dy, , is a maximum., 3. Let C be a simple closed piecewise-smooth space curve that lies in a plane with unit normal, , vector n 苷 具a, b, c典 and has positive orientation with respect to n. Show that the plane area, enclosed by C is, 1, 2, , y, , C, , 共bz ⫺ cy兲 dx ⫹ 共cx ⫺ az兲 dy ⫹ 共ay ⫺ bx兲 dz, , ; 4. Investigate the shape of the surface with parametric equations x 苷 sin u, y 苷 sin v,, , z 苷 sin共u ⫹ v兲. Start by graphing the surface from several points of view. Explain the, appearance of the graphs by determining the traces in the horizontal planes z 苷 0, z 苷 ⫾1,, and z 苷 ⫾ 12., , 5. Prove the following identity:, , ⵜ共F ⴢ G兲 苷 共F ⴢ ⵜ兲G ⫹ 共G ⴢ ⵜ兲F ⫹ F ⫻ curl G ⫹ G ⫻ curl F, , ;, , Graphing calculator or computer required, , 1163, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97817_16_ch16_p1158-1164.qk_97817_16_ch16_p1158-1164 11/9/10 9:26 AM Page 1164, , 6. The figure depicts the sequence of events in each cylinder of a four-cylinder internal combus-, , stio, n, Ex, hau, , Ex, plo, sio, n, , pre, ssi, on, Co, m, , Int, ake, , tion engine. Each piston moves up and down and is connected by a pivoted arm to a rotating, crankshaft. Let P共t兲 and V共t兲 be the pressure and volume within a cylinder at time t, where, a 艋 t 艋 b gives the time required for a complete cycle. The graph shows how P and V vary, through one cycle of a four-stroke engine., , P, , $, , Water, , #, , C, %, , Crankshaft, Connecting rod, Flywheel, , !, 0, , @, V, , During the intake stroke (from ① to ②) a mixture of air and gasoline at atmospheric pressure is drawn into a cylinder through the intake valve as the piston moves downward. Then, the piston rapidly compresses the mix with the valves closed in the compression stroke (from, ② to ③) during which the pressure rises and the volume decreases. At ③ the sparkplug ignites, the fuel, raising the temperature and pressure at almost constant volume to ④. Then, with, valves closed, the rapid expansion forces the piston downward during the power stroke (from, ④ to ⑤). The exhaust valve opens, temperature and pressure drop, and mechanical energy, stored in a rotating flywheel pushes the piston upward, forcing the waste products out of the, exhaust valve in the exhaust stroke. The exhaust valve closes and the intake valve opens., We’re now back at ① and the cycle starts again., (a) Show that the work done on the piston during one cycle of a four-stroke engine is, W 苷 xC P dV, where C is the curve in the PV-plane shown in the figure., [Hint: Let x共t兲 be the distance from the piston to the top of the cylinder and note that, the force on the piston is F 苷 AP共t兲 i, where A is the area of the top of the piston. Then, W 苷 xC F ⴢ dr, where C1 is given by r共t兲 苷 x共t兲 i, a 艋 t 艋 b. An alternative approach is, to work directly with Riemann sums.], (b) Use Formula 16.4.5 to show that the work is the difference of the areas enclosed by the, two loops of C., 1, , 1164, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Ans7eMV_Ans7eMV_pA024-A033.qk_97879_Ans7eMV_Ans7eMV_pA024-A033 11/10/10 3:04 PM Page 29, , ANSWERS TO ODD-NUMBERED EXERCISES, , APPENDIX H, , 17. v共t兲 苷 共1  ln t兲 i  j  et k,, , 9. 具2t  1, 2t  1, 3t 2 典 , 具2, 2, 6t典 , s9t 4  8t 2  2, , ⱍ v共t兲 ⱍ 苷 s2  2 ln t  共ln t兲, , 11. s2 i  e t j  et k, e t j  et k, e t  et, 13. e t 关共cos t  sin t兲 i  共sin t  cos t兲 j  共t  1兲 k兴,, ,  e2t , a共t兲 苷 共1兾t兲 i  et k, 19. (a) About 3.8 ft above the ground, 60.8 ft from the athlete, (b) ⬇21.4 ft, (c) ⬇64.2 ft from the athlete, 21. (c) 2et vd  et R, 2, , e t 关2 sin t i  2 cos t j  共t  2兲 k兴, e tst 2  2t  3, 1, 15. v共t兲 苷 t i  2t j  k, r共t兲 苷 ( 2 t 2  1兲 i  t 2 j  t k, 1 3, 1, 1, 17. (a) r共t兲 苷 ( 3 t  t ) i  共t  sin t  1兲 j  ( 4  4 cos 2t) k, (b), , 23. (a) v 苷 R共sin  t i  cos  t j兲, PROBLEMS PLUS, , z, , N, , (c) a 苷  2 r, , PAGE 900, , 1. (a) 90, v02兾共2t兲, 3. (a) ⬇0.94 ft to the right of the table’s edge, ⬇15 ft兾s, , 0.6, 0.4, 0.2, 0, , _10, 200, x, , 19., 23., 25., 29., 31., 33., , A29, , 0, , 0 y, _200, , 10, , ⱍ, , ⱍ, , t苷4, 21. r共t兲 苷 t i  t j  2 t 2 k, v共t兲 苷 s25t 2  2, (a) ⬇3535 m, (b) ⬇1531 m, (c) 200 m兾s, 27. ⬇10.2, ⬇79.8, 30 m兾s, 13.0, 36.0, 55.4, 85.5, 共250, 50, 0兲; 10s93 ⬇ 96.4 ft兾s, (a) 16 m, (b) ⬇23.6 upstream, 5, , 12, , 20, , 40, , 0, 40, , 0, , _12, , _4, , 35. The path is contained in a circle that lies in a plane perpen-, , dicular to c with center on a line through the origin in the direction, of c., 37. 6t, 6, 39. 0, 1, 41. e t  et, s2, 2, 2, 43. 4.5 cm兾s , 9.0 cm兾s, 45. t 苷 1, CHAPTER 13 REVIEW, , True-False Quiz, 1. True, 3. False, 9. True, 11. False, Exercises, 1. (a), , N, , PAGE 897, , 5. False, 13. True, , 7. False, , z, , (b) ⬇7.6 (c) ⬇2.13 ft to the right of the table’s edge, 5. 56, 7. r共u, v兲 苷 c  u a  v b where a 苷 具 a1, a 2 , a 3 典,, b 苷 具b1, b 2 , b 3 典, c 苷 具 c1, c 2 , c 3 典, CHAPTER 14, EXERCISES 14.1, , N, , PAGE 912, , 1. (a) 27; a temperature of 15C with wind blowing at, 40 km兾h feels equivalent to about 27C without wind., (b) When the temperature is 20C, what wind speed gives a wind, chill of 30C ? 20 km兾h, (c) With a wind speed of 20 km兾h, what temperature gives a wind, chill of 49C ? 35C, (d) A function of wind speed that gives wind-chill values when the, temperature is 5C, (e) A function of temperature that gives wind-chill values when the, wind speed is 50 km兾h, 3. ⬇94.2; the manufacturer’s yearly production is valued at $94.2, million when 120,000 labor hours are spent and $20 million in, capital is invested., 5. (a) ⬇20.5; the surface area of a person 70 inches tall who, weighs 160 pounds is approximately 20.5 square feet., 7. (a) 25; a 40-knot wind blowing in the open sea for 15 h will, create waves about 25 ft high., (b) f 共30, t兲 is a function of t giving the wave heights produced by, 30-knot winds blowing for t hours., (c) f 共v, 30兲 is a function of v giving the wave heights produced by, winds of speed v blowing for 30 hours., (b) ⺢ 2 (c) 关1, 1兴, 9. (a) 1, (b) 兵共x, y, z兲 x 2  y 2  z 2 4, x  0, y  0, z  0其,, 11. (a) 3, interior of a sphere of radius 2, center the origin, in the first octant, 13. 