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11. F ⫽ x i ⫹ x3y2 j ⫹ z k; C the boundary of the semi-ellipsoid, , z, , z ⫽ "4 2 4x 2 2 y 2 in the plane z ⫽ 0, , z=1–y, , 12. F ⫽ z i ⫹ x j ⫹ y k; C the curve of intersection of the plane, , y, , x ⫹ y ⫹ z ⫽ 0 and the sphere x 2 ⫹ y2 ⫹ z2 ⫽ 1 [Hint: Recall, that the area of an ellipse x 2 /a2 ⫹ y2 /b2 ⫽ 1 is pab.], , C, x, , In Problems 13–16, use Stokes’ theorem to evaluate, eeS (curl F) ⭈ n dS. Assume that the surface S is oriented upward., , (2, 0, 0), , 2, , 13. F ⫽ 6yz i ⫹ 5x j ⫹ yze x k; S that portion of the paraboloid, , FIGURE 9.14.7 Curve C for Problem 6, 7. F ⫽ xy i ⫹ 2yz j ⫹ xz k; C the boundary given in Problem 6, 8. F ⫽ (x ⫹ 2 z) i ⫹ (3x ⫹ y) j ⫹ (2y ⫺ z) k; C the curve of, , intersection of the plane x ⫹ 2y ⫹ z ⫽ 4 with the coordinate, planes, 9. F ⫽ y3 i ⫺ x 3 j ⫹ z3 k; C the trace of the cylinder x 2 ⫹ y2 ⫽ 1, in the plane x ⫹ y ⫹ z ⫽ 1, 10. F ⫽ x 2y i ⫹ (x ⫹ y2) j ⫹ xy2z k; C the boundary of the surface, shown in FIGURE 9.14.8, , 14., 15., , 16., 17., , z ⫽ 14 x2 ⫹ y2 for 0 ⱕ z ⱕ 4, F ⫽ y i ⫹ (y ⫺ x) j ⫹ z2 k; S that portion of the sphere, x 2 ⫹ y2 ⫹ (z ⫺ 4)2 ⫽ 25 for z ⱖ 0, F ⫽ 3x 2 i ⫹ 8x 3y j ⫹ 3x 2y k; S that portion of the plane z ⫽ x, that lies inside the rectangular cylinder defined by the planes, x ⫽ 0, y ⫽ 0, x ⫽ 2, y ⫽ 2, F ⫽ 2xy2z i ⫹ 2x 2yz j ⫹ (x 2y2 ⫺ 6x) k; S that portion of the, plane z ⫽ y that lies inside the cylinder x 2 ⫹ y2 ⫽ 1, Use Stokes’ theorem to evaluate, , BC, 䉲, , z, , where C is the circle x2 ⫹ y2 ⫽ 9, by finding a surface S with, C as its boundary and such that the orientation of C is counterclockwise as viewed from above., 18. Consider the surface integral eeS (curl F) ⭈ n dS, where, F ⫽ xyz k and S is that portion of the paraboloid, z ⫽ 1 ⫺ x 2 ⫺ y2 for z ⱖ 0 oriented upward., (a) Evaluate the surface integral by the method of Section, 9.13; that is, do not use Stokes’ theorem., (b) Evaluate the surface integral by finding a simpler surface, that is oriented upward and has the same boundary as the, paraboloid., (c) Use Stokes’ theorem to verify the result in part (b)., , z = 9 – y2, C, , y, y = 2x, , y=3, , x, , FIGURE 9.14.8 Curve C for Problem 10, , 9.15, , D, , z, , y, , 2, , z 2 e x dx ⫹ xy2 dy ⫹ tan⫺1 y dz, , Triple Integrals, , INTRODUCTION The steps leading to the definition of the three-dimensional definite integral, or triple integral are quite similar to the steps leading to the definition of the double integral., Obvious differences: instead of a function of two variables we are integrating a function f of three, variables, not over a region R in a coordinate plane, but over a region D of 3-space., 1. Let w ⫽ F(x, y, z) be defined over a closed and bounded region D of space., 2. By means of a three-dimensional grid of vertical and horizontal planes parallel to the coordinate planes, form a partition P of D into n subregions (boxes) Dk of volumes ⌬Vk that lie, entirely in D., 3. Let iPi be the norm of the partition or the length of the longest diagonal of the Dk., 4. Choose a sample point (x *k , y *k , z *k ) in each subregion Dk. See FIGURE 9.15.1., 5. Form the sum a F (x *k , y *k , z *k ) ⌬Vk., n, , (x*k, y*k, z*k), x, , FIGURE 9.15.1 Sample point in kth, subregion, , 564, , |, , A sum of the form g k ⫽ 1 F(x *k , y *k , z *k ) ⌬Vk, where (x *k , y *k , z *k ) is an arbitrary point within each, Dk and ⌬Vk denotes the volume of each Dk, is called a Riemann sum. The type of partition used, in step 2, where all the Dk lie completely within D, is called an inner partition of D., , CHAPTER 9 Vector Calculus, , k⫽1, n
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Definition 9.15.1, , The Triple Integral, , Let F be a function of three variables defined over a closed region D of 3-space. Then the, triple integral of F over D is given by, lim a F(x *k , y *k , z *k ) DVk., 9 F(x, y, z) dV iPiS0, n, , (1), , k1, , D, , As in our previous discussions on the integral, when F is continuous over D, the limit in, (1) exists; that is, F is integrable over D., , Evaluation by Iterated Integrals If the region D is bounded above by the graph, of z f2(x, y) and bounded below by the graph of z f1(x, y), then it can be shown that the, f2(x, y), triple integral (1) can be expressed as a double integral of the partial integral ef1(x, y) F(x, y, z) dz;, that is,, 9 F(x, y, z) dV 6 c, D, , R, , #, , f2(x, y), , f1(x, y), , F(x, y, z) dz d dA,, , where R is the orthogonal projection of D onto the xy-plane. In particular, if R is a Type I region,, then, as shown in FIGURE 9.15.2, the triple integral of F over D can be written as an iterated integral:, 9 F(x, y, z) dV , D, , g2(x), , b, , ## #, a, , g1(x), , f2 (x, y), , F(x, y, z) dz dy dx., , (2), , f1(x, y), , z = f2(x, y), D, , z, , z = f1(x, y), , y, , a, y = g1(x), , y = g2(x), R, , b, x, , FIGURE 9.15.2 Geometric interpretation of (2), , To evaluate the iterated integral in (2) we begin by evaluating the partial integral, , #, , f2(x, y), , F(x, y, z) dz,, , f1(x, y), , in which both x and y are held fixed., In a double integral there are only two possible orders of integration: dy dx and dx dy. The, triple integral in (2) illustrates one of six possible orders of integration:, dz dy dx,, , dz dx dy,, , dy dx dz,, , dx dy dz,, , dx dz dy,, , dy dz dx., , The last two differentials tell the coordinate plane in which the region R is situated. For, example, the iterated integral corresponding to the order of integration dx dz dy must have, 9.15 Triple Integrals, , |, , 565
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the form, , 9 F(x, y, z) dV , , d, , k2 (y), , ## #, k1(y), , c, , D, , h2 (y, z), , F(x, y, z) dx dz dy., , h1 (y, z), , The geometric interpretation of this integral and the region R of integration in the yz-plane are, shown in FIGURE 9.15.3., z = k 2(y), z, , R, , x = h1(y, z), , D, , z = k1(y), , c, x, , d, , y, , x = h2(y, z), , FIGURE 9.15.3 Integration: first x, then z, then y, , Applications A list of some of the standard applications of the triple integral follows:, Volume: If F(x, y, z) 1, then the volume of the solid D is, V 9 dV., D, , Mass: If r(x, y, z) is density, then the mass of the solid D is given by, m 9 r(x, y, z) dV., D, , First Moments: The first moments of the solid about the coordinate planes indicated by the, subscripts are given by, Mxy 9 zr(x, y, z) dV,, D, , Mxz 9 yr(x, y, z) dV,, D, , Myz 9 xr(x, y, z) dV., D, , Center of Mass: The coordinates of the center of mass of D are given by, x, , Myz, m, , ,, , y, , Mxz, ,, m, , z, , Mxy, m, , ., , Centroid: If r (x, y, z) a constant, the center of mass is called the centroid of the solid., Second Moments: The second moments, or moments of inertia of D about the coordinate, axes indicated by the subscripts, are given by, , Ix 9 (y 2 z 2)r(x, y, z) dV, Iy 9 (x 2 z 2)r(x, y, z) dV, Iz 9 (x 2 y 2)r(x, y, z) dV., D, , D, , D, , Radius of Gyration: As in Section 9.10, if I is a moment of inertia of the solid about a given, axis, then the radius of gyration is, Rg , EXAMPLE 1, , Volume of a Solid, , I, ., Åm, , Find the volume of the solid in the first octant bounded by the graphs of z 1 y2, y 2x,, and x 3., 566, , |, , CHAPTER 9 Vector Calculus
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SOLUTION As indicated in FIGURE 9.15.4(a), the first integration with respect to z is from 0 to, 1 y 2. Furthermore, from Figure 9.15.4(b) we see that the projection of the solid D in the, z, , D, , y, , y, , y=1, z=1–, x, , x=, , y = 2x, , y2, , R, , y, 2, , x=3, x, , x=3, (a), , (b), , FIGURE 9.15.4 Solid D and region R of integration in Example 1, , xy-plane is a region of Type II. Hence, we next integrate, with respect to x, from y/2 to 3. The, last integration is with respect to y from 0 to 1. Thus,, V 9 dV , , 1, , ## #, , 1, , ##, 0, , (1 2 y 2) dx dy, , y>2, , # cx 2 xy d, 2, , 0, , 3, , dy, y>2, , 1, , # a3 2 3y, , , , dz dx dy, , 3, , 1, , , , 1 2 y2, , y>2 0, , 0, , D, , , , 3, , 2, , 2, , 0, , c3y 2 y 3 2, , 1, 1, y y 3 b dy, 2, 2, , 1, 1 2, 1, 15, y y4 d ., 4, 8, 8, 0, , Changing the Order of Integration, , EXAMPLE 2, , Change the order of integration in, 6, , ##, 0, , 4 2 2x>3, , 0, , #, , 3 2 x>2 2 3y>4, , F(x, y, z) dz dy dx, , 0, , to dy dx dz., SOLUTION As seen in FIGURE 9.15.5(a), the region D is the solid in the first octant bounded by, the three coordinate planes and the plane 2x 3y 4z 12. Referring to Figure 9.15.5(b), and the table, we conclude that, 6, , ##, 0, , 4 2 2x>3, , 0, , #, , 3 2 x>2 2 3y>4, , 0, , 3, , F(x, y, z) dz dy dx , , 6 2 2z, , ## #, 0, , 0, , 4 2 2x>3 2 4z>3, , F(x, y, z) dy dx dz., , 0, , Order of, Integration, , First, Integration, , Second, Integration, , Third, Integration, , dz dy dx, dy dx dz, , 0 to 3 x/2 3y/4, 0 to 4 2x/3 4z /3, , 0 to 4 2x/3, 0 to 6 2z, , 0 to 6, 0 to 3, , 9.15 Triple Integrals, , |, , 567
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z, , z, z=3– 1 x– 3 y, 2, 4, x=0, , y=0, y, y=0, y=4–, , y, x = 6 – 2z, , 2x, 3, , y=4–, , z=0, , x, , x, , (a), , 2 x– 4 z, 3, 3, , (b), , FIGURE 9.15.5 Changing order of integration in Example 2, , Depending on the geometry of a region in 3-space, the evaluation of a triple integral over that, region may be made easier by utilizing a new coordinate system., , Cylindrical Coordinates The cylindrical coordinate system combines the polar description of a point in the plane with the rectangular description of the z-component of a point in space., As seen in FIGURE 9.15.6(a), the cylindrical coordinates of a point P are denoted by the ordered triple, (r, u, z). The word cylindrical arises from the fact that a point P in space is determined by the intersection of the planes z constant and u constant with a cylinder r constant. See Figure 9.15.6(b)., z, z, , O, x, , θ, y, , (x, y, z), or, (r, θ, z), P, , r, , z = constant (plane), , P, , y, , y, x, (r, θ ), x, , (a), , θ = constant, (plane), (b), , r = constant, (cylinder), , FIGURE 9.15.6 Cylindrical coordinates, , Conversion of Cylindrical Coordinates to Rectangular Coordinates From, Figure 9.15.6(a) we also see that the rectangular coordinates (x, y, z) of a point can be obtained, from the cylindrical coordinates (r, u, z) by means of, x r cos u,, EXAMPLE 3, , y r sin u,, , z z., , (3), , Cylindrical to Rectangular Coordinates, , Convert (8, p/3, 7) in cylindrical coordinates to rectangular coordinates., SOLUTION, , From (3),, x 8 cos, , p, 4,, 3, , y 8 sin, , p, 4"3,, 3, , z 7., , Thus, (8, p/3, 7) is equivalent to (4, 4 "3, 7) in rectangular coordinates., , Conversion of Rectangular Coordinates to Cylindrical Coordinates To, , express rectangular coordinates (x, y, z) as cylindrical coordinates, we use, r 2 x 2 y2 ,, , 568, , |, , CHAPTER 9 Vector Calculus, , y, tan u ,, x, , z z., , (4)