兵共x, y兲 y  2x其, y, , ⱍ, , (0, 1, 0), , ⱍ, , y, x, , (2, 1, 0), 0, , (b) r 共t兲 苷 i   sin  t j   cos  t k,, r 共t兲 苷  2 cos  t j   2 sin  t k, 3. r共t兲 苷 4 cos t i  4 sin t j  共5  4 cos t兲 k, 0  t  2, 1, 5. 3 i  共2兾 2兲 j  共2兾 兲 k, 7. 86.631, 9. 兾2, 11. (a) 具 t 2, t, 1 典兾st 4  t 2  1, (b) 具t 3  2t, 1  t 4, 2t 3  t典兾st 8  5t 6  6t 4  5t 2  1, (c) st 8  5t 6  6t 4  5t 2  1兾共t 4  t 2  1兲2, 13. 12兾17 3兾2, 15. x  2y  2 苷 0, , x, , y=2x, , 15. 兵共x, y兲, , ⱍ, , 1, 9, , x2  y2, , 1其, 共, ln 9兴, , y, 1, 9 ≈+¥=1, , 0, , x, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Ans7eMV_Ans7eMV_pA024-A033.qk_97879_Ans7eMV_Ans7eMV_pA024-A033 11/10/10 3:04 PM Page 30, , A30, , APPENDIX H, , ANSWERS TO ODD-NUMBERED EXERCISES, , ⱍ, , 17. 兵共x, y兲 1  x  1, 1  y  1其, , 29. z 苷 9  x 2  9y 2,, , y, , z, , elliptic paraboloid, , (0, 0, 9), , 1, , _1, , 0, , 1, , x, , (0, 1, 0), , (3, 0, 0), _1, , ⱍ, , 19. 兵共x, y兲 y  x 2, x 苷 1其, , x, , y, , y, , 31. z 苷 s4  4x 2  y 2 ,, , y=≈, , z, , top half of ellipsoid, _1, , ⱍ, , 21. 兵共x, y, z兲 x 2  y 2  z 2  1其, , 0, , (0, 0, 2), , x, , 1, , (0, 2, 0), , (1, 0, 0), , z, , y, , x, , 0, y, , 33. ⬇56, ⬇35, , 35. 11C, 19.5C, , 37. Steep; nearly flat, , x, , 39., 23. z 苷 1  y, plane parallel to x-axis, , 41., , z, , z, 5, , z, , 14, , (0, 0, 1), , y, x, , 0, , y, , x, , (0, _1, 0), , x, , y, , 25. 4x  5y  z 苷 10, plane, , 43. 共 y  2x兲 2 苷 k, , z, , 45. y 苷 sx  k, y, , y, (0, 0, 10), , x, 2, 1, , x, , 0, _1, 43 2 1, , 0, , 0, , _2, , 1 234, , (0, 2, 0), , (2.5, 0, 0), y, , x, , 47. y 苷 kex, , 49. y 2  x 2 苷 k 2, , y, , 27. z 苷 y 2  1 , parabolic cylinder, , z, 1 2, , y, 3, , 3, , 2, 1, 0, , 0, 0, , _2, _1, x, , y, , x, , 0, , x, , 1, 2, _3, , 3, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Ans7eMV_Ans7eMV_pA024-A033.qk_97879_Ans7eMV_Ans7eMV_pA024-A033 11/10/10 3:04 PM Page 31, , ANSWERS TO ODD-NUMBERED EXERCISES, , APPENDIX H, , 51. x 2  9y 2 苷 k, , 73., , 10, , z, , y, , 5, , z=4, 2, , 1, , 3, , 4, , 0, , z, x, , z=3, , 0, _5, , _10, 2, , z=2, , x 0, , y, x, y, , 0, , x, , _2, y, , The function values approach 0 as x, y become large; as 共x, y兲, approaches the origin, f approaches  or 0, depending on the, direction of approach., 75. If c 苷 0, the graph is a cylindrical surface. For c  0, the level, curves are ellipses. The graph curves upward as we leave the origin, and the steepness increases as c increases. For c 0, the level, curves are hyperbolas. The graph curves upward in the y-direction, and downward, approaching the xy-plane, in the x-direction giving, a saddle-shaped appearance near (0, 0, 1)., 77. c 苷 2, 0, 2, 79. (b) y 苷 0.75x  0.01, EXERCISES 14.2, , 55., , 0, , _2 2, , z=1, , 53., , A31, , PAGE 923, , N, , 1. Nothing; if f is continuous, f 共3, 1兲 苷 6, 3.  2, 2, 7. 7, 9. Does not exist, 11. Does not exist, 5. 1, 13. 0, 15. Does not exist, 17. 2, 19. s3, 21. Does not exist, 23. The graph shows that the function approaches different num5, , z 0, , _2, , y, , 0, , 2 2, , 0, , _2, , x, , bers along different lines., 25. h共x, y兲 苷 共2 x  3y  6兲 2  s2x  3y  6 ;, 兵共x, y兲 2x  3y  6其, 27. Along the line y 苷 x, 29. ⺢ 2, 31. 兵共x, y兲 x 2  y 2 苷 1其, , 57., , ⱍ, , 1.0, , ⱍ, ⱍ, , ⱍ, , 33. 兵共x, y兲 x 2  y 2  4其, , z 0.5, , 37. 兵共x, y兲 共x, y兲 苷 共0, 0兲其, , 0.0, , ⱍ, , 35. 兵共x, y, z兲 x 2  y 2  z 2  1其, 39. 0, , 41. 1, , 43., , _4, y 0, , 0, , 4 4, , _4, x, , z, , (a) C (b) II, (b) I, 61. (a) F, (a) B (b) VI, 65. Family of parallel planes, Family of circular cylinders with axis the x-axis 共k  0兲, (a) Shift the graph of f upward 2 units, (b) Stretch the graph of f vertically by a factor of 2, (c) Reflect the graph of f about the xy-plane, (d) Reflect the graph of f about the xy-plane and then shift it, upward 2 units, , 59., 63., 67., 69., , 71., 20, 0, z, _20, _40, , _5, y, , 0, , 5, , 5, , 0x, , _5, , f appears to have a maximum value of about 15. There are two, local maximum points but no local minimum point., , 2, 1, 0, _1, , _2, _2, y, , 0, , 2, , 2, , 0x, , f is continuous on ⺢ 2, EXERCISES 14.3, , N, , PAGE 935, , 1. (a) The rate of change of temperature as longitude varies, with, , latitude and time fixed; the rate of change as only latitude varies;, the rate of change as only time varies., (b) Positive, negative, positive, 3. (a) fT 共15, 30兲 ⬇ 1.3; for a temperature of 15C and wind, speed of 30 km兾h, the wind-chill index rises by 1.3C for each, degree the temperature increases. fv 共15, 30兲 ⬇ 0.15; for a, temperature of 15C and wind speed of 30 km兾h, the wind-chill, index decreases by 0.15C for each km兾h the wind speed, increases., (b) Positive, negative (c) 0, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Ans7eMV_Ans7eMV_pA034-A042.qk_97879_Ans7eMV_Ans7eMV_pA034-A043 11/10/10 3:04 PM Page 38, , A38, 1, 15, , 45. (a), 47., , ANSWERS TO ODD-NUMBERED EXERCISES, , APPENDIX H, , 1, 0, , 1, 3, , (b), , 1z, 0, , x x, , 3., , 13. abc, , 冉, , 1, 2, , 19., , 1, 45, , f 共x, y, z兲 dx dy dz, , x, , sy, sy, , PROBLEMS PLUS, , 1. 30, , (c), , N, , 49. ln 2, , 51. 0, 4.5, , PAGE 1077, , sin 1, , 2, 8, , 3, 9s3, , 冊, , The line y 苷 2x, , 4.5, , 4.5, , 7. (b) 0.90, 4.5, , 21. f 共x, y兲 苷 共xy  1兲e xy i  x 2e xy j, , x, i, sx 2  y 2  z 2, y, z, , j, k, sx 2  y 2  z 2, sx 2  y 2  z 2, 25. f 共x, y兲 苷 2x i  j, 23.  f 共x, y, z兲 苷, , CHAPTER 16, EXERCISES 16.1, , N, , 1., , y, , PAGE 1085, , 2, , 2, , _6, , _4, , _2, , 0, , _1, , 0, , 4, , 6, , x, , _2, , 1, , _2, , y, , x, , 1, , 27., , 4, , _1, , _4, , 4, , 3., y, , 2, , _4, , 29. III, 35. (a), _2, , 2, , 33. 共2.04, 1.03兲, , 31. II, , (b) y 苷 1兾x, x  0, , y, , x, , _2, x, , 0, , 5., , y, , y 苷 C兾x, 0, , x, , EXERCISES 16.2, , 7., , 9., , z, , z, , 27. 3, , 11. IV, , 13. I, , 15. IV, , PAGE 1096, , 3兾2, , 3, 2, , 35, 3, , 2.5, , x, , y, , x, , N, , 243, 5, 共145  1兲, 3. 1638.4, 5. 8, 7. 2, 1, 2, 6, 9. s5, 11. 12 s14 共e  1兲, 13. 5 共e  1兲, 15., 17. (a) Positive, (b) Negative, 19. 45, 6, 21. 5  cos 1  sin 1, 23. 1.9633, 25. 15.0074, , 1., , 1, 54, , y, , 17. III, , 2.5, , 2, , .5, , 2.5, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Ans7eMV_Ans7eMV_pA034-A042.qk_97879_Ans7eMV_Ans7eMV_pA034-A043 11/10/10 3:04 PM Page 39, , ANSWERS TO ODD-NUMBERED EXERCISES, , APPENDIX H, , 29. (a), , 11, 8, ,  1兾e, , (b), , EXERCISES 16.6, , 2.1, , PAGE 1132, , 1. P: no; Q: yes, 3. Plane through 共0, 3, 1兲 containing vectors 具 1, 0, 4典 , 具 1, 1, 5典, 5. Hyperbolic paraboloid, 7., , F{r(1)}, , 1, , F ”r” œ„2 ’’, , 2, , F{r(0)}, , 0, , N, , A39, , √ constant, , 2.1, , _0.2, z 0, , 31., , 172,704, 5,632,705, , 14, , s2 共1  e, , 兲, , 0, , 33. 2 k, 共4兾 , 0兲, , 35. (a) x 苷 共1兾m兲 xC x 共x, y, z兲 ds,, , _2, , y 苷 共1兾m兲 xC y 共x, y, z兲 ds,, z 苷 共1兾m兲 xC z 共x, y, z兲 ds, where m 苷, 37. Ix 苷 k (, , y, , u constant, , 1, , 9., , , , 4, 3, , ), Iy 苷 k (, , , , 1, 2, , 2, 3, , ), , 1, , xC  共x, y, z兲 ds, , (b) 共0, 0, 3 兲, 1, 2, , x, 0, , 39. 2, , 2, , 41., , 43. (a) 2ma i  6mbt j, 0, t 1, (b) 2ma 2  2 mb, 4, 45. ⬇1.67  10 ft-lb, 47. (b) Yes, 51. ⬇22 J, 9, , 7, 3, 2, , u constant, , 1, , √ constant, z 0, , EXERCISES 16.3, , N, , PAGE 1106, , 1. 40, 3. f 共x, y兲 苷 x 2  3xy  2y 2  8y  K, 5. Not conservative, 7. f 共x, y兲 苷 ye x  x sin y  K, 2 3, 9. f 共x, y兲 苷 x ln y  x y  K, 1, 13. (a) f 共x, y兲 苷 2 x 2 y 2, (b) 2, 11. (b) 16, 2, 15. (a) f 共x, y, z兲 苷 xyz  z, (b) 77, 17. (a) f 共x, y, z兲 苷 ye xz, 19. 4兾e, (b) 4, 21. It doesn’t matter which curve is chosen., 23. 30, 25. No, 27. Conservative, (b) Yes (c) Yes, 31. (a) Yes, (b) Yes (c) Yes, 33. (a) No, , _1, , _1, _1, y, , 0, , 0, , 1 1, , x, , 11., 1, , z 0, , √ constant, _1, _1, , EXERCISES 16.4, , 1. 8, , 3., , N, , 2, 3, , PAGE 1113, , 5. 12, , 7., , 15. 8e  48e1, , 13. 4, , y, 1, 3, , 9. 24, 17.  12, 1, , 11. , 19. 3, , _1, , 0, , 0, 1 1, , u constant, , 16, 3, , 21. (c), , 23. 共4a兾3 , 4a兾3 兲 if the region is the portion of the disk, x 2  y 2 苷 a 2 in the first quadrant, 27. 0, , x, , 9, 2, , IV, 15. II, 17. III, x 苷 u, y 苷 v  u, z 苷 v, y 苷 y, z 苷 z, x 苷 s1  y 2  14 z 2, x 苷 2 sin  cos , y 苷 2 sin  sin ,, z 苷 2 cos , 0 , 兾4, 0  2, or x 苷 x, y 苷 y, z 苷 s4  x 2  y 2, x 2  y 2 2, 25. x 苷 x, y 苷 4 cos , z 苷 4 sin , 0, x 5, 0 , 13., 19., 21., 23., , [, EXERCISES 16.5, , N, , PAGE 1121, , 1. (a) 0, (b) 3, 3. (a) ze x i  共xye z  yze x 兲 j  xe z k, , (b) 2兾sx 2  y 2  z 2, 5. (a) 0, 7. (a) 具e y cos z, e z cos x, e x cos y典, , (b) y共e z  e x 兲, , (b) e sin y  e sin z  e sin x, 9. (a) Negative, (b) curl F 苷 0, (b) curl F points in the negative z-direction, 11. (a) Zero, 13. f 共x, y, z兲 苷 xy 2z 3  K, 15. Not conservative, 17. f 共x, y, z兲 苷 xe yz  K, 19. No, x, , y, , ], , 29. x 苷 x, y 苷 ex cos ,, , z 苷 ex sin , 0, 0  2, , x, , 2, , 1, , 3,, z 0, , z, , 1, 1, y, , 31. (a) Direction reverses, , 0, , 1 0, , x, , 2, , (b) Number of coils doubles, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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Index, RP, , denotes Reference Page numbers., , absolute maximum and minimum values, 970,, 975, absolute value, A6, absolutely convergent series, 756, acceleration of a particle, 887, components of, 890, as a vector, 887, addition of vectors, 816, 819, Airy, Sir George, 770, Airy function, 770, alternating harmonic series, 753, 756, alternating series, 751, Alternating Series Estimation Theorem, 754, Alternating Series Test, 751, angle(s),, between planes, 845, between vectors, 825, 826, angular momentum, 895, angular speed, 888, aphelion, 707, apolune, 701, approximation, linear, 941, 945, linear, to a tangent plane, 941, by Taylor polynomials, 792, by Taylor’s Inequality, 780, 793, Archimedes’ Principle, 1158, arc curvature, 877, arc length, 878, of a parametric curve, 672, of a polar curve, 691, of a space curve, 877, 878, area,, by Green’s Theorem, 1111, enclosed by a parametric curve, 671, in polar coordinates, 678, 689, of a sector of a circle, 689, surface, 674, 1038, 1128, 1130, argument of a complex number, A7, arithmetic-geometric mean, 726, astroid, 669, asymptote of a hyperbola, 698, auxiliary equation, 1167, complex roots of, 1169, real roots of, 1168, average rate of change, 886, , average value of a function, 1003, 1051, axes, coordinate, 810, axis of a parabola, 694, basis vectors, 820, Bernoulli, John, 664, 778, Bessel, Friedrich, 766, Bessel function, 766, 770, Bézier, Pierre, 677, Bézier curves, 663, 677, binomial coefficients, 784, binomial series, 784, discovery by Newton, 791, binormal vector, 882, blackbody radiation, 801, boundary curve, 1146, boundary-value problem, 1171, bounded sequence, 721, bounded set, 975, brachistochrone problem, 664, Brahe, Tycho, 891, branches of a hyperbola, 698, , C 1 tansformation, 1064, calculator, graphing, 662, 685. See also computer algebra system, Cantor, Georg, 737, Cantor set, 737, cardioid, 682, Cassini, Giovanni, 689, CAS. See computer algebra system, Cauchy, Augustin-Louis, 1008, Cauchy-Schwarz Inequality, 831, center of gravity. See center of mass, center of mass, 1028, 1089, of a lamina, 1029, of a solid, 1047, of a surface, 1136, of a wire, 1089, centripetal force, 899, centroid of a solid, 1047, Chain Rule for several variables, 948,, 950, 951, change of variable(s), in a double integral, 1023, 1065, 1068, in a triple integral, 1053, 1058, 1070, , characteristic equation, 1167, charge, electric, 1027, 1028, 1047, 1184, charge