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(–√2 , √2 , 1), or, (2, 3π /4, 1), , Convert (2!2, !2, 1) in rectangular coordinates to cylindrical coordinates., , z, , SOLUTION From (4) we see that, , (–√2 , √2 , 0), z=1, , x, , Rectangular to Cylindrical Coordinates, , EXAMPLE 4, , r=2, , r 2 ("2 )2 ( "2 )2 4,, , y, , 3, θ= π, 4, , FIGURE 9.15.7 Converting rectangular to, cylindrical coordinates in Example 4, , "2, , tan u , , 1, z 1., "2, If we take r 2, then, consistent with the fact that x 0 and y 0, we take u 3p/4.*, Consequently, ("2, "2, 1) is equivalent to (2, 3p/4, 1) in cylindrical coordinates. See, FIGURE 9.15.7., , Triple Integrals in Cylindrical Coordinates Recall from Section 9.11 that the area, of a polar rectangle is A r* r u, where r* is the average radius. From FIGURE 9.15.8(a) we, see that the volume of a cylindrical wedge is simply V (area of base)(height) r* r u, z. Thus, if F(r, u, z) is a continuous function over the region D, as shown in Figure 9.15.8(b),, then the triple integral of F over D is given by, 9 F(r, u, z) dV 6 c, D, , R, , #, , f2(r, u), , f1(r, u), , b, , F(r, u, z) dz d dA , , z, , g2(u), , f2 (r, u), , ## #, g1(u), , a, , z, , F(r, u, z) r dz dr du., , f1(r, u), , z = f2(r, θ ), , D, , Δr, Δz, , z = f1(r, θ ), Δθ, , y, , y, , θ =β, , r*, , r = g1(θ ), , θ =α, x, , R, , x, r = g2(θ ), (b), , (a), , FIGURE 9.15.8 Cylindrical wedge in (a); region D in (b), , EXAMPLE 5, , A solid in the first octant has the shape determined by the graph of the cone z "x 2 y 2, and the planes z 1, x 0, and y 0. Find the center of mass if the density is given by, r(r, u, z) r., , z, D, z=1, , θ=, , SOLUTION, 9.15.9 that, , π, 2, y, , z=r, , θ =0, , Center of Mass, , In view of (4), the equation of the cone is z r. Hence, we see from FIGURE, m 9 rdV , D, p>2, , r=1, , , x, , FIGURE 9.15.9 Solid in Example 5, , # # r zd, 2, , 0, , p>2, , , , 1, , 0, , # #, 0, , 1, , p>2, , 1, , # # # r (r dz dr du), 0, , 0, , r, , 1, , dr du, r, , 1, , 0, , (r 2 2 r 3) dr du , , p, 24, , *If we use u tan1(1) p/4, then we can use r 2. Notice that the combinations r 2,, u p/4 and r 2, u 3p/4 are inconsistent., , 9.15 Triple Integrals, , |, , 569
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Mxy 9 zr dV , , 1, , p>2, , # ##, 0, , D, , 0, , p>2, , 1, , zr 2 dz dr du, , r, , 1, , , , # #, 0, , 0, , 1, 2, , p>2, , , , z2 2 1, r d dr du, 2, r, , # #, 0, , 1, , 0, , (r 2 2 r 4) dr du , , p, ., 30, , In the integrals for Mxz and Myz we substitute y r sin u and x r cos u:, Mxz 9 r 2 sin u dV , D, p>2, , , , , # #, , 1, , p>2, , 1, , 0, , 0, , # ##, , x, , m, , , , 1>20, p>24, , r 3 sin u dz dr du, , r, , r, , 0, , 3, , 2 r 4) sin u dr du , , D, , Myz, , 0, , 1, , Myz 9 r 2 cos u dV , Hence,, , 0, , 1, , r 3 z sin u d dr du, , # # (r, 0, , 1, , p>2, , < 0.38,, , y, , p>2, , 1, , # ##, 0, , 0, , r, , 1, 20, , 1, , r 3 cos u dz dr du , , Mxz, 1>20, , < 0.38,, m, p>24, , z, , Mxy, m, , 1, ., 20, p>30, , , , p>24, , < 0.8., , The center of mass has the approximate coordinates (0.38, 0.38, 0.8)., , Spherical Coordinates As seen in FIGURE 9.15.10(a), the spherical coordinates of a, point P are given by the ordered triple (r, f, u), where r is the distance from the origin to P, f, !, is the angle between the positive z-axis and the vector OP , and u is the angle measured from, !, !, the positive x-axis to the vector projection OQ of OP .* Figure 9.15.10(b) shows that a point P, in space is determined by the intersection of a cone f constant, a plane u constant, and a, sphere r constant; whence arises the name “spherical” coordinates., φ = constant, (cone), , z, P, , φ, x, , ρ, , z, , θ = constant, (plane), , (x, y, z), or, ( ρ, φ, θ ), , P, , z, , O, , y, , y, y, , θ, , Q, , x, , x, (a), , (b), , ρ = constant, (sphere), , FIGURE 9.15.10 Spherical coordinates, , Conversion of Spherical Coordinates to Rectangular and Cylindrical, Coordinates To transform from spherical coordinates (r, f, u) to rectangular coordinates, (x, y, z), we observe from Figure 9.15.10(a) that, !, !, x iOQ i cos u,, y iOQ i sin u,, , *u is the same angle as in polar and cylindrical coordinates., , 570, , |, , CHAPTER 9 Vector Calculus, , !, z iOP i cos f.
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!, !, Since u OQ u r sin f and u OP u r, the foregoing equations become, x r sin f cos u,, It is customary to take r, , y r sin f sin u,, , z r cos f., (5), !, 0 and 0 f p. Also, since u OQ u r sin f r, the formulas, , r r sin f,, , u u,, , z r cos f,, , (6), , enable us to transform from spherical coordinates ( r, f, u) to cylindrical coordinates (r, u, z)., , Spherical to Rectangular and Cylindrical Coordinates, , EXAMPLE 6, , Convert (6, p/4, p/3) in spherical coordinates to rectangular coordinates and cylindrical, coordinates., Identifying r 6, f p/4, and u p/3, we find from (5) that, , SOLUTION, , p, p, 3"2, cos , ,, 4, 3, 2, , x 6 sin, , y 6 sin, , p, p, 3"6, sin , ,, 4, 3, 2, , z 6 cos, , p, 3"2., 4, , The rectangular coordinates of the point are (3 "2>2, 3"6>2, 3"2)., From (6) we obtain, r 6 sin, , p, 3"2,, 4, , u, , p, ,, 3, , z 6 cos, , p, 3"2., 4, , Thus, the cylindrical coordinates of the point are (3 "2, p>3, 3"2)., , Conversion of Rectangular Coordinates to Spherical Coordinates To trans-, , form from rectangular coordinates to spherical coordinates, we use, z, , y, tan u ,, x, , r2 x 2 y2 z 2,, , z, , cos f , , 2, , "x y 2 z 2, , sin, , ., , (7), , Triple Integrals in Spherical Coordinates As seen in FIGURE 9.15.11, the volume, , of a spherical wedge is given by the approximation, sin, , V < r2 sin f r f u., y, , Thus, in a triple integral of a continuous spherical coordinate function F(r, f, u), the differential, of volume dV is given by, dV r2 sin f dr df du., , x, , A typical triple integral in spherical coordinates has the form, 9 F(r, f, u) dV , , FIGURE 9.15.11 Spherical wedge, , D, , EXAMPLE 7, , b, , g2(u), , ## #, a, , g1(u), , f2(f, u), , F(r, f, u) r2 sin f dr df du., , f1(f, u), , Moment of Inertia, , Find the moment of inertia about the z-axis of the homogeneous solid bounded between the, spheres, x2 y 2 z 2 a 2, SOLUTION, , and, , x2 y 2 z 2 b 2 ,, , a b., , If d(r, f, u) k is the density,* then, Iz 9 (x 2 y 2) k dV., D, , *We must use a different symbol to denote density to avoid confusion with the symbol r of spherical, coordinates., , 9.15 Triple Integrals, , |, , 571
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z, , φ varies from, 0 to π, , From (5) we find x 2 y2 r2 sin2 f and x 2 y2 z 2 r2. Thus the equations of the spheres, are simply r a and r b. See FIGURE 9.15.12. Consequently, in spherical coordinates the, foregoing integral becomes, , ρ varies from, a to b, , 2p, , y, , Iz k, , θ varies from, 0 to 2 π, , x, , p, , # ##, 0, , 0, , 2p, , k, , FIGURE 9.15.12 Limits of integration in, Example 7, , p, , 0, , 2p, , k, , # #, 0, , p, , 0, , r2 sin2f (r2 sin f dr df du), , a, , # ##, 0, , b, , b, , r4 sin3f dr df du, , a, , r5 3 b, sin f d df du, 5, a, 2p, , k, (b 5 2 a 5), 5, , # #, , k, (b 5 2 a 5), 5, , 2p, , 0, , #, , prime, meridian, , φ = constant, θ = constant, P, it u, , l, , at, , de, l o n g it u d e, , equator, , FIGURE 9.15.13 Parallels and great circles, , Exercises, , 9.15, , 1., , 2, , ## #, 2, , 3, , 2., , x, , ###, 1, , 1, , 6, , 3., , 0, , 1, , 4., , "y, , 4x 2z 3 dz dy dx, , y, , # # # cos a y b dz dx dy, 0, , 0, , 6., , dy dz dx, , 0, , y2, , "2, , # # #, 0, , 572, , x, , 0, , 2, , 0, , 2 2 x2 2 y2, , xye z dz dx dy, , 0, , 1>2, , ## #, 0, , 0, , x2, , 0, , 1, 2, , "x 2 y 2, , dy dx dz, , bounded by the graphs of y x, y x 2, y 1, y 3,, z 0, and z 5., 10. Evaluate eeeD (x 2 y2) dV, where D is the region bounded, by the graphs of y x2, z 4 y, and z 0., In Problems 11 and 12, change the indicated order of integration, to each of the other five orders., 2, , 2, , ex, , x dz dx dy, , "y 0, , |, , 1, , 9. Evaluate eeeD z dV, where D is the region in the first octant, , 62x2z, , 0, , 0, , p>2, , 5., , 2, , ## #, 0, , 4, , 24xy dz dy dx, , 12x, , 0, , ###, 0, , 8., , xy, , 0, , 1, , (x y z) dx dy dz, , ## #, , 8pk 5, (b 2 a 5)., 15, , du , , Answers to selected odd-numbered problems begin on page ANS-24., 7., , 62x, , #, , 2p, , p, 1, cos 3 f d du, 3, 0, , Spherical coordinates are used in navigation. If we think of the Earth as a sphere of fixed, radius centered at the origin, then a point P can be located by specifying two angles u and f., As shown in FIGURE 9.15.13, when f is held constant, the resulting curve is called a parallel., Fixed values of u result in curves called great circles. Half of one of these great circles joining, the north and south poles is called a meridian. The intersection of a parallel and a meridian, gives the position of a point P. If 0 f 180 and 180 u 180, the angles 90 f, and u are said to be the latitude and longitude of P, respectively. The prime meridian corresponds to a longitude of 0. The latitude of the equator is 0; the latitudes of the north and, south poles are, in turn, 90 (or 90 North) and 90 (or 90 South)., , 1, , 2 1, , c cos f , , REMARKS, , In Problems 1–8, evaluate the given iterated integral., 4, , (1 2 cos 2 f) sin f df du, , 0, , 0, , 4k 5, , (b 2 a 5), 15, , p, , CHAPTER 9 Vector Calculus, , 11., , 4 2 2y, , ## #, 0, , 0, , 4, , x 2y, , F(x, y, z) dz dx dy
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2, , 12., , ##, 0, , "36 2 9x2>2, , #, , 0, , 3, , F(x, y, z) dz dy dx, , 1, , In Problems 13 and 14, consider the solid given in the figure., Set up, but do not evaluate, the integrals giving the volume V of, the solid using the indicated orders of integration., z, , 13., , z=4, , y, y=8, , y = x3, , x, , FIGURE 9.15.14 Solid for Problem 13, , (a) dz dy dx, , (b) dx dz dy, , (c) dy dx dz, , x+z=2, , y, y=3, , x, , FIGURE 9.15.15 Solid for Problem 14, , (a) dx dz dy, , (b) dy dx dz, , (c) dz dx dy, , [Hint: Part (c) will require two integrals.], In Problems 15–20, sketch the region D whose volume V is, given by the iterated integral., 15., , 3, , ###, 0, , 16. 4, , dx dz dy, , 0, , 3, , ##, , "9 2 y 2, , 1, , "1 2 x2, , 0, , 0, , 17., , 2 2 2z>3, , 0, , # #, , #, , 1 "1 2 x, 2, , 18., , ##, 0, , 2, , 19., , 3, , 20., , "4 2 x2, , 0, 22y, , ## #, 0, , dz dx dy, , #, , #, , 5, , 1>x, , 0, , In Problems 35–38, convert the point given in cylindrical, coordinates to rectangular coordinates., 3p, 5p, , 5b, 36. a2,, , 3b, 4, 6, p, 7p, 37. a "3, , 4b, 38. a4,, , 0b, 3, 4, In Problems 39–42, convert the point given in rectangular, coordinates to cylindrical coordinates., 35. a10,, , 40. (2"3, 2, 17), 42. (1, 2, 7), , dz dy dx, , 41. ("2, "6, 2), , In Problems 4346, convert the given equation to cylindrical, coordinates., 43. x 2 y2 z 2 25, 45. x 2 y2 z2 1, , 44. x y z 1, 46. x2 z2 16, , 0, , x 2 y2, , dx dz dy, , 3, , dy dz dx, , 0, , In Problems 21–24, find the volume of the solid bounded by the, graphs of the given equations., 2, , the y-axis if the density is as given in Problem 25. Find the, radius of gyration., 32. Find the moment of inertia of the solid in Figure 9.15.15 about, the x-axis if the density is constant. Find the radius of gyration., 33. Find the moment of inertia about the z-axis of the solid in the, first octant that is bounded by the coordinate planes and the, graph of x y z 1 if the density is constant., 34. Find the moment of inertia about the y-axis of the solid, bounded by the graphs of z y, z 4 y, z 1, z 0, x 2,, and x 0 if the density at a point P is directly proportional to, the distance from the yz-plane., , 39. (1, 1, 9), , 4, , "y, , 29. x 2 y 2 1, z y 8, z 2y 2; r(x, y, z) x y 4, 30. x 2 y2 z2 1, z 1, z 2; r(x, y, z) z2 [Hint: Do, , dz dy dx, , "y, , 0, , ## #, 1, , "25 2 x2 2 y2, , 4, , 2, , In Problems 29 and 30, set up, but do not evaluate, the iterated, integrals giving the mass of the solid that has the given shape, and density., , 31. Find the moment of inertia of the solid in Figure 9.15.14 about, , z = √x, , 4, , the density at a point P is directly proportional to the distance, from the xy-plane., 26. Find the centroid of the solid in FIGURE 9.15.15 if the density is, constant., 27. Find the center of mass of the solid bounded by the graphs, of x 2 z 2 4, y 0, and y 3 if the density at a point P is, directly proportional to the distance from the xz-plane., 28. Find the center of mass of the solid bounded by the graphs of, y x 2, y x, z y 2, and z 0 if the density at a point P, is directly proportional to the distance from the xy-plane., , not use dz dy dx.], , z, , 14., , 24. x 2, y x, y 0, z x 2 y2, z 0, 25. Find the center of mass of the solid given in FIGURE 9.15.14 if, , 2, , 21. x y , 4 x y , z 0, z 3, 22. x 2 y2 4, z x y, the coordinate planes, first octant, 23. y x 2 z 2, y 8 x 2 z 2, , In Problems 47–50, convert the given equation to rectangular, coordinates., 47. z r 2, 49. r 5 sec u, , 48. z 2r sin u, 50. u p/6, , In Problems 51–58, use triple integrals and cylindrical, coordinates. In Problems 51–54, find the volume of the solid, that is bounded by the graphs of the given equations., 51. x 2 y2 4, x 2 y2 z 2 16, z 0, , 9.15 Triple Integrals, , |, , 573
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52. z 10 x 2 y2, z 1, 53. z x y , x y 25, z 0, , In Problems 67–70, convert the given equation to spherical, coordinates., , 54. y x 2 z 2, 2y x 2 z 2 4, , 67. x 2 y2 z 2 64, , 2, , 2, , 2, , 2, , 55. Find the centroid of the homogeneous solid that is bounded, , by the hemisphere z "a 2 2 x 2 2 y 2 and the plane z 0., 56. Find the center of mass of the solid that is bounded by the, graphs of y2 z 2 16, x 0, and x 5 if the density at a, point P is directly proportional to distance from the yz-plane., 57. Find the moment of inertia about the z-axis of the solid that, is bounded above by the hemisphere z "9 2 x 2 2 y 2, and below by the plane z 2 if the density at a point P is, inversely proportional to the square of the distance from the, z-axis., 58. Find the moment of inertia about the x-axis of the solid that, is bounded by the cone z "x 2 2 y 2 and the plane z 1 if, the density at a point P is directly proportional to the distance, from the z-axis., In Problems 59–62, convert the point given in spherical, coordinates to (a) rectangular coordinates and (b) cylindrical, coordinates., 2 p p, 59. a , , b, 3 2 6, 61. a8,, , p 3p, ,, b, 4 4, , 65. a, , !3 1, , , 1b, 2 2, , 1 5p p, , b, 3 3 6, , 62. a ,, , 64. (1, !3, 1), , !3, 1, , 0, b, 2, 2, , 9.16, , 69. z 3x 3y, , 68. x 2 y2 z 2 4z, , 2, , 70. x 2 y2 z 2 1, , In Problems 71–74, convert the given equation to rectangular, coordinates., 71. r 10, , 72. f p/3, , 73. r 2 sec f, , 74. r sin2 f cos f, , In Problems 75–82, use triple integrals and spherical, coordinates. In Problems 75–78, find the volume of the solid, that is bounded by the graphs of the given equations., 75. z "x 2 y 2, x 2 y2 z2 9, 76., 77., 78., 79., , x 2 y2 z 2 4, y x, y "3 x, z 0, first octant, z 2 3x 2 3y2, x 0, y 0, z 2, first octant, Inside x2 y2 z2 1 and outside z2 x2 y2, Find the centroid of the homogeneous solid that is bounded, , hemisphere z "1 2 x 2 2 y 2 and the plane z 0 if the, density at a point P is directly proportional to the distance, from the xy-plane., 81. Find the mass of the solid that is bounded above by the hemi-, , 5p 2p, 60. a5,, ,, b, 4 3, , 66. a, , 2, , by the cone z "x 2 y 2 and the sphere x2 y2 z2 2z., 80. Find the center of mass of the solid that is bounded by the, , In Problems 63–66, convert the points given in rectangular coordinates to spherical coordinates., 63. (5, 5, 0), , 2, , sphere z "25 2 x 2 2 y 2 and below by the plane z 4 if, the density at a point P is inversely proportional to the distance, from the origin. [Hint: Express the upper f limit of integration, as an inverse cosine.], 82. Find the moment of inertia about the z-axis of the solid that, is bounded by the sphere x 2 y2 z2 a2 if the density at a, point P is directly proportional to the distance from the origin., , Divergence Theorem, , INTRODUCTION In Section 9.14 we saw that Stokes’ theorem was a three-dimensional, generalization of a vector form of Green’s theorem. In this section we present a second vector, form of Green’s theorem and its three-dimensional analogue., , Another Vector Form of Green’s Theorem Let F(x, y) P(x, y) i Q(x, y) j be, a two-dimensional vector field, and let T (dx/ds) i (dy/ds) j be a unit tangent to a simple, closed plane curve C. In (1) of Section 9.14 we saw that 养C (F T) ds can be evaluated by a, double integral involving curl F. Similarly, if n (dy/ds) i (dx/ds) j is a unit normal to C, (check T n), then 养C (F n) ds can be expressed in terms of a double integral of div F. From, Green’s theorem,, 䉲, , 䉲, , BC, 䉲, , (F n) ds , , that is,, , 574, , |, , CHAPTER 9 Vector Calculus, , BC, 䉲, , 0Q, 0Q, 0P, 0P, P dy 2 Q dx 6 c, 2 a b d dA 6 c, , d dA;, 0x, 0y, 0x, 0y, , BC, 䉲, , R, , R, , (F n) ds 6 div F dA., R, , (1)
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The result in (1) is a special case of the divergence or Gauss’ theorem. The following is a, generalization of (1) to 3-space:, , Theorem 9.16.1, , Divergence Theorem, , Let D be a closed and bounded region in 3-space with a piecewise-smooth boundary S that is, oriented outward. Let F(x, y, z) P(x, y, z) i Q(x, y, z) j R(x, y, z) k be a vector field for, which P, Q, and R are continuous and have continuous first partial derivatives in a region of, 3-space containing D. Then, 6 (F n) dS 9 div F dV., S, , z, , PARTIAL PROOF: We will prove (2) for the special region D shown in FIGURE 9.16.1 whose, surface S consists of three pieces:, , n, S2, , S3, , D, n, , x, , (bottom), , S1: z f1(x, y),, , (x, y) in R, , (top), , S2: z f2(x, y),, , (x, y) in R, , (side), , S3: f1(x, y) z f2(x, y),, , n, , S1, , (2), , D, , y, R, , (x, y) on C,, , where R is the projection of D onto the xy-plane and C is the boundary of R. Since, , C, , div F , , FIGURE 9.16.1 Region D used in proof of, Theorem 9.16.1, , 0Q, 0R, 0P, , , 0x, 0y, 0z, , and, , F n P(i n) Q( j n) R(k n),, , we can write, 6 (F n) dS 6 P(i n) dS 6 Q( j n) dS 6 R( k n) dS, S, , S, , S, , S, , 0Q, 0P, 0R, 9 div F dV 9 0x dV 9 0y dV 9 0z dV., , and, , D, , D, , D, , D, , To prove (2) we need only establish that, 0P, 6 P (i n) dS 9 0x dV, , (3), , 0Q, 6 Q ( j n) dS 9 0y dV, , (4), , 0R, 6 R (k n) dS 9 0z dV., , (5), , S, , D, , S, , D, , S, , D, , Indeed, we shall prove only (5), since the proofs of (3) and (4) follow in a similar manner. Now,, 0R, 9 0z dV 6 c, D, , R, , #, , f2(x, y), , f1(x, y), , 0R, dz d dA 6 fR (x, y, f2(x, y)) 2 R(x, y, f1(x, y))g dA., 0z, R, , 9.16 Divergence Theorem, , |, , 575, , (6)
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Next we write, 6 R(k n) dS 6 R(k n) dS 6 R(k n) dS 6 R(k n) dS., S, , S1, , S2, , S3, , On S1: Since the outward normal points downward, we describe the surface as g(x, y, z) , f1(x, y) z 0. Thus,, , n, , 0f1, 0f1, i, j2k, 0x, 0y, 2, , 0f1, 0f1, 1 a b a b, Å, 0x, 0y, , 2, , so that k n , , From the definition of dS we then have, , 1, 0f1 2, 0f1 2, 1 a b a b, Å, 0x, 0y, , ., , 6 R(k n) dS 6 R(x, y, f1(x, y)) dA., S1, , (7), , R, , On S2: The outward normal points upward, so, 0f2, 0f2, i2, jk, 0x, 0y, , n, , 2, , Å, , 1 a, , 0f2, 0f2, b a b, 0x, 0y, , 1, , so that k n , , 2, , Å, , from which we find, , 1 a, , 0f2 2, 0f2 2, b a b, 0x, 0y, , 6 R(k n) dS 6 R(x, y, f2(x, y)) dA., S2, , z, , R, , On S3: Since this side is vertical, k is perpendicular to n. Consequently, k n 0 and, , n, , S, , (8), , 6 R(k n) dS 0., , D, , (9), , S3, , Finally, adding (7), (8), and (9), we get, y, , 6 fR(x, y, f2(x, y)) R(x, y, f1(x, y))] dA,, , n, x, , R, , FIGURE 9.16.2 Region D with no vertical, side, z, , D, , Sb, y, , Sa, x, , which is the same as (6)., Although we proved (2) for a special region D that has a vertical side, we note that this type, of region is not required in Theorem 9.16.1. A region D with no vertical side is illustrated in, FIGURE 9.16.2; a region bounded by a sphere or an ellipsoid also does not have a vertical side. The, divergence theorem also holds for the region D bounded between two closed surfaces, such as, the concentric spheres Sa and Sb shown in FIGURE 9.16.3; the boundary surface S of D is the union, of Sa and Sb. In this case S (F n) dS D div F dV becomes, , n, , FIGURE 9.16.3 Region D is bounded, between two concentric spheres, , 576, , |, , CHAPTER 9 Vector Calculus, , 6 (F n) dS 6 (F n) dS 9 div F dV,, Sb, , Sa, , D
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where n points outward from D; that is, n points away from the origin on Sb and n points toward, the origin on Sa., , Verifying Divergence Theorem, , EXAMPLE 1, , Let D be the region bounded by the hemisphere x 2 y2 (z 1)2 9, 1 z 4, and the, plane z 1. Verify the divergence theorem if F x i y j (z 1) k., SOLUTION The closed region is shown in FIGURE 9.16.4., , z, S1: x2 + y2 + (z – 1)2 = 9, 1≤z≤4, , n1, , n2, , Triple Integral: Since F x i y j (z 1) k, we see div F 3. Hence,, 9 div F dV 9 3 dV 3 9 dV 54p., , D, S2: z = 1, y, R, x2 + y2 = 9, , D, , D, , (10), , D, , x, , FIGURE 9.16.4 Hemispherical region D in, Example 1, , In the last calculation, we used the fact that D dV gives the volume of the hemisphere ( 23 p33)., Surface Integral: We write S S1 S2, where S1 is the hemisphere and S2 is the plane, z 1. If S1 is a level surface of g(x, y, z) x2 y2 (z 1)2, then a unit outer normal is, n, , =g, x i y j (z 2 1) k, y, x, z21, , i j, k., 2, 2, 2, 3, 3, 3, i=gi, "x y (z 2 1), y2, (z 2 1)2, x2, 1, 1, , , (x 2 y 2 (z 2 1)2) 9 3, 3, 3, 3, 3, 3, , Now, , Fn, , and so, , 6 (F n) dS 6 (3) a, S1, , R, , 2p, , 9, , # #, 0, , 3, , 0, , 3, "9 2 x 2 2 y 2, , dAb, , (9 2 r 2)1>2r dr du 54p. d polar coordinates, , On S2, we take n k so that F n z 1. But since z 1, S2 (z 1) dS 0., Hence, we see that S (F n) dS 54p 0 54p agrees with (10)., , Using Divergence Theorem, , EXAMPLE 2, 2, , If F xy i y z j z3 k, evaluate S (F n) dS, where S is the unit cube defined by 0 x 1,, 0 y 1, 0 z 1., SOLUTION See Figure 9.13.14 and Problem 38 in Exercises 9.13. Rather than evaluate six, surface integrals, we apply the divergence theorem. Since div F F y 2yz 3z2,, we have from (2), 2, 6 (F n) dS 9 ( y 2yz 3z ) dV, S, , D, 1, , , , ###, 0, , 0, , 1, , , , 1, , ##, 0, , 1, , 0, , ( y 2yz 3z 2) dx dy dz, , 1, , 0, , ( y 2yz 3z 2) dy dz, , 9.16 Divergence Theorem, , |, , 577