density, 1028, 1047, circle of curvature, 883, circular paraboloid, 856, circulation of a vector field, 1150, cissoid of Diocles, 668, 687, Clairaut, Alexis, 931, Clairaut’s Theorem, 931, clipping planes, 850, closed curve, 1101, Closed Interval Method, for a function, of two variables, 976, closed set, 975, closed surface, 1140, Cobb, Charles, 903, Cobb-Douglas production function, 904,, 934, 987, cochleoid, 710, coefficient(s), binomial, 784, of a power series, 765, of static friction, 861, comets, orbits of, 708, common ratio, 729, Comparison Test for series, 746, complementary equation, 1173, Completeness Axiom, 722, complex conjugate, A5, complex exponentials, A11, complex number(s), A5, addition and subtraction of, A5, argument of, A7, division of, A6, A8, equality of, A5, imaginary part of, A5, modulus of, A5, multiplication of, A5, A8, polar form, A7, powers of, A9, principal square root of, A6, real part of, A5, roots of, A10, component function, 864, 1081, components of acceleration, 890, components of a vector, 817, 828, , A43, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:56 PM Page A44, , A44, , INDEX, , composition of functions, continuity of, 922, computer algebra system, 662, for integration, 775, computer algebra system, graphing with,, function of two variables, 906, level curves, 910, parametric equations, 662, parametric surface, 1126, partial derivatives, 931, polar curve, 685, sequence, 719, space curve, 867, vector field, 1082, conchoid, 665, 687, conditionally convergent series, 757, conductivity (of a substance), 1144, cone, 694, 854, parametrization of, 1126, conic section, 694, 702, directrix, 694, 702, eccentricity, 702, focus, 694, 696, 702, polar equation, 704, shifted, 699, vertex (vertices), 694, conjugates, properties of, A6, connected region, 1101, conservation of energy, 1105, conservative vector field, 1085, 1106, constant force, 829, constraint, 981, 985, continued fraction expansion, 726, continuity, of a function, 865, of a function of three variables, 922, of a function of two variables, 920, contour curves, 907, contour map, 907, 933, convergence, absolute, 756, conditional, 757, interval of, 767, radius of, 767, of a sequence, 716, of a series, 729, convergent sequence, 716, convergent series, 729, properties of, 733, conversion, cylindrical to rectangular, coordinates, 1052, cooling tower, hyperbolic, 856, coordinate axes, 810, coordinate planes, 810, coordinate system,, cylindrical, 1052, polar, 678, spherical, 1057, three-dimensional rectangular, 810, , coplanar vectors, 837, Coriolis acceleration, 898, Cornu’s spiral, 676, cosine function, power series for, 782, critical point(s), 970, 980, critically damped vibration, 1182, cross product, 832, direction of, 834, geometric characterization of, 835, magnitude of, 835, properties of, 836, cross-section, of a surface, 851, curl of a vector field, 1115, curvature, 677, 879, curve(s), Bézier, 663, 677, boundary, 1146, cissoid of Diocles, 687, closed, 1101, Cornu’s spiral, 676, dog saddle, 915, epicycloid, 669, equipotential, 914, grid, 1124, helix, 865, length of, 877, level, 907, monkey saddle, 915, orientation of, 1092, 1108, ovals of Cassini, 689, parametric, 660 865, piecewise-smooth,1088, polar, 680, serpentine, 137, simple, 1102, space, 864, 865, strophoid, 693, 711, swallotail catastrophe, 668, toroidal spiral, 867, trochoid, 667, twisted cubic, 867, witch of Maria Agnesi, 667, cusp, 665, cycloid, 663, cylinder, 851, parabolic, 851, parametrization of, 1126, cylindrical coordinate system, 1052, conversion equations for, 1052, triple integrals in, 1053, cylindrical coordinates, 1054, damped vibration, 1181, damping constant, 1181, decreasing sequence, 720, definite integral, 998, of a vector function, 875, del (ⵜ), 960, , De Moivre, Abraham, A9, De Moivre’s Theorem, A9, density, of a lamina, 1027, of a solid, 1047, dependent variable, 902, 950, derivative(s),, directional, 957, 958, 961, higher partial, 930, normal, 1122, notation for partial, 927, partial, 926, of a power series, 772, second, 874, second directional, 968, second partial, 930, of a vector function, 871, determinant, 832, differentiable function, 942, differential, 943, 945, differential equation,, homogeneous, 1166, linearly independent solutions, 1167, logistic, 727, nonhomogeneous, 1166, 1173, partial, 932, second-order, 1166, differentiation,, formulas for, RP5, formulas for vector functions, 874, implicit, 929, 952, partial, 924, 929, 930, of a power series, 772, term-by-term, 772, of a vector function, 874, directed line segment, 815, direction numbers, 842, directional derivative, 957, 958, 961, maximum value of, 962, of a temperature function, 957, 958, second, 958, directrix, 694, 702, displacement vector, 815, 829, distance, between lines, 847, between planes, 847, between point and line in space, 839, between point and plane, 839, between points in space, 812, distance formula in three dimensions, 812, divergence, of an infinite series, 729, of a sequence, 716, of a vector field, 1118, Divergence, Test for, 733, Divergence Theorem, 1153, divergent sequence, 716, divergent series, 729, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:56 PM Page A45, , INDEX, , division of power series, 787, DNA, helical shape of, 866, dog saddle, 915, domain of a function, 902, domain sketching, 902, Doppler effect, 956, dot product, 824, in component form, 824, properties of, 825, double integral, 998, 1000, change of variable in, 1065, 1068, over general regions, 1012, 1013, Midpoint Rule for, 1002, in polar coordinates, 1021, 1022, 1023, properties of, 1005, 1017, over rectangles, 998, double Riemann sum, 1001, Douglas, Paul, 903, Dumpster design, minimizing cost of, 980, e (the number) as a sum of an infinite, series, 781, eccentricity, 702, electric charge, 1027, 1028, 1047, electric circuit, analysis of, 1184, electric field (force per unit charge), 1084, electric flux, 1143, electric force, 1084, ellipse, 696, 702, A19, directrix, 702, eccentricity, 702, foci, 696, 702, major axis, 696, 707, minor axis, 696, polar equation, 704, 707, reflection property, 697, vertices, 696, ellipsoid, 852, 854, elliptic paraboloid, 852, 854, energy, conservation of, 1105, kinetic, 1105, potential, 1105, epicycloid, 669, epitrochoid, 676, equation(s), differential (see differential equation), of an ellipse, 696, 704, heat conduction, 937, of a hyperbola, 697, 698, 699, 704, Laplace’s, 932, 1119, of a line in space, 840, 841, of a line through two points, 842, linear, 844, logistic difference, 727, of a parabola, 694, 704, parametric, 660, 841, 865, 1123, of a plane, 843, , of a plane through three points, 845, polar, 680, 704, of a space curve, 865, of a sphere, 813, symmetric, 842, van der Waals, 938, vector, 840, wave, 932, equipotential curves, 914, equivalent vectors, 816, error in Taylor approximation, 793, error estimate for alternating series, 754, estimate of the sum of a series, 742, 749,, 754, 759, Euler, Leonhard, 739, 745, 781, Euler’s formula, A11, expected values, 1035, exponential function(s),, integration of, 786, 787, power series for, 779, Extreme Value Theorem, 975, family, of epicycloids and hypocycloids, 668, of parametric curves, 664, Fibonacci, 715, 726, Fibonacci sequence, 715, 726, field, conservative, 1085, electric, 1084, force, 1084, gradient, 966, 1084, gravitational, 1084, incompressible, 1119, irrotational, 1118, scalar, 1081, vector, 1080, 1081, velocity, 1080, 1083, first octant, 810, first-order optics, 798, flow lines, 1086, fluid flow, 1083, 1119, 1142, flux, 1141, 1143, flux integral, 1141, foci, 696, focus, 694, 702, of a conic section, 702, of an ellipse, 696, 702, of a hyperbola, 697, of a parabola, 694, folium of Descartes, 711, force,, centripetal, 899, constant, 829, resultant, 821, torque, 837, force field, 1080, 1084, forced vibrations, 1183, , A45, , four-leaved rose, 682, Frenet-Serret formulas, 886, Fubini, Guido, 1008, Fubini’s Theorem, 1008, 1041, function(s), 902, Airy, 770, arc length, 877, average value of, 1003, 1051, Bessel, 766, 770, Cobb-Douglas production, 904, 934, 987, component, 864, 1081, composite, 922, continuity of, 920, 922, continuous, 865, differentiability of, 942, domain of, 902, gradient of, 960, 962, graph of, 904, harmonic, 932, homogeneous, 956, integrable, 1000, joint density, 1032, 1047, limit of, 917, 922, linear, 905, maximum and minimum values of, 970, of n variables, 911, polynomial, 921, potential, 1085, probability density, 1032, range of, 902, rational, 921, representation as a power series, 770, of several variables, 902, 910, of three variables, 910, of two variables, 902, vector, 826, Fundamental Theorem of Calculus,, higher-dimensional versions, 1159, for line integrals, 1099, for vector functions, 875, Galileo, 664, 671, 694, Gauss, Karl Friedrich, 1153, Gaussian optics, 798, Gauss’s Law, 1143, Gauss’s Theorem, 1153, geometric series, 729, geometry of a tetrahedron, 840, Gibbs, Joseph Willard, 821, gradient, 960, 962, gradient vector, 960, 962, interpretations of, 1066, gradient vector field, 1066, 1084, graph(s), of equations in three dimensions, 811, of a function of two variables, 904, of a parametric curve, 660, of a parametric surface, 1136, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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See computer algebra system, gravitational field, 1084, great circle, 1063, Green, George, 1109, 1152, Green’s identities, 1122, Green’s Theorem, 1108, 1152, vector forms, 1120, Gregory, James, 774, 778, Gregory’s series, 774, grid curves, 1124, half-space, 911, harmonic function, 932, harmonic series, 732, 741, alternating, 753, heat conduction equation, 937, heat conductivity, 1144, heat flow, 1143, heat index, 924, Hecht, Eugene, 797, helix, 865, hidden line rendering, 850, higher partial derivatives, 930, homogeneous differential equation, 1166, homogeneous function, 956, Hooke’s Law, 1180, horizontal plane, 811, Huygens, Christiaan, 664, hydro-turbine optimization, 990, hyperbola, 697, 702, asymptotes, 698, branches, 698, directrix, 702, eccentricity, 702, equation, 698, 699, 704, foci, 697, 702, polar equation, 704, reflection property, 702, vertices, 698, hyperbolic paraboloid, 853, 854, hyperboloid, 854, hypersphere, 1051, hypocycloid, 668, i (imaginary number), A5, i (standard basis vector), 820, ideal gas law, 938, image of a point, 1065, image of a region, 1065, implicit differentiation, 929, 952, Implicit Function Theorem, 953, 954, incompressible velocity field, 1119, increasing sequence, 720, increment, 945, independence of path, 1100, independent random variable, 1034, , independent variable, 902, 950, inertia (moment of), 1030, 1047, 1098, infinite sequence. See sequence, infinite series. See series, initial point, of a parametric curve, 661, of a vector, 815, 1170, inner product, 824, integrable function, 1000, integral(s), change of variables in, 1023, 1064, 1068,, 1070, conversion to cylindrical coordinates, 1053, conversion to polar coordinates, 1022, conversion to spherical coordinates, 1058, definite, 998, double (see double integral), iterated, 1006, 1007, line (see line integral), surface, 1134, 1141, table of, RP6 –10, triple, 1041, 1042, Integral Test, 740, integrand, discontinuous, 547, integration,, formulas, RP6 –10, partial, 1007, of a power series, 772, reversing order of, 1009, 1017, over a solid, 1054, term-by-term, 772, of a vector function, 871, intermediate variable, 950, intersection, of planes, 845, of polar graphs, area of, 690, of three cylinders, 1056, interval of convergence, 767, inverse transformation, 1065, irrotational vector field, 1118, isothermal, 907, 914, iterated integral, 1006, 1007, , j (standard basis vector), 820, Jacobi, Carl, 1067, Jacobian of a transformation, 1067, 1070, joint density function, 1032, 1047, , k (standard basis vector), 820, Kepler, Johannes, 706, 891, Kepler’s Laws, 706, 891, 892, 896, kinetic energy, 1105, Kirchhoff’s Laws, 1184, Kondo, Shigeru, 781, Lagrange, Joseph-Louis, 982, Lagrange multiplier, 981, 982, lamina, 1027, 1029, Laplace, Pierre, 932, 1119, Laplace operator, 1119, , Laplace’s equation, 932, 1119, law of conservation of angular, momentum, 895, Law of Conservation of Energy, 1106, least squares method, 979, least upper bound, 722, Leibniz, Gottfried Wilhelm, 791, length, of a parametric curve, 672, of a polar curve, 691, of a space curve, 877, of a vector, 818, level curve(s), 907, 910, level surface, 911, tangent plane to, 964, limaçon, 686, limit(s),, of a function of three variables, 922, of a function of two variables, 917, of a sequence, 716, of a vector function, 864, Limit Comparison Test, 748, Limit Laws,, for functions of two variables, 920, for sequences, 717, linear approximation, 941, 945, linear combination, 1166, linear differential equation, 1166, linear equation of a plane, 844, linear function, 905, linearity of an integral, 1005, linearization, 941, linearly independent solutions, 1167, line(s) in the plane, equation of, through, two points, 842, line(s) in space, normal, 965, parametric equations of, 841, skew, 843, symmetric equations of, 842, tangent, 872, vector equation of, 840, 841, line integral, 1087, Fundamental Theorem for, 1099, for a plane curve, 1087, with respect to arc length, 1090, for a space curve, 1092, work defined as, 1094, of vector fields, 1094, 1095, Lissajous figure, 662, 668, lithotripsy, 697, local maximum and minimum values, 970, logistic difference equation, 727, logistic sequence, 727, LORAN system, 701, Maclaurin, Colin, 745, Maclaurin series, 777, 778, table of, 785, magnitude of a vector, 818, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:57 PM Page A47, , INDEX, , major axis of ellipse, 696, marginal productivity, 934, marginal propensity to consume or save, 736, mass, of a lamina, 1027, of a solid, 1047, of a surface, 1136, of a wire, 1089, mass, center of. See center of mass, mathematical induction, 723, mathematical model. See model(s),, mathematical, maximum and minimum values, 970, Mean Value Theorem for, double integrals, 1076, method of Lagrange multipliers, 981, 982, 985, method of least squares, 979, method of undetermined coefficients, 1173,, 1177, Midpoint Rule,, for double integrals, 1002, for triple integrals, 1049, minor axis of ellipse, 696, Möbius, August, 1139, Möbius strip, 1133, 1139, model(s), mathematical,, Cobb-Douglas, for production costs, 904,, 934, 987, for vibration of membrane, 766, von Bertalanffy, 655, modulus, A6, moment, about an axis, 1029, of inertia, 1030, 1047, 1098, of a lamina, 1029, about a plane, 1047, polar, 1031, second, 1030, of a solid, 1047, monkey saddle, 915, monotonic sequence, 720, Monotonic Sequence Theorem, 722, motion of a projectile, 888, motion in space, 886, motion of a spring, force affecting, damping, 1181, resonance, 1184, restoring, 1180, multiple integrals. See double integral;, triple integral(s), multiplication of power series, 787, multiplier (Lagrange), 981, 982, 985, multiplier effect, 736, natural exponential function, power series, for, 778, n-dimensional vector, 819, Newton, Sir Isaac, 791, 892, 896, Newton’s Law of Gravitation, 892, 1083, Newton’s Second Law of Motion, 892, 1180, , Nicomedes, 665, nonhomogeneous differential equation,, 1166, 1173, nonparallel planes, 845, normal component of acceleration, 890, 891, normal derivative, 1122, normal line, 965, normal plane, 883, normal vector, 844, 882, nth-degree Taylor polynomial, 779, number, complex, A5, , O (origin), 810, octant, 810, one-to-one transformation, 1065, open region, 1101, optics, first-order, 798, Gaussian, 798, third-order, 798, orbit of a planet, 892, order of integration, reversed, 1009, 1017, ordered triple, 810, Oresme, Nicole, 732, orientation of a curve, 1092, 1108, orientation of a surface, 1139, oriented surface, 1139, origin, 810, orthogonal projection, 831, orthogonal surfaces, 969, orthogonal vectors, 826, osculating circle, 883, osculating plane, 883, Ostrogradsky, Mikhail, 1153, ovals of Cassini, 689, overdamped vibration, 1182, parabola, 694, 702, axis, 694, directrix, 694, equation, 694, 695, focus, 694, 702, polar equation, 704, vertex, 694, parabolic cylinder, 851, paraboloid, 852, 856, parallel planes, 845, parallel vectors, 817, parallelepiped, volume of, 837, Parallelogram Law, 816, 831, parameter, 660, 841, 865, parametric curve, 660, 865, arc length of, 672, area under, 671, slope of tangent line to, 669, parametric equations, 660, 841, 865, of a line in space, 841, of a space curve, 865, of a surface, 1123, of a trajectory, 889, , A47, , parametric surface, 1123, graph of, 1136, surface area of, 1128, 1129, surface integral over, 1135, tangent plane to, 1127, parametrization of a space curve, 878, with respect to arc length, 879, smooth, 879, partial derivative(s), 926, of a function of more than three, variables, 929, interpretations of, 927, notations for, 927, as a rate of change, 926, rules for finding, 927, second, 930, as slopes of tangent lines, 927, partial differential equation, 932, partial integration, 1007, partial sum of a series, 728, particle, motion of, 886, path, 1100, perihelion, 707, perilune, 701, perpendicular vectors, 826, piecewise-smooth curve, 1088, Planck’s Law, 801, plane region of type I, 1013, plane region of type II, 1014, plane(s), angle between, 845, coordinate, 810, equation(s) of, 840, 843, 844, equation of, through three points, 845, horizontal, 811, line of intersection, 845, normal, 883, osculating, 883, parallel, 845, tangent to a surface, 939, 964, 1127, vertical, 902, planetary motion, 891, laws of, 706, planimeter, 1111, point(s) in space, coordinates of, 810, distance between, 812, projection of, 811, polar axis, 678, polar coordinate system, 678, conic sections in, 702, conversion of double integral to, 1021, conversion equations for Cartesian, coordinates, 680, polar curve, 680, arc length of, 691, graph of, 680, symmetry in, 683, tangent line to, 683, polar equation, graph of, 680, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:57 PM Page A48, , A48, , INDEX, , polar equation of a conic, 704, polar form of a complex number, A7, polar graph, 680, polar moment of inertia, 1031, polar rectangle, 1021, polar region, area of, 689, pole, 678, polynomial function of two variables, 921, position vector, 818, positive orientation, of a boundary curve, 1146, of a closed curve, 1188, of a surface, 1140, potential energy, 1105, potential function, 1085, power, 1110, power series, 765, coefficients of, 765, for cosine and sine, 782, differentiation of, 772, division of, 787, for exponential function, 782, integration of, 772, interval of convergence, 767, multiplication of, 787, radius of convergence, 767, representations of functions as, 771, principal square root of a complex, number, A6, principal unit normal vector, 882, principle of superposition, 1175, probability, 1032, probability density function, 1032, product, cross, 832 (see also cross product), dot, 824 (see also dot product), scalar, 824, scalar triple, 836, triple, 836, projectile, path of, 668, 888, projection, 811, 828, orthogonal, 831, p-series, 741, quadratic approximation, 980, quadric surface(s), 851, cone, 854, cylinder, 851, ellipsoid, 854, hyperboloid, 854, paraboloid, 852, 853, 854, table of graphs, 854, quaternion, 821, radiation from stars, 801, radius of convergence, 767, radius of gyration, 1032, range of a function, 902, rational function, 921, , Ratio Test, 758, Rayleigh-Jeans Law, 801, rearrangement of a series, 761, rectangular coordinate system, 811, conversion to cylindrical coordinates, 1052, conversion to spherical coordinates, 1057, recursion relation, 1189, reflection property, of an ellipse, 697, of a hyperbola, 702, region, connected, 1101, open, 1101, plane, of type I or II, 1013, 1014, simple plane, 1109, simple solid, 1153, simply-connected, 1102, solid (of type 1, 2, or 3), 1042, 1043, 1044, remainder estimates, for the Alternating Series, 754, for the Integral Test, 742, remainder of the Taylor series, 779, representation of a function, as a power, series, 770, resonance, 1184, restoring force, 1180, resultant force, 821, reversing order of integration, 1009, 1017, Riemann sums for multiple, integrals, 1001, 1041, right-hand rule, 810, 834, Roberval, Gilles de, 671, rocket science, 988, roller derby, 1063, Root Test, 760, roots of a complex number, A10, rubber membrane, vibration of, 766, ruling of a surface, 851, saddle point, 971, sample point, 999, satellite dish, parabolic, 856, scalar, 817, scalar equation of a plane, 844, scalar field, 1081, scalar multiple of a vector, 817, scalar product, 824, scalar projection, 828, scalar triple product, 836, geometric characterization of, 837, secant vector, 872, second derivative, 874, of a vector function, 874, Second Derivatives Test, 971, second directional derivative, 968, second moment of inertia, 1030, second-order differential equation,, solutions of, 1166, 1171, second partial derivative, 930, , sector of a circle, area of, 689, sequence,, bounded, 721, convergent, 716, decreasing, 720, divergent, 716, Fibonacci, 715, graph of, 719, increasing, 720, limit of, 716, logistic, 727, monotonic, 720, of partial sums, 728, term of, 714, series, 728, absolutely convergent, 756, alternating, 751, alternating harmonic, 753, 756, 757, binomial, 784, coefficients of, 765, conditionally convergent, 757, convergent, 729, divergent, 729, geometric, 729, Gregory’s, 774, harmonic, 732, 741, infinite, 728, Maclaurin, 777, 778, p-, 741, partial sum of, 728, power, 765, rearrangement of, 761, strategy for testing, 763, sum of, 729, Taylor, 777, 778, term of, 728, trigonometric, 765, series solution of a differential, equation, 1188, set, bounded or closed, 975, shifted conics, 699, shock absorber, 1181, Sierpinski carpet, 737, simple curve, 1102, simple plane region, 1109, simple solid region, 1153, simply-connected region, 1102, Simpson, Thomas, 996, sine function, power series for, 782, sink, 1157, skew lines, 843, smooth curve, 879, smooth parametrization, 879, smooth surface, 1128, snowflake curve, 806, solid, volume of, 1042, 1043, solid angle, 1163, solid region, 1153, source, 1157, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:58 PM Page A49, , INDEX, , space, three-dimensional, 810, space curve, 864, 865, 866, 867, arc length of, 877, speed of a particle, 886, sphere, equation of, 813, flux across, 1141, parametrization of, 1125, surface area of, 1129, spherical coordinate system, 1057, conversion equations for, 1057, triple integrals in, 1058, spherical wedge, 1058, spring constant, 1180, Squeeze Theorem, for sequences, 718, standard basis vectors, 820, stationary points, 970, steady state solution, 1186, Stokes, Sir George, 1147, 1152, Stokes’ Theorem, 1146, strategy for testing series, 763, streamlines, 1086, strophoid, 693, 711, sum,, of a geometric series, 730, of an infinite series, 729, telescoping, 732, of vectors, 816, surface(s), closed, 1140, graph of, 1136, level, 911, oriented, 1139, parametric, 1123, positive orientation of, 1140, quadric, 851, smooth, 1128, surface area,, of a parametric surface, 674, 1128, 1129, of a sphere, 1129, of a surface z 苷 f 共x, y兲, 1037, 1038, 1130, surface integral, 1134, over a parametric surface, 1135, of a vector field, 1140, surface of revolution, parametric, representation of, 1127, swallowtail catastrophe curve, 668, symmetric equations of a line, 842, symmetry in polar graphs, 683, , T and T ⫺1 transformations, 1064, 1065, table of differentiation formulas, RP5, tables of integrals, RP6–10, tangent line(s),, to a parametric curve, 669, 670, to a polar curve, 683, to a space curve, 873, tangent plane, to a level surface, 939, 964, , to a parametric surface, 1127, to a surface F共x, y, z兲 苷 k, 940, 964, to a surface z 苷 f 共x, y兲, 939, tangent plane approximation, 941, tangent vector, 872, tangential component of acceleration, 890, tautochrone problem, 664, Taylor, Brook, 778, Taylor polynomial, 779, 980, applications of, 792, Taylor series, 777, 778, Taylor’s Inequality, 780, telescoping sum, 732, temperature-humidity index, 912, 924, term of a sequence, 714, term of a series, 728, term-by-term differentiation and, integration, 772, terminal point of a parametric curve, 661, terminal point of a vector, 815, Test for Divergence, 733, tests for convergence and divergence, of series, Alternating Series Test, 751, Comparison Test, 746, Integral Test, 740, Limit Comparison Test, 748, Ratio Test, 758, Root Test, 760, summary of tests, 763, tetrahedron, 840, third-order optics, 798, Thomson, William (Lord Kelvin), 1109,, 1147, 1152, three-dimensional coordinate systems,, 810, 811, TNB frame, 882, toroidal spiral, 867, torque, 895, Torricelli, Evangelista, 671, torsion of a space curve, 885, torus, 1134, total differential, 944, total electric charge, 1029, 1047, trace of a surface, 851, trajectory, parametric equations for, 889, transfer curve, 899, transformation, 1064, inverse, 1065, Jacobian of, 1067, 1070, one-to-one, 1065, tree diagram, 932, trefoil knot, 867, Triangle Inequality for vectors, 831, Triangle Law, 816, trigonometric series, 765, triple integral(s), 1041, 1042, applications of, 1046, in cylindrical coordinates, 1053, , A49, , over a general bounded region, 1042, Midpoint Rule for, 1049, in spherical coordinates, 1058, 1059, triple product, 836, triple Riemann sum, 1041, trochoid, 667, twisted cubic, 867, type I or type II plane region, 1013, 1014, type 1, 2, or 3 solid region, 1042, 1043, 1044, ultraviolet catastrophe, 801, underdamped vibration, 1182, undetermined coefficients, method of,, 1173, 1177, uniform circular motion, 888, unit normal vector, 882, unit tangent vector, 872, unit vector, 821, van der Waals equation, 938, variable(s), dependent, 902, 950, independent, 902, 950, independent random, 1034, intermediate, 950, variables, change of. See change of variable(s), variation of parameters, method of, 1177, 1178, vector(s), 815, acceleration, 887, addition of, 816, 818, algebraic, 818, 819, angle between, 825, basis, 820, binormal, 882, combining speed, 823, components of, 828, coplanar, 837, cross product of, 832, difference, 818, displacement, 829, dot product, 825, equality of, 816, force, 1083, geometric representation of, 818, gradient, 960, 962, i, j, and k, 820, length of, 818, magnitude of, 818, multiplication of, 817, 819, n-dimensional, 819, normal, 844, orthogonal, 826, parallel, 817, perpendicular, 826, position, 818, properties of, 819, representation of, 818, scalar mulitple of, 817, standard basis, 820, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com

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97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:58 PM Page A50, , A50, , INDEX, , vector(s) (continued), tangent, 872, three-dimensional, 818, triple product, 837, two-dimensional, 818, unit, 821, unit normal, 882, unit tangent, 872, velocity, 886, zero, 816, vector equation, of a line, 840, 841, of a plane, 844, vector field, 1080, 1081, conservative, 1085, curl of, 1115, divergence of, 1118, electric flux of, 1143, flux of, 1141, force, 1080, 1084, gradient, 1084, gravitational, 1084, incompressible, 1119, irrotational, 1118, line integral of, 1094, 1095, potential function, 1104, , surface integral of, 1141, velocity, 1080, 1083, vector function, 864, continuity of, 865, derivative of, 871, integration of, 875, limit of, 864, vector product, 832, properties of, 836, vector projection, 828, vector triple product, 837, vector-valued function. See vector function, continuous, 865, limit of, 864, velocity field, 1083, airflow, 1080, ocean currents, 1080, wind patterns, 1080, velocity vector, 886, velocity vector field, 1080, vertex of a parabola, 694, vertices of an ellipse, 696, vertices of a hyperbola, 698, vibration of a rubber membrane, 766, vibration of a spring, 1180, vibrations, 1180, 1181, 1183, , volume, 353, by double integrals, 998, of a hypersphere, 1051, by polar coordinates, 1024, of a solid, 1000, by triple integrals, 1046, wave equation, 932, wind-chill index, 903, wind patterns in San Francisco Bay area, 1080, witch of Maria Agnesi, 667, work (force), defined as a line integral, 1094, Wren, Sir Christopher, 674, , x-axis, 810, x-coordinate, 810, X-mean, 1035, y-axis, 810, y-coordinate, 810, Y-mean, 1035, z-axis, 810, z-coordinate, 810, zero vectors, 816, , Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)., Editorial review has deemed that any suppressed content does not materially affect the overall, learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., www.TechnicalPdf.com