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, , Sr, , n, , P0, Dr, n, , FIGURE 9.16.5 Region Dr in (11), , #, , 1, , 0, , #, , 1, , 0, , a, , 1, y2, y 2z 3yz 2 b d dz, 2, 0, , 1, 1, 1, 1, a z 3z 2 b dz a z z 2 z 3 b d 2., 2, 2, 2, 0, , Physical Interpretation of Divergence In Section 9.14 we saw that we could, express the normal component of the curl of a vector field F at a point as a limit involving the, circulation of F. In view of (2), it is possible to interpret the divergence of F at a point as a limit, involving the flux of F. Recall from Section 9.7 that the flux of the velocity field F of a fluid is, the rate of fluid flow—that is, the volume of fluid flowing through a surface per unit time. In, Section 9.7 we saw that the divergence of F is the flux per unit volume. To reinforce this last idea, let us suppose P0(x0, y0, z 0) is any point in the fluid and Sr is a small sphere of radius r centered, at P0. See FIGURE 9.16.5. If Dr is the sphere Sr and its interior, then the divergence theorem gives, 6 (F n) dS 6 div F dV., Sr, , (11), , Dr, , If we take the approximation div F(P) div F(P0) at all points P(x, y, z) within the small sphere,, then (11) gives, 6 (F n) dS < 9 div F(P0) dV, Sr, , Dr, , div F(P0) 9 dV, , (12), , Dr, , div F(P0)Vr ,, where Vr is the volume ( 43 pr 3) of the spherical region Dr. By letting r S 0, we see from (12), that the divergence of F is the limiting value of the ratio of the flux of F to the volume of the, spherical region:, div F(P0) lim, rS0, , 1, (F n) dS., Vr 6, Sr, , Hence, divergence F is flux per unit volume., The divergence theorem is extremely useful in the derivation of some of the famous equations, in electricity and magnetism and hydrodynamics. In the discussion that follows we shall consider, an example from the study of fluids., , Continuity Equation At the end of Section 9.7 we mentioned that one interpretation, of div F was a measure of the rate of change of the density of a fluid at a point. To see why, this is so, let us suppose that F is a velocity field of a fluid and that r(x, y, z, t) is the density, of the fluid at a point P(x, y, z) at a time t. Let D be the closed region consisting of a sphere S, and its interior. We know from Section 9.15 that the total mass m of the fluid in D is given by, m D r(x, y, z, t) dV. The rate at which the mass increases in D is given by, 0r, dm, d, , r(x, y, z, t) dV 9, dV., dt, dt 9, 0t, D, , (13), , D, , Now from Figure 9.7.3 we saw that the volume of fluid flowing through an element of surface, area S per unit time is approximated by (F n) S. The mass of the fluid flowing through an, element of surface area S per unit time is then ( r F n) S. If we assume that the change in, 578, , |, , CHAPTER 9 Vector Calculus
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mass in D is due only to the flow in and out of D, then the volume of fluid flowing out of D per, unit time is given by (10) of Section 9.13, S (F n) dS, whereas the mass of the fluid flowing, out of D per unit time is S ( r F n) dS. Hence, an alternative expression for the rate at which, the mass increases in D is, 6 (rF n) dS., , (14), , S, , By the divergence theorem, (14) is the same as, 9 div(rF) dV., , (15), , D, , Equating (13) and (15) then yields, 0r, 9 0t dV 9 div(rF) dV or, D, , D, , 0r, 9 a 0t div( rF)b dV 0., D, , Since this last result is to hold for every sphere, we obtain the equation of continuity for fluid, flows:, 0r, div(rF) 0., 0t, , (16), , On page 514 we stated that if div F F 0, then a fluid is incompressible. This fact, follows immediately from (16). If a fluid is incompressible (such as water), then r is constant,, so consequently ( r F) r F. But in addition r/ t 0 and so (16) implies F 0., , 9.16, , Exercises, , Answers to selected odd-numbered problems begin on page ANS-24., , In Problems 1 and 2, verify the divergence theorem., 1. F xy i yz j xz k; D the region bounded by the unit cube, , defined by 0 x 1, 0 y 1, 0 z 1, , 2. F 6xy i 4yz j xey k; D the region bounded by the three, , 10. F 2yz i x 3j xy2 k; D the region bounded by the ellipsoid, , x 2 /a 2 y2 /b 2 z 2 /c 2 1, 11. F 2xz i 5y2 j z2 k; D the region bounded by z y,, z 4 y, z 2 12 x 2, x 0, z 0. See FIGURE 9.16.6., , coordinate planes and the plane x y z 1, , z, , In Problems 3–14, use the divergence theorem to find the, outward flux S (F n) dS of the given vector field F., , x=0, , 3. F x 3 i y 3 j z 3 k; D the region bounded by the sphere, , z=y, , 4., , x, , 5., 6., 7., , x 2 y2 z 2 a2, F 4x i y j 4z k; D the region bounded by the sphere, x 2 y2 z 2 4, F y2 i xz3 j (z 1)2 k; D the region bounded by the, cylinder x 2 y2 16 and the planes z 1, z 5, F x2 i 2yz j 4z 3 k; D the region bounded by the parallelepiped defined by 0 x 1, 0 y 2, 0 z 3, F y 3 i x3 j z3 k; D the region bounded within by, , z "4 2 x 2 2 y 2 , x2 y2 3, z 0, 8. F (x 2 sin y) i z 2 j xy3 k; D the region bounded by, y x 2, z 9 y, z 0, 9. F (x i y j z k)/(x 2 y2 z 2); D the region bounded by, the concentric spheres x 2 y2 z 2 a2, x 2 y2 z 2 b2,, b a, , y, z=4–y, z = 2 – 1 x2, 2, , FIGURE 9.16.6 Region D for Problem 11, 12. F 15x 2y i x 2z j y4 k; D the region bounded by x y 2,, , z x y, z 3, x 0, y 0, , 13. F 3x2y2 i y j 6z xy2 k; D the region bounded by the, , paraboloid z x 2 y2 and the plane z 2y, 14. F xy2 i x 2y j 6 sin x k; D the region bounded by the, cone z "x 2 y 2 and the planes z 2, z 4, , 9.16 Divergence Theorem, , |, , 579
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15. The electric field at a point P(x, y, z) due to a point charge q, , located at the origin is given by the inverse square field, Eq, , In Problems 17–21, assume that S forms the boundary of a, closed and bounded region D., 17. If a is a constant vector, show that S (a n) dS 0., 18. If F P i Q j R k and P, Q, and R have continuous second, , r, ,, iri3, , partial derivatives, prove that, , where r x i y j z k., (a) Suppose S is a closed surface, Sa is a sphere x 2 y2 z 2 a2, lying completely within S, and D is the region bounded, between S and Sa. See FIGURE 9.16.7. Show that the outward, flux of E for the region D is zero., (b) Use the result of part (a) to prove Gauss’ law:, 6 (E n) dS 4pq;, , 6 (curl F n) d S 0., S, , In Problems 19 and 20, assume that f and g are scalar functions, with continuous second partial derivatives. Use the divergence, theorem to establish Green’s identities., 19. 6 ( f =g) n dS 9 ( f =2g =f =g) dV, , S, , that is, the outward flux of the electric field E through, any closed surface (for which the divergence theorem, applies) containing the origin is 4pq., z, , S, , D, , 20. 6 ( f =g 2 g=f ) n dS 9 ( f =2g 2 g=2f ) dV, S, , D, , 21. If f is a scalar function with continuous first partial derivatives,, , prove that, 6 f n dS 9 =f dV., , Sa, , S, , y, D, x, , S, , FIGURE 9.16.7 Region D for Problem 15(a), 16. Suppose there is a continuous distribution of charge through-, , out a closed and bounded region D enclosed by a surface S., Then the natural extension of Gauss’ law is given by, 6 (E n) dS 9 4pr dV,, S, , D, , [Hint: Use (2) on f a, where a is a constant vector, and Problem 27, in Exercises 9.7.], 22. The buoyancy force on a floating object is B S p n dS,, where p is the fluid pressure. The pressure p is related to, the density of the fluid r (x, y, z) by a law of hydrostatics:, p r (x, y, z)g, where g is the constant acceleration of gravity. If the weight of the object is W mg, use the result of, Problem 21 to prove Archimedes’ principle, B W 0. See, FIGURE 9.16.8., , D, , B, , where r(x, y, z) is the charge density or charge per unit, volume., (a) Proceed as in the derivation of the continuity equation, (16) to show that div E 4pr., (b) Given that E is an irrotational vector field, show that the, potential f for E satisfies Poisson’s equation 2f 4pr., , 9.17, , W, , FIGURE 9.16.8 Floating object in Problem 22, , Change of Variables in Multiple Integrals, , INTRODUCTION In many instances it is either a matter of convenience or of necessity to, b, , make a substitution, or change of variable, in a definite integral ea f (x) dx in order to evaluate it., If f is continuous on [a, b], x g (u) has a continuous derivative, and dx g (u) du, then, , #, , b, , a, , 580, , |, , CHAPTER 9 Vector Calculus, , d, , f (x) dx , , # f (g(u)) g (u) du,, c, , (1)
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If the function g is one-to-one,, then it has an inverse and so, c g1(a) and d g1(b)., , where the u-limits of integration c and d are defined by a g(c) and b g(d). There are three, things that bear emphasizing in (1). To change the variable in a definite integral we replace x, where it appears in the integrand by g(u), we change the interval of integration [a, b] on the x-axis, to the corresponding interval [c, d] on the u-axis, and we replace dx by a function multiple (namely,, the derivative of g) of du. If we write J(u) dx/du, then (1) has the form, , #, , b, , f (x) dx , , a, , #, , d, , f ( g(u)) J(u) du., , (2), , c, , For example, using x 2 sin u, p/2 u p/2, we get, x-limits, , T, , #, , 2, , 0, , (2, 2), y=x, , "4 2 x 2 dx , , #, , p>2, , 0, , 2 cos u (2 cos u) du 4, , #, , p>2, , 0, , cos 2 u du p., , Double Integrals Although changing variables in a multiple integral is not as straightforward as the procedure in (1), the basic idea illustrated in (2) carries over. To change variables, in a double integral we need two equations such as, , y, R, , u-limits, T f (2 sin u) J (u), , f (x), , x f (u, v),, , x, , y g(u, v)., , (3), , To be analogous with (2), we expect that a change of variables in a double integral would take, the form, x2 + y2 = 8, , 6 F( x, y) dA 6 F( f (u, v), g(u, v)) J(u, v) dA ,, , (a) Region R in xy-plane, , R, , θ, π /2, π /4, , r, , √8, , (b) Region S in r θ -plane, , FIGURE 9.17.1 Region S is used to, evaluate (6), v, , x r cos u,, 2, , 5, 3, ( √ ,√ ), 2, 2, , u2 + v2 = 4, , S, , (1, 0), , S1, , 0, , S2, u, , (2, 0), , (a), y, , (4, 4), , x=4, , y=x, R, (1, 1), , y=1, , ##, , led to, , u2 – v 2 = 1, S3, , S, , where S is the region in the uv-plane corresponding to the region R in the xy-plane and J(u, v) is, some function that depends on the partial derivatives of the equations in (3). The symbol dA on, the right side of (4) represents either du dv or dv du., In Section 9.11 we briefly discussed how to change a double integral R F(x, y) dA from, rectangular coordinates to polar coordinates. Recall that in Example 2 of that section the, substitutions, , S, 0, , (4), , x, , "8 2 x 2, , y r sin u, , (5), , "8, , (6), , 1, dy dx , 5 x 2 y2, , p>2, , # #, p>4, , 0, , 1, r dr du., 5 r2, , As we see in FIGURE 9.17.1, the introduction of polar coordinates changes the original region, of integration R in the xy-plane to the more convenient rectangular region of integration S in, the ru-plane. We note, too, that by comparing (4) with (6), we can identify J(r, u) r and, dA dr du., The change-of-variable equations in (3) define a transformation or mapping T from the, uv-plane to the xy-plane. A point (x0, y0) in the xy-plane determined from x0 f (u0, v0), y0 g(u0, v0), is said to be an image of (u0, v0)., EXAMPLE 1, , Image of a Region, , Find the image of the region S shown in FIGURE 9.17.2(a) under the transformation x u2 v2,, y u2 v2 ., , (4, 1), x, , (b), , FIGURE 9.17.2 Region R is the image of, region S in Example 1, , SOLUTION We begin by finding the images of the sides of S that we have indicated by S1,, S2, and S3., S1: On this side v 0 so that x u2, y u2. Eliminating u then gives y x. Now, imagine moving along the boundary from (1, 0) to (2, 0) (that is, 1 u 2). The, equations x u2, y u2 then indicate that x ranges from x 1 to x 4 and y ranges, 9.17 Change of Variables in Multiple Integrals, , |, , 581
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simultaneously from y 1 to y 4. In other words, in the xy-plane the image of S1 is, the line segment y x from (1, 1) to (4, 4)., S2: On this boundary u2 v2 4 and so x 4. Now as we move from the point, (2, 0) to (#52 , #32 ), the remaining equation y u2 v2 indicates that y ranges from, y 22 02 4 to y (#52 )2 2 ( #32 )2 1. In this case the image of S2 is the vertical line segment x 4 starting at (4, 4) and going down to (4, 1)., , S3: Since u2 v2 1, we get y 1. But as we move on this boundary from ( #52 , #32 ),, to (1, 0), the equation x u2 v2 indicates that x ranges from x 4 to x 1. The image, of S3 is the horizontal line segment y 1 starting at (4, 1) and ending at (1, 1)., The image of S is the region R given in Figure 9.17.2(b)., Observe in Example 1 that as we traverse the boundary of S in the counterclockwise direction, the boundary of R is traversed in a clockwise manner. We say that the transformation of, the boundary of S has induced an orientation on the boundary of R., While a proof of the formula for changing variables in a multiple integral is beyond the level, of this text, we will give some of the underlying assumptions that are made about the equations, (3) and the regions R and S. We assume that, • The functions f and g have continuous first partial derivatives on S., • The transformation is one-to-one., • Each of the regions R and S consists of a piecewise-smooth simple closed curve and its, interior., • The determinant, 0x, 0u, 4, 0y, 0u, , 0x, 0v, 0x 0y, 0x 0y, 4 , 2, 0y, 0u 0v, 0v 0u, 0v, , (7), , is not zero on S., , v, , A transformation T is said to be one-to-one if each point (x0, y0) in R is the image under T of, a unique point (u0, v0) in S. Put another way, no two points in S have the same image in R. With, the restrictions that r 0 and 0 u 2p, the equations in (5) define a one-to-one transformation, from the ru-plane to the xy-plane. The determinant in (7) is called the Jacobian determinant, or, simply Jacobian, of the transformation T and is the key to changing variables in a multiple integral. The Jacobian of the transformation defined by the equations in (3) is denoted by the symbol, , y, T, , S, , T –1, , 0(x, y), ., 0(u, v), , R, (x0, y0), , (u0, v0), u, , FIGURE 9.17.3 Transformation T and its, inverse, , x, , Similar to the notion of a one-to-one function, a one-to-one transformation T has an inverse, transformation T 1 such that (u0, v0) is the image under T 1 of (x0, y0). See FIGURE 9.17.3. If it, is possible to solve (3) for u and v in terms of x and y, then the inverse transformation is defined, by a pair of equations, u h(x, y),, , v k(x, y)., , (8), , The Jacobian of the inverse transformation T 1 is, 0u, 0(u, v), 0x, 4, 0(x, y), 0v, 0x, 582, , |, , CHAPTER 9 Vector Calculus, , 0u, 0y, 4, 0v, 0y, , (9)
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and is related to the Jacobian of the transformation T by, 0(x, y) 0(u, v), 1., 0(u, v) 0(x, y), , (10), , Jacobian, , EXAMPLE 2, , The Jacobian of the transformation x r cos u, y r sin u is, 0x, 0(x, y), 0r, 4, 0y, 0(r, u), 0r, , 0x, 0u, cos u, 4 2, 0y, sin u, 0u, , r sin u, 2 r ( cos 2 u sin2 u) r., r cos u, , We now turn our attention to the main point of this discussion: how to change variables in a, multiple integral. The idea expressed in (4) is valid; the function J(u, v) turns out to be | (x, y) (u, v)|., Under the assumptions made above, we have the following result:, Theorem 9.17.1, , Change of Variables in a Double Integral, , If F is continuous on R, then, 0(x, y), 6 F(x, y) dA 6 F( f (u, v), g(u, v)) 2 0(u, v) 2 dA9., R, , (11), , S, , Formula (3) of Section 9.11 for changing a double integral to polar coordinates is just a special, case of (11) with, 2, , since r 0. In (6) then we have J(r, u) | (x, y)/ (r, u)| r., A change of variables in a multiple integral can be used for either a simplification of the, integrand or a simplification of the region of integration. The actual change of variables used is, often inspired by the structure of the integrand F(x, y) or by equations that define the region R. As, a consequence, the transformation is then defined by equations of the form given in (8); that is,, we are dealing with the inverse transformation. The next two examples will illustrate these ideas., , y, (0, π ), S2, (0, 0), , R, , x + 2y = 2 π, S3, (2π, 0), , S1, , x, , EXAMPLE 3, , (a), v, , Changing Variables in a Double Integral, , Evaluate R sin(x 2y) cos(x 2y) dA over the region R shown in FIGURE 9.17.4(a)., (2π, 2π), , v=u, , 0(x, y), 2 ZrZ r,, 0(r, u), , u = 2π, S, , u, , (0, 0), v = –u, (2π, –2π), (b), , FIGURE 9.17.4 Region S is the image of, region R in Example 3, , SOLUTION The difficulty in evaluating this double integral is clearly the integrand. The presence of the terms x 2y and x 2y prompts us to define the change of variables u x 2y,, v x 2y. These equations will map R onto a region S in the uv-plane. As in Example 1, we, transform the sides of the region., S1: y 0 implies u x and v x or v u. As we move from (2p, 0) to (0, 0), we see, that the corresponding image points in the uv-plane lie on the line segment v u from, (2p, 2p) to (0, 0)., S2: x 0 implies u 2y and v 2y, or v u. As we move from (0, 0) to (0, p),, the corresponding image points in the uv-plane lie on the line segment v u from, (0, 0) to (2p, 2p)., S3: x 2y 2p implies u 2p. As we move from (0, p) to (2p, 0), the equation, v x 2y shows that v ranges from v 2p to v 2p. Thus, the image of S3 is the, 9.17 Change of Variables in Multiple Integrals, , |, , 583
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vertical line segment u 2p starting at (2p, 2p) and going up to (2p, 2p). See, Figure 9.17.4(b)., Now, solving for x and y in terms of u and v gives, x, , 1, 1, (u v), y (u 2 v)., 2, 4, , 0x, 0(x, y), 0u, 4, 0y, 0(u, v), 0u, , Therefore,, , 0x, 1, 0v, 2, 4 4, 0y, 1, 0v, 4, , 1, 2, 1, 4 ., 1, 4, , 4, , Hence, from (11) we find that, 1, 6 sin (x 2y) cos (x 2 2y) dA 6 sin u cos v 2 4 2 dA9, R, , y, , S, , y=, , # #, , 1, 4, , #, , 2p, , , , 1, 2, , #, , 2p, , , , 1, 4, , #, , 2p, , , , , R, , EXAMPLE 4, xy = 5, xy = 1, x, (a), , (1, 5), , u, , sin u sin vR, , 0, , du, u, , sin2 u du, , 0, , (1 2 cos 2u) du, , 0, , 2p, 1, p, 1, cu 2 sin 2ud ., 4, 2, 2, 0, , SOLUTION In this case the integrand is fairly simple, but integration over the region R would, be tedious since we would have to express R xy dA as the sum of three integrals. (Verify this.), The equations of the boundaries of R suggest the change of variables, u, , y, ,, x2, , v xy., , (12), , Obtaining the image of R is a straightforward matter in this case, since the images of the curves, that make up the four boundaries are simply u 1, u 4, v 1, and v 5. In other words, the, image of the region R is the rectangular region S: 1 u 4, 1 v 5. See Figure 9.17.5(b)., Now instead of trying to solve the equations in (12) for x and y in terms of u and v, we can, compute the Jacobian (x, y)/ (u, v) by computing (u, v)/ (x, y) and using (10). We have, , S, , (4, 1), u, (b), , FIGURE 9.17.5 Region S is image of, region R in Example 4, |, , sin u cos v dv du, , u, , 0, , Evaluate R xy dA over the region R shown in FIGURE 9.17.5(a)., , (4, 5), , (1, 1), , u, , Changing Variables in a Double Integral, , v, , 584, , 1, 4, , y = 4x2, x2, , 2p, , , , CHAPTER 9 Vector Calculus, , 0u, 0(u, v), 0x, 4, 0(x, y), 0v, 0x, , 0u, 2y, 0y, , 4 3 x3, 0v, y, 0y, , 1, 3y, x2 3 2 ,, x, x
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and so from (10),, 0(x, y), 1, x2, 1, , ., 0(u, v), 0(u, v), 3y, 3u, 0(x, y), 1, 6 xy dA 6 v 2 3u 2 dA9, , Hence,, , R, , S, 4, , , , 1, 3, , ##, , 1, 3, , 4, , , , 4, , 1, , #, , 1, , #, , 4, , 1, , 5, , 1, , v, dv du, u, , v2 5, R du, 2u 1, 4, 1, du 4 ln uR 4 ln 4., u, 1, , Triple Integrals To change variables in a triple integral, let, x f (u, v, w),, , y g(u, v, w),, , z h(u, v, w),, , be a one-to-one transformation T from a region E in uvw-space to a region D in xyz-space. If F, is continuous on D, then, 0(x, y, z), 9 F(x, y, z) dV 9 F( f (u, v, w), g(u, v, w), h(u, v, w)) 2 0(u, v, w) 2 dV9, D, , E, , 0x, 0u, 0(x, y, z), 0y, 6, 0(u, v, w), 0u, 0z, 0u, , where, , 0x, 0v, 0y, 0v, 0z, 0v, , 0x, 0w, 0y, 6., 0w, 0z, 0w, , We leave it as an exercise for the reader to show that if T is the transformation from spherical to, rectangular coordinates defined by, x r sin f cos u,, , Exercises, , z r cos f,, , (13), , 0(x, y, z), r2 sin f., 0(r, f, u), , then, , 9.17, , y r sin f sin u,, , Answers to selected odd-numbered problems begin on page ANS-24., , 1. Consider a transformation T defined by x 4u v, y 5u 4v., , Find the images of the points (0, 0), (0, 2), (4, 0), and (4, 2) in, the uv-plane under T., , 2. Consider a transformation T defined by x "v 2 u,, , y v u. Find the images of the points (1, 1), (1, 3), and, ( "2, 2) in the xy-plane under T 1., , In Problems 3–6, find the image of the set S under the given, transformation., 3., 4., 5., 6., , S: 0 u 2, 0 v u; x 2u v, y u 3v, S: 1 u 4, 1 v 5; u x y, v x 2y, S: 0 u 1, 0 v 2; x u2 v2, y uv, S: 1 u 2, 1 v 2; x uv, y v2, 9.17 Change of Variables in Multiple Integrals, , |, , 585
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In Problems 7–10, find the Jacobian of the transformation, T from the uv-plane to the xy-plane., 7. x veu, y veu, 2, , 8. x e3u sin v, y e3u cos v, , y, y, 2y, 2x, 9. u 2 , v , 10. u 2, ,v 2, 2, x, x, x y, x y2, 11. (a) Find the image of the region S: 0 u 1, 0 v 1, under the transformation x u uv, y uv., (b) Explain why the transformation is not one-to-one on the, boundary of S., 12. Determine where the Jacobian (x, y)/ (u, v) of the transformation in Problem 11 is zero., In Problems 13–22, evaluate the given integral by means of the, indicated change of variables., 13. R (x y) dA, where R is the region bounded by the graphs of, , x 2y 6, x 2y 6, x y 1, x y 3; u x 2y,, vxy, cos 12(x 2 y), 14. 6, dA, where R is the region bounded by the, 3x y, R, , 15., , 16., , 17., , 18., , 19., , graphs of y x, y x p, y 3x 3, y 3x 6;, u x y, v 3x y, y2, 2, 6 x dA, where R is the region bounded by the graphs y x ,, R, y2, x2, y 12 x2, x y2, x 12 y2; u , v , y, x, 2, 2 3, R (x y ) dA, where R is the region bounded by the circles, x2 y2 2x, x2 y2 4x, x2 y2 2y, x2 y2 6y;, 2y, 2x, u 2, ,v 2, [Hint: Form u2 v2.], 2, x y, x y2, R (x2 y2) dA, where R is the region in the first quadrant, bounded by the graphs of x2 y2 a, x2 y2 b, 2xy c,, 2xy d, 0 a b, 0 c d; u x2 y2, v 2xy, R (x2 y2) sin xy dA, where R is the region bounded by the, graphs of x2 y2 1, x2 y2 9, xy 2, xy 2;, u x2 y2, v xy, x, 6 y x 2 dA, where R is the region in the first quadrant, R, , bounded by the graphs of x 1, y x 2 , y 4 x 2 ;, x "v 2 u, y v u, , Chapter in Review, , 9, , 1. A particle whose position vector is r(t) cos t i cos t j , 2. The path of a moving particle whose position vector is, , r(t) (t2 1) i 4 j t4 k lies in a plane. _____, 3. The binormal vector is perpendicular to the osculating plane., _____, 586, , |, , CHAPTER 9 Vector Calculus, , (2, 3) and (4, 1); x 2u 4v, y 3u v, , 21. R y4 dA, where R is the region in the first quadrant bounded, , by the graphs of xy 1, xy 4, y x, y 4x; u xy, v y/x, , 22. D (4z 2x 2y) dV, where D is the parallelepiped, , 1 y z 3, 1 y z 1, 0 x y 3; u y z ,, v y z, w x y, , In Problems 23–26, evaluate the given double integral by means, of an appropriate change of variables., 2, , 2, , 23. 10 01x e(yx)/(yx) dy dx, 24. 02 0x2 e y 2 2xy x dy dx, 25. R (6x 3y) dA, where R is the trapezoidal region in the first, , quadrant with vertices (1, 0), (4, 0), (2, 4), and ( 12 , 1), 26. R (x y)4 exy dA, where R is the square region with vertices, (1, 0), (0, 1), (1, 2), and (2, 1), 27. A problem in thermodynamics is to find the work done by an, ideal Carnot engine. This work is defined to be the area of, the region R in the first quadrant bounded by the isothermals, xy a, xy b, 0, a, b, and the adiabatics xy1.4 c,, 1.4, xy d, 0 c d. Use A R dA and an appropriate, substitution to find the area shown in FIGURE 9.17.6., y, xy = a, xy = b, , R, , xy1.4 = d, xy1.4 = c, x, , FIGURE 9.17.6 Region R for Problem 27, 28. Use V D dV and the substitutions u x /a, v y/b, w z /c, , to show that the volume of the ellipsoid x2/a2 y2/b2 z2/c2 1, is V 43 pabc., y2, x2, 29. Evaluate the double integral 6 a, b dA, where R is the, 25, 9, R, , elliptical region whose boundary is the graph of x2/25 y2/9 1., Use the substitutions u x/5, v y/3, and polar coordinates., 30. Verify that the Jacobian of the transformation given in (13), is (x, y, z)/ (r, f, u) r2 sin f., , Answers to selected odd-numbered problems begin on page ANS-24., , Answer Problems 1–20 without referring back to the text. Fill in, the blank or answer true/false. Where appropriate, assume continuity of P, Q, and their first partial derivatives., "2 sin t k moves with constant speed. _____, , 20. R y dA, where R is the triangular region with vertices (0, 0),, , 4. If r(t) is the position vector of a moving particle, then the veloc-, , ity vector v(t) r (t) and the acceleration vector a(t) r(t), are orthogonal. _____, 5. z is perpendicular to the graph of z f (x, y). _____, 6. If f 0, then f constant. _____, 7. The integral C (x2 y2) dx 2xy dy, where C is given by, y x3 from (0, 0) to (1, 1), has the same value on the curve, y x6 from (0, 0) to (1, 1). _____
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8. The value of the integral C 2xy dx x2 dy between two points, 9., , 10., 11., 12., 13., 14., , 15., 16., 17., 18., , A and B depends on the path C. _____, If C1 and C2 are two smooth curves such that eC1Pdx Q dy , eC2P dx Q dy, then eCP dx Q dy is independent of the, path. _____, If the work C F d r depends on the curve C, then F is nonconservative. _____, If P/ x Q/ y, then C P dx Q dy is independent of the, path. _____, In a conservative force field F, the work done by F around a, simple closed curve is zero. _____, Assuming continuity of all partial derivatives, f 0., _____, The surface integral of the normal component of the curl of a, conservative vector field F over a surface S is equal to zero., _____, Work done by a force F along a curve C is due entirely to the, tangential component of F. _____, For a two-dimensional vector field F in the plane z 0, Stokes’, theorem is the same as Green’s theorem. _____, If F is a conservative force field, then the sum of the potential, and kinetic energies of an object is constant. _____, If C P dx Q dy is independent of the path, then F P i Q j, is the gradient of some function f. _____, , 19. If f 1> "x 2 y 2 is a potential function for a conservative, , force field F, then F _____., , 20. If F f (x) i g( y) j h(z) k, then curl F _____., 21. Find the velocity and acceleration of a particle whose position, , 22., , 23., , 24., , 25., 26., , vector is r(t) 6t i t j t 2 k as it passes through the plane, x y z 4., The velocity of a moving particle is v(t) 10t i , (3t2 4t) j k. If the particle starts at t 0 at (1, 2, 3), what, is its position at t 2?, The acceleration of a moving particle is a(t) "2 sin t i , "2 cos t j. Given that the velocity and position of the particle, at t p/4 are v(p/4) i j k and r(p/4) i 2 j (p/4) k,, respectively, what was the position of the particle at t 3p/4?, Given that r(t) 12 t 2 i 13 t 3 j 12 t 2 k is the position vector of, a moving particle, find the tangential and normal components, of the acceleration at any t. Find the curvature., Sketch the curve traced by r(t) cosh t i sinh t j t k., Suppose that the vector function of Problem 25 is the position, vector of a moving particle. Find the vectors T, N, and B at, t 1. Find the curvature at that point., , In Problems 27 and 28, find the directional derivative of the, given function in the indicated direction., 2, , 2, , 27. f (x, y) x y y x; Du f in the direction of 2 i 6 j, 28. F(x, y, z) ln(x 2 y 2 z 2 ); D uF in the direction of, , 2 i j 2 k, , 29. Consider the function f (x, y) x2y4. At (1, 1) what is, , (a) The rate of change of f in the direction of i?, (b) The rate of change of f in the direction of i j?, (c) The rate of change of f in the direction of j?, , 30. Let w "x 2 y 2 z 2., , (a) If x 3 sin 2t, y 4 cos 2t, and z 5t3, find dw/dt., r, t, (b) If x 3 sin 2 , y 4 cos 2 , and z 5t3r 3, find w/ t., r, t, , 31. Find an equation of the tangent plane to the graph of, , z sin (xy) at (12, 23p, 12 "3 )., , 32. Determine whether there are any points on the surface z2 xy , , 2x y2 1 at which the tangent plane is parallel to z 2., 33. Express the volume of the solid shown in FIGURE 9.R.1 as one or, more iterated integrals using the order of integration (a) dy dx and, (b) dx dy. Choose either part (a) or part (b) to find the volume., z, z = √1 – x2, , y, y=x, , y = 2x, , x, , x=1, , FIGURE 9.R.1 Solid for Problem 33, 34. A lamina has the shape of the region in the first quadrant, , bounded by the graphs of y x2 and y x3. Find the center, of mass if the density at a point P is directly proportional to, the square of the distance from the origin., 35. Find the moment of inertia of the lamina described in Problem, 34 about the y-axis., 36. Find the volume of the sphere x2 y2 z2 a2 using a triple, integral in (a) rectangular coordinates, (b) cylindrical coordinates, and (c) spherical coordinates., 37. Find the volume of the solid that is bounded between the cones, z "x 2 y 2 and z "9x 2 9y 2 and the plane z 3., 38. Find the volume of the solid shown in FIGURE 9.R.2., z, , z = √4 – x2 – y2, , z = √3x2 + 3y2, z = √1 – x2 – y2, , y, , x, , FIGURE 9.R.2 Solid for Problem 38, , In Problems 39–42, find the indicated expression for the vector, field F x2 y i xy2 j 2xyz k., 39. F, 41. ( F), 43. Evaluate, , #, , C, , 40. F, 42. ( F), , z2, ds, where C is given by, x 2 y2, , x cos 2t,, , y sin 2t,, , z 2t,, , p t 2p., , CHAPTER 9 in Review, , |, , 587
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44. Evaluate C (xy 4x) ds, where C is given by 2x y 2, 45., , 46., 47., 48., 49., , 50., , from (1, 0) to (0, 2)., Evaluate C 3x2y2 dx (2x3y 3y2) dy, where C is given by, y 5x4 7x2 14x from (0, 0) to (1, 2)., y dx x dy, Show that, 2p , where C is the circle, BC x 2 y 2, x2 y2 a2., Evaluate C y sin pz dx x2 ey dy 3xyz dz, where C is given, by x t, y t 2, z t 3 from (0, 0, 0) to (1, 1, 1)., If F 4y i 6x j and C is given by x2 y2 1, evaluate C F dr, in two different ways., Find the work done by the force F x sin y i y sin x j, acting along the line segments from (0, 0) to (p/2, 0) and from, (p/2, 0) to (p/2, p)., 2, 1, Find the work done by F 2, i 2, j from (12 , 12 ), 2, x y, x y2, to (1, "3) acting on the path shown in FIGURE 9.R.3., 䉲, , 䉲, , S (F n) dS without the aid of the divergence theorem., [Hint: The lateral surface area of the cylinder is 2pac.], 3, , 63. Evaluate the integral R (x2 y2) "x 2 2 y 2 dA, where R is, , the region bounded by the graphs of x 0, x 1, y 0, and, y 1 by means of the change of variables u 2xy, v x2 y2., 64. Evaluate the integral, 6, R, , 1, 2, , "(x 2 y) 2(x y) 1, , dA,, , where R is the region bounded by the graphs of y x, x 2,, and y 0 by means of the change of variables x u uv,, y v uv., 65. As shown in FIGURE 9.R.4, a sphere of radius 1 has its center, on the surface of a sphere of radius a 1. Find the surface, area of that portion of the larger sphere cut out by the smaller, sphere., , y, , (1, √3), , (–1, 1), , ( (, , (1, 1), , 1 1, – ,, 2 2, , x, , FIGURE 9.R.4 Spheres in Problem 65, , FIGURE 9.R.3 Path for Problem 50, 51. Evaluate S (z/xy) dS, where S is that portion of the cylinder, , 52., 53., 54., 55., , 56., , 57., 58., , 59., , 60., 61., , 62., , z x2 in the first octant that is bounded by y 1, y 3, z 1,, z 4., If F i 2 j 3 k, find the flux of F through the square, defined by 0 x 1, 0 y 1, z 2., If F c(1/r), where c is constant and ||r|| r, r x i y j z k,, find the flux of F through the sphere x2 y2 z2 a2., Explain why the divergence theorem is not applicable in, Problem 53., Find the flux of F c(1/r) through any surface S that forms, the boundary of a closed bounded region of space not containing the origin., If F 6x i 7z j 8y k, use Stokes’ theorem to evaluate, S (curl F n) dS, where S is that portion of the paraboloid, z 9 x2 y2 within the cylinder x2 y2 4., Use Stokes’ theorem to evaluate C 2y dx 3x dy 10z dz,, where C is the circle (x 1)2 ( y 3)2 25, z 3., Find the work C F dr done by the force F x2 i y2 j z2 k, around the curve C that is formed by the intersection of the, plane z 2 y and the sphere x2 y2 z2 4z., If F x i y j z k, use the divergence theorem to evaluate, S (F n) dS, where S is the surface of the region bounded, by x2 y2 1, z 0, z 1., Repeat Problem 59 for F 13 x3 i 13 y3 j 13 z3 k., If F (x2 ey tan1z) i (x y)2 j (2yz x10) k, use the, divergence theorem to evaluate S (F n) dS, where S is the, surface of the region in the first octant bounded by z 1 x2,, z 0, z 2 y, y 0., Suppose F x i y j (z2 1) k and S is the surface of the, region bounded by x2 y2 a2 , z 0, z c. Evaluate, 䉲, , 66. On the surface of a globe or, more precisely, on the surface of, , the Earth, the boundaries of the states of Colorado and Wyoming, are both “spherical rectangles.” (In this problem we assume that, the Earth is a perfect sphere.) Colorado is bounded by the lines of, longitude 102W and 109W and the lines of latitude 37N and, 41N. Wyoming is bounded by longitudes 104W and 111W, and latitudes 41N and 45N. See FIGURE 9.R.5., (a) Without explicitly computing their areas, determine which, state is larger and explain why., (b) By what percentage is Wyoming larger (or smaller) than, Colorado? [Hint: Suppose the radius of the Earth is R., Project a spherical rectangle in the Northern Hemisphere, that is determined by latitudes u1 and u2 and longitudes, f1 and f2 onto the xy-plane.], (c) One reference book gives the areas of the two states as, 104,247 and 97,914 mi2. How does this answer compare, with the answer in part (b)?, , 䉲, , 588, , |, , CHAPTER 9 Vector Calculus, , WY, CO, , FIGURE 9.R.5 States WY and CO are spherical rectangles, in Problem 66
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17.1, , Complex Numbers, , INTRODUCTION You have undoubtedly encountered complex numbers in your earlier, , courses in mathematics. When you first learned to solve a quadratic equation ax2 bx c 0, by the quadratic formula, you saw that the roots of the equation are not real, that is, complex,, whenever the discriminant b2 4ac is negative. So, for example, simple equations such as, x2 5 0 and x2 x 1 0 have no real solutions. For example, the roots of the last equation, "3, 1, "3, 1, and , . If it is assumed that "3 "3"1, then the roots, are , 2, 2, 2, 2, 1, "3, 1, "3, are written , "1 and 2, "1., 2, 2, 2, 2, A Definition Two hundred years ago, around the time that complex numbers were gaining some respectability in the mathematical community, the symbol i was originally used as a, , disguise for the embarrassing symbol "1. We now simply say that i is the imaginary unit, and define it by the property i2 1. Using the imaginary unit, we build a general complex, number out of two real numbers., Definition 17.1.1, , Complex Number, , A complex number is any number of the form z a ib where a and b are real numbers, and i is the imaginary unit., , Note: The imaginary part of, z 4 9i is 9, not 9i., , Terminology The number i in Definition 17.1.1 is called the imaginary unit. The real, number x in z x iy is called the real part of z; the real number y is called the imaginary, part of z. The real and imaginary parts of a complex number z are abbreviated Re(z) and Im(z),, respectively. For example, if z 4 9i, then Re(z) 4 and Im(z) 9. A real constant multiple, of the imaginary unit is called a pure imaginary number. For example, z 6i is a pure imaginary, number. Two complex numbers are equal if their real and imaginary parts are equal. Since this, simple concept is sometimes useful, we formalize the last statement in the next definition., Definition 17.1.2, , Equality, , Complex numbers z1 x1 iy1 and z2 x2 iy2 are equal, z1 z2, if, Re(z1) Re(z2), , and, , Im(z1) Im(z2)., , A complex number x iy 0 if x 0 and y 0., , Arithmetic Operations Complex numbers can be added, subtracted, multiplied, and, divided. If z1 x1 iy1 and z2 x2 iy2, these operations are defined as follows., Addition:, , z1 z2 (x1 iy1) (x2 iy2) (x1 x2) i( y1 y2), , Subtraction:, , z1 z2 (x1 iy1) (x2 iy2) (x1 x2) i( y1 y2), , Multiplication:, , z1z2 (x1 iy1)(x2 iy2), x1x2 y1 y2 i( y1x2 x1 y2), , Division:, , x1 iy1, z1, , z2, x2 iy2, , , 820, , |, , CHAPTER 17 Functions of a Complex Variable, , x1x2 y1y2, y1x2 2 x1y2, i, x 22 y 22, x 22 y 22
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The familiar commutative, associative, and distributive laws hold for complex numbers., Commutative laws:, , e, , Associative laws:, , e, , Distributive law:, , z1 z2 z2 z1, z1z2 z2z1, z1 (z2 z3) (z1 z2) z3, z1(z2z3) (z1z2)z3, , z1(z2 z3) z1z2 z1z3, , In view of these laws, there is no need to memorize the definitions of addition, subtraction,, and multiplication. To add (subtract) two complex numbers, we simply add (subtract) the corresponding real and imaginary parts. To multiply two complex numbers, we use the distributive, law and the fact that i 2 1., EXAMPLE 1, , Addition and Multiplication, , If z1 2 4i and z2 3 8i, find (a) z1 z2 and (b) z1z2., SOLUTION, , (a) By adding the real and imaginary parts of the two numbers, we get, (2 4i) (3 8i) (2 3) (4 8)i 1 12i., , (b) Using the distributive law, we have, (2 4i)(3 8i) (2 4i)(3) (2 4i)(8i), 6 12i 16i 32i 2, (6 32) (16 12)i 38 4i., There is also no need to memorize the definition of division, but before discussing that we, need to introduce another concept., , Conjugate If z is a complex number, then the number obtained by changing the sign of, its imaginary part is called the complex conjugate or, simply, the conjugate of z. If z x iy,, then its conjugate is, z x 2 iy., For example, if z 6 3i, then z 6 3i; if z 5 i, then z 5 i. If z is a real number,, say z 7, then z 7. From the definition of addition it can be readily shown that the conjugate, of a sum of two complex numbers is the sum of the conjugates:, z1 z2 z1 z2 ., Moreover, we have the additional three properties, z1 2 z2 z1 2 z2 ,, , z1z2 z1z2 ,, , z1, z1, a b ., z2, z2, , The definitions of addition and multiplication show that the sum and product of a complex, number z and its conjugate z are also real numbers:, z z (x iy) (x iy) 2x, , (1), , zz (x iy)(x iy) x2 i2y2 x2 y2., , (2), , The difference between a complex number z and its conjugate z is a pure imaginary number:, z z (x iy) (x iy) 2iy., , (3), , Since x Re(z) and y Im(z), (1) and (3) yield two useful formulas:, Re(z) , , zz, 2, , and, , Im(z) , , z2z, ., 2i, , 17.1 Complex Numbers, , |, , 821
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However, (2) is the important relationship that enables us to approach division in a more practical, manner: To divide z1 by z2, we multiply both numerator and denominator of z1/z2 by the conjugate, of z2. This procedure is illustrated in the next example., , Division, , EXAMPLE 2, , If z1 2 3i and z2 4 6i, find (a), , z1, 1, and (b) ., z2, z1, , SOLUTION In both parts of this example we shall multiply both numerator and denominator, by the conjugate of the denominator and then use (2)., 2 2 3i, 2 2 3i 4 2 6i, 8 2 12i 2 12i 18i 2, , , 4 6i, 4 6i 4 2 6i, 16 36, , (a), , , , 10 2 24i, 5, 6, , 2, i., 52, 26, 13, , 1, 1 2 3i, 2 3i, 2, 3, , , , , i., 2 2 3i, 2 2 3i 2 3i, 49, 13, 13, , (b), y, z = x + iy, , x, , FIGURE 17.1.1 z as a position vector, , Geometric Interpretation A complex number z x iy is uniquely determined by an, ordered pair of real numbers (x, y). The first and second entries of the ordered pairs correspond,, in turn, with the real and imaginary parts of the complex number. For example, the ordered pair, (2, 3) corresponds to the complex number z 2 3i. Conversely, z 2 3i determines the, ordered pair (2, 3). In this manner we are able to associate a complex number z x iy with, a point (x, y) in a coordinate plane. But, as we saw in Section 7.1, an ordered pair of real numbers, can be interpreted as the components of a vector. Thus, a complex number z x iy can also, be viewed as a vector whose initial point is the origin and whose terminal point is (x, y). The, coordinate plane illustrated in FIGURE 17.1.1 is called the complex plane or simply the z-plane. The, horizontal or x-axis is called the real axis and the vertical or y-axis is called the imaginary axis., The length of a vector z, or the distance from the origin to the point (x, y), is clearly "x 2 y 2 ., This real number is given a special name., Modulus or Absolute Value, , Definition 17.1.3, , The modulus or absolute value of z x iy, denoted by ZzZ, is the real number, ZzZ "x 2 y 2 "z z., , (4), , Modulus of a Complex Number, , EXAMPLE 3, , If z 2 3i, then Zz Z "22 (3)2 "13., , y, , As FIGURE 17.1.2 shows, the sum of the vectors z1 and z2 is the vector z1 z2. For the triangle, given in the figure, we know that the length of the side of the triangle corresponding to the vector, z1 z2 cannot be longer than the sum of the remaining two sides. In symbols this is, , z1 + z2, z1, , z1, , Zz1 z2 Z Zz1 Z Zz2 Z., , z2, x, , FIGURE 17.1.2 Sum of vectors, , (5), , The result in (5) is known as the triangle inequality and extends to any finite sum:, Zz1 z2 z3 p zn Z Zz1 Z Zz2 Z Zz3 Z p Zzn Z., , (6), , Using (5) on z1 z2 (z2), we obtain another important inequality:, Zz1 z2 Z Zz1 Z Zz2 Z., 822, , |, , CHAPTER 17 Functions of a Complex Variable, , (7)
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REMARKS, Many of the properties of the real system hold in the complex number system, but there are, some remarkable differences as well. For example, we cannot compare two complex numbers, z1 x1 iy1, y1 0, and z2 x2 iy2, y2 0, by means of inequalities. In other words, statements such as z1 z2 and z2 z1 have no meaning except in the case when the two numbers, z1 and z2 are real. We can, however, compare the absolute values of two complex numbers., Thus, if z1 3 4i and z2 5 i, then |z1| 5 and |z2| !26, and consequently |z1| |z2|., This last inequality means that the point (3, 4) is closer to the origin than is the point (5, 1)., , Exercises, , 17.1, , Answers to selected odd-numbered problems begin on page ANS-39., , In Problems 1–26, write the given number in the form a ib., 3, , 1. 2i 3i 5i, 2. 3i 5 i 4 7i 3 10i 2 9, 3. i 8, , 4. i 11, , 5. (5 9i) (2 4i), , 6. 3(4 i) 3(5 2i), , 7. i (5 7i), , 8. i (4 i) 4i(1 2i), , 9. (2 3i)(4 i), , 10. ( 12 14 i)( 23 53 i), , 11. (2 3i)2, , 12. (1 i)3, , 2, 13., i, 2 2 4i, 15., 3 5i, , i, 14., 1i, 10 2 5i, 16., 6 2i, , 17., , (3 2 i) (2 3i), 1i, , 18., , 24. (2 3i) a, 25. a, , 22i 2, b, 1 2i, , i, 1, b a, b, 32i, 2 3i, , 33. 2z i(2 9i), 35. z 2 i, , 34. z 2z 7 6i 0, 36. z 2 4z, , 39. 10 8i, 11 6i, 40. 12 14 i, 23 16 i, 41. Prove that |z1 z2| is the distance between the points z1 and, , (1 i) (1 2 2i), (2 i) (4 2 3i), , z2 in the complex plane., , 42. Show for all complex numbers z on the circle x 2 y 2 4 that, , Zz 6 8iZ 12., , Discussion Problems, 43. For n a nonnegative integer, in can be one of four values: i, 1,, , 1, (1 i) (1 2 2i) (1 3i), , 17.2, , In Problems 33–38, use Definition 17.1.2 to find a complex, number z satisfying the given equation., , 22i, z, 38., 3 4i, 1 3i, 1z, In Problems 39 and 40, determine which complex number is, closer to the origin., , 20., , 26., , 28. Re(z 2), 30. Im( z 2 z 2), 32. Zz 5z Z, , 37. z 2z , , (4 5i) 2i 3, (2 i)2, 22. (1 i )2 (1 i)3, 1, 23. (3 6i) (4 i )(3 5i ) , 22i, (5 2 4i) 2 (3 7i), (4 2i) (2 2 3i), 21. i (1 i)(2 i)(2 6i), 19., , In Problems 27–32, let z x iy. Find the indicated expression., 27. Re(1/z), 29. Im(2z 4z 4i), 31. Zz 1 3iZ, , 2, , i, and 1. In each of the following four cases express the integer, exponent n in terms of the symbol k, where k 0, 1, 2, . . . ., (a) in i (b) in 1 (c) in i (d) in 1, 44. (a) Without doing any significant work such as multiplying, out or using the binomial theorem, think of an easy way, of evaluating (1 i)8., (b) Use your method in part (a) to evaluate (1 i)64., , Powers and Roots, , INTRODUCTION Recall from calculus that a point (x, y) in rectangular coordinates can also, , be expressed in terms of polar coordinates (r, u). We shall see in this section that the ability to, express a complex number z in terms of r and u greatly facilitates finding powers and roots of z., , Polar Form Rectangular coordinates (x, y) and polar coordinates (r, u) are related by the, equations x r cos u and y r sin u (see Section 14.1). Thus a nonzero complex number, z x iy can be written as z (r cos u) i(r sin u) or, z r (cos u i sin u)., 17.2 Powers and Roots, , (1), |, , 823
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y, z = x + iy, r, , r sin θ, , θ, , x, , r cos θ, , FIGURE 17.2.1 Polar coordinates, , We say that (1) is the polar form of the complex number z. We see from FIGURE 17.2.1 that the, polar coordinate r can be interpreted as the distance from the origin to the point (x, y). In other, words, we adopt the convention that r is never negative so that we can take r to be the modulus, of z; that is, r ZzZ. The angle u of inclination of the vector z measured in radians from the positive, real axis is positive when measured counterclockwise and negative when measured clockwise., The angle u is called an argument of z and is written u arg z. From Figure 17.2.1 we see that, an argument of a complex number must satisfy the equation tan u y/x. The solutions of this, equation are not unique, since if u0 is an argument of z, then necessarily the angles u0 2p,, u0 4p, . . ., are also arguments. The argument of a complex number in the interval p u p, is called the principal argument of z and is denoted by Arg z. For example, Arg(i) p/2., , A Complex Number in Polar Form, , EXAMPLE 1, , Express 1 "3i in polar form., , y, 5π /3, – π /3, , 1 – √3i, , x, , SOLUTION With x 1 and y "3, we obtain r ZzZ #(1)2 ("3)2 2. Now since, the point (1, "3) lies in the fourth quadrant, we can take the solution of tan u "3/1 "3, to be u arg z 5p/3. It follows from (1) that a polar form of the number is, z 2 acos, , As we see in FIGURE 17.2.2, the argument of 1 "3i that lies in the interval (p, p], the principal argument of z, is Arg z p/3. Thus, an alternative polar form of the complex number is, p, p, z 2 ccos a b i sin a b d ., 3, 3, , FIGURE 17.2.2 Two arguments of, z 1 "3i in Example 1, , 5p, 5p, i sin, b., 3, 3, , Multiplication and Division The polar form of a complex number is especially convenient to use when multiplying or dividing two complex numbers. Suppose, z1 r1(cos u1 i sin u1), , z2 r2(cos u2 i sin u2),, , and, , where u1 and u2 are any arguments of z1 and z2, respectively. Then, z1z2 r1r2[(cos u1 cos u2 sin u1 sin u2) i(sin u1 cos u2 cos u1 sin u2)], , (2), , and for z2 0,, z1, r1, [(cos u1 cos u2 sin u1 sin u2) i(sin u1 cos u2 cos u1 sin u2)]., z2, r2, , (3), , From the addition formulas from trigonometry, (2) and (3) can be rewritten, in turn, as, , and, , z1z2 r1r2 [cos(u1 u2) i sin(u1 u2)], , (4), , z1, r1, [cos(u1 u2) i sin(u1 u2)]., z2, r2, , (5), , Inspection of (4) and (5) shows that, Zz1z2 Z Zz1 Z Zz2 Z ,, arg(z1z2) arg z1 arg z2,, , and, , EXAMPLE 2, , 2, , Zz1 Z, z1, 2 , ,, z2, Zz2 Z, , z1, arga b arg z1 arg z2., z2, , (6), , (7), , Argument of a Product and of a Quotient, , We have seen that Arg z1 p/2 for z1 i. In Example 1 we saw that Arg z2 p/3 for, z2 1 "3i. Thus, for, z1z2 i(1 2 "3i ) "3 i, , 824, , |, , CHAPTER 17 Functions of a Complex Variable, , and, , z1, "3, i, 1, , , i, z2, 4, 4, 1 2 "3i
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it follows from (7) that, arg(z1z2) , , p, p, p, 2, , 2, 3, 6, , and, , z1, p, p, 5p, arga b 2 a b , ., z2, 2, 3, 6, , In Example 2 we used the principal arguments of z1 and z2 and obtained arg(z1z2) Arg(z1z2) and, arg(z1/z2) Arg(z1/z2). It should be observed, however, that this was a coincidence. Although (7), is true for any arguments of z1 and z2, it is not true, in general, that Arg(z1z2) Arg z1 Arg z2, and Arg(z1/z2) Arg z1 Arg z2. See Problem 39 in Exercises 17.2., , Integer Powers of z We can find integer powers of the complex number z from the, results in (4) and (5). For example, if z r (cos u i sin u), then with z1 z and z2 z, (4) gives, z 2 r 2 [cos (u u) i sin (u u)] r 2 (cos 2u i sin 2u)., Since z 3 z2z, it follows that, z3 r3 (cos 3u i sin 3u)., Moreover, since arg(1) 0, it follows from (5) that, 1, z2 r2 [cos(2u) i sin(2u)]., z2, Continuing in this manner, we obtain a formula for the nth power of z for any integer n:, z n r n (cos nu i sin nu)., EXAMPLE 3, , (8), , Power of a Complex Number, , 3, , Compute z for z 1 "3i., , SOLUTION In Example 1 we saw that, p, p, z 2 c cos a b i sin a b d ., 3, 3, , Hence from (8) with r 2, u p/3, and n 3, we get, , p, p, (1 2 "3i)3 23 c cos a3a b b i sin a3a b b d, 3, 3, 8[cos(p) i sin(p)] 8., , DeMoivre’s Formula When z cos u i sin u, we have ZzZ r 1 and so (8) yields, (cos u i sin u)n cos nu i sin nu., , (9), , This last result is known as DeMoivre’s formula and is useful in deriving certain trigonometric, identities. See Problems 37 and 38 in Exercises 17.2., , Roots A number w is said to be an nth root of a nonzero complex number z if w n z . If, , we let w r(cos f i sin f) and z r (cos u i sin u) be the polar forms of w and z , then in, view of (8) , w n z becomes, rn (cos nf i sin nf) r (cos u i sin u)., From this we conclude that rn r or r r1/n and, cos nf i sin nf cos u i sin u., By equating the real and imaginary parts, we get from this equation, cos nf cos u, , and, , sin nf sin u., 17.2 Powers and Roots, , |, , 825
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These equalities imply that nf u 2kp, where k is an integer. Thus,, f, , u 2kp, ., n, , As k takes on the successive integer values k 0, 1, 2, . . ., n 1, we obtain n distinct roots with, the same modulus but different arguments. But for k n we obtain the same roots because the, sine and cosine are 2p-periodic. To see this, suppose k n m, where m 0, 1, 2, . . . . Then, f, , u 2(n m)p, u 2mp, , 2p, n, n, , sin f sin a, , and so, , u 2mp, b,, n, , cos f cos a, , u 2mp, b., n, , We summarize this result. The n nth roots of a nonzero complex number z r (cos u i sin u), are given by, wk r 1>n c cos a, , where k 0, 1, 2, . . ., n 1., EXAMPLE 4, , u 2kp, u 2kp, b i sin a, bd,, n, n, , (10), , Roots of a Complex Number, , Find the three cube roots of z i., SOLUTION With r 1, u arg z p/2, the polar form of the given number is, z cos(p/2) i sin(p/2). From (10) with n 3 we obtain, wk (1)1>3 ccos a, , p>2 2kp, 3, , Hence, the three roots are, , p>2 2kp, 3, , b d , k 0, 1, 2., , k 0, w0 cos, , p, p, "3, 1, i sin , i, 6, 6, 2, 2, , k 1, w1 cos, , 5p, 5p, "3, 1, i sin, , i, 6, 6, 2, 2, , k 2, w2 cos, , 3p, 3p, i sin, i., 2, 2, , The root w of a complex number z obtained by using the principal argument of z with, k 0 is sometimes called the principal nth root of z. In Example 4, since Arg(i) p/2,, , y, w1, , b i sin a, , w0, x, w2, , FIGURE 17.2.3 Three cube roots of i, , w0 "3/2 (1/2)i is the principal third root of i., Since the roots given by (8) have the same modulus, the n roots of a nonzero complex, number z lie on a circle of radius r1/n centered at the origin in the complex plane. Moreover,, since the difference between the arguments of any two successive roots is 2p/n, the nth roots, of z are equally spaced on this circle. FIGURE 17.2.3 shows the three cube roots of i equally spaced, on a unit circle; the angle between roots (vectors) wk and wk 1 is 2p/3., As the next example will show, the roots of a complex number do not have to be “nice”, numbers as in Example 3., EXAMPLE 5, , Roots of a Complex Number, , Find the four fourth roots of z 1 i., SOLUTION In this case, r "2 and u arg z p/4. From (10) with n 4, we obtain, wk ( "2)1>4 ccos a, , 826, , |, , CHAPTER 17 Functions of a Complex Variable, , p>4 2kp, 4, , b i sin a, , p>4 2kp, 4, , b d , k 0, 1, 2, 3.
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The roots, rounded to four decimal places, are, k 0, w0 ( "2 )1>4 ccos, k 1, w1 ( "2)1>4 ccos, k 2, w2 ( "2)1>4 ccos, k 3, w3 ( "2)1>4 ccos, , 17.2, , Exercises, , 1. 2, , 2. 10, , 3. 3i, , 4. 6i, , 5. 1 i, , 6. 5 5i, , 7. "3 i, , 8. 2 2 "3i, , 10., , 12, , "3 i, , In Problems 11–14, write the number given in polar form in the, form a ib., 11. z 5 acos, , 7p, 7p, i sin, b, 6, 6, , 12. z 8 "2 acos, 13. z 6 acos, , 11p, 11p, i sin, b, 4, 4, , p, p, i sin b, 8, 8, , 14. z 10 acos, , p, p, i sin b, 5, 5, , In Problems 15 and 16, find z1z2 and z1/z2. Write the number in, the form a ib., 15. z1 2 acos, , p, p, 3p, 3p, i sin b, z2 4 acos, i sin, b, 8, 8, 8, 8, , 16. z1 "2 acos, , z2 "3 acos, , p, p, i sin b,, 4, 4, , p, p, i sin b, 12, 12, , In Problems 17–20, write each complex number in polar form., Then use either (4) or (5) to obtain a polar form of the given, number. Write the polar form in the form a ib., 17. (3 3i)(5 5 "3i), , 9p, 9p, i sin, d 0.2127 1.0696i, 16, 16, , 17p, 17p, i sin, d 1.0696 2 0.2127i, 16, 16, 25p, 25p, i sin, d 0.2127 2 1.0696i., 16, 16, , Answers to selected odd-numbered problems begin on page ANS-39., , In Problems 1–10, write the given complex number in polar form., , 3, 9., 1 i, , p, p, i sin d 1.0696 0.2127i, 16, 16, , 18. (4 4i)(1 i), , 19., , i, 2 2 2i, , 20., , "2 "6i, 1 "3i, , In Problems 21–26, use (8) to compute the indicated power., 21. (1 "3i)9, 23. ( 12 12 i)10, 25. acos, , p, p 12, i sin b, 8, 8, , 26. c "3 acos, , 22. (2 2i)5, 24. ("2 "6i)4, , 2p, 2p 6, i sin, bd, 9, 9, , In Problems 27–32, use (10) to compute all roots. Sketch these, roots on an appropriate circle centered at the origin., 27. (8)1/3, , 28. (1)1/8, , 29. (i)1/2, , 30. (1 i)1/3, , 31. (1 "3i)1/2, , 32. (1 "3i)1/4, , 33. z4 1 0, , 34. z8 2z4 1 0, , In Problems 33 and 34, find all solutions of the given equation., , In Problems 35 and 36, express the given complex number first, in polar form and then in the form a ib., 35. acos, , p, p, p 12, p 5, i sin b c2 acos i sin b d, 9, 9, 6, 6, , 3p, 3p 3, i sin, bd, 8, 8, 36., p, p 10, c2 acos, i sin b d, 16, 16, c8 acos, , 37. Use the result (cos u i sin u)2 cos 2u i sin 2u to find, , trigonometric identities for cos 2u and sin 2u., , 38. Use the result (cos u i sin u)3 cos 3u i sin 3u to find, , trigonometric identities for cos 3u and sin 3u., , 17.2 Powers and Roots, , |, , 827
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39. (a) If z1 1 and z2 5i, verify that, , 40. For the complex numbers given in Problem 39, verify in both, , parts (a) and (b) that, , Arg(z1z2) Arg(z1) Arg(z2)., , arg(z1z2) arg(z1) arg(z2), , (b) If z1 1 and z2 5i, verify that, Arg(z1/z2) Arg(z1) Arg(z2)., , 17.3, , z1, arg a b arg(z1) arg(z2)., z2, , and, , Sets in the Complex Plane, , INTRODUCTION In the preceding sections we examined some rudiments of the algebra and, , geometry of complex numbers. But we have barely scratched the surface of the subject known, as complex analysis; the main thrust of our study lies ahead. Our goal in the sections and chapters, that follow is to examine functions of a single complex variable z x iy and the calculus of, these functions., Before introducing the notion of a function of a complex variable, we need to state some, essential definitions and terminology about sets in the complex plane., , Terminology Before discussing the concept of functions of a complex variable, we need, to introduce some essential terminology about sets in the complex plane., z0, , ρ, , Suppose z0 x0 iy0. Since Zz z0 Z "(x 2 x0)2 ( y 2 y0)2 is the distance between the, points z x iy and z0 x0 iy0, the points z x iy that satisfy the equation, Zz z0 Z r,, , |z – z0| = ρ, , FIGURE 17.3.1 Circle of radius r, , r, , 0, lie on a circle of radius r centered at the point z0. See FIGURE 17.3.1., , EXAMPLE 1, , Circles, , (a) Zz Z 1 is the equation of a unit circle centered at the origin., (b) Zz 1 2i Z 5 is the equation of a circle of radius 5 centered at 1 2i., , z0, , FIGURE 17.3.2 Open set, , The points z satisfying the inequality Zz z0 Z r, r 0, lie within, but not on, a circle of, radius r centered at the point z0. This set is called a neighborhood of z0 or an open disk. A, point z0 is said to be an interior point of a set S of the complex plane if there exists some neighborhood of z0 that lies entirely within S. If every point z of a set S is an interior point, then S is, said to be an open set. See FIGURE 17.3.2. For example, the inequality Re(z) 1 defines a right, half-plane, which is an open set. All complex numbers z x iy for which x 1 are in this, set. If we choose, for example, z0 1.1 2i, then a neighborhood of z0 lying entirely in the set, is defined by Zz (1.1 2i)Z 0.05. See FIGURE 17.3.3. On the other hand, the set S of points in, the complex plane defined by Re(z) 1 is not open, since every neighborhood of a point on the, line x 1 must contain points in S and points not in S. See FIGURE 17.3.4., |z – (1.1 + 2i)| < 0.05, y, , y, , in S, , z = 1.1 + 2i, , not in S, x, , x, x=1, , FIGURE 17.3.3 Open set magnified view, of a point near x 1, , 828, , |, , CHAPTER 17 Functions of a Complex Variable, , x=1, , FIGURE 17.3.4 Set S is not open
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EXAMPLE 2, , Open Sets, , FIGURE 17.3.5 illustrates some additional open sets., y, , y, x, , x, , Im(z) < 0, lower half-plane, (a), , –1 < Re(z) < 1, infinite strip, (b), , y, , y, , x, , x, , |z| > 1, exterior of unit circle, (c), , 1 < |z| < 2, circular ring, (d), , FIGURE 17.3.5 Four examples of open sets, , The set of numbers satisfying the inequality, r1, , z2, , z1, , FIGURE 17.3.6 Connected set, , Zz z0 Z, , r2,, , such as illustrated in Figure 17.3.5(d), is called an open annulus., If every neighborhood of a point z0 contains at least one point that is in a set S and at least, one point that is not in S, then z0 is said to be a boundary point of S. The boundary of a set S, is the set of all boundary points of S. For the set of points defined by Re(z) 1, the points on, the line x 1 are boundary points. The points on the circle Z z i Z 2 are boundary points for, the disk Z z i Z 2., If any pair of points z1 and z2 in an open set S can be connected by a polygonal line that lies, entirely in the set, then the open set S is said to be connected. See FIGURE 17.3.6. An open connected, set is called a domain. All the open sets in Figure 17.3.5 are connected and so are domains. The, set of numbers satisfying Re(z) 4 is an open set but is not connected, since it is not possible, to join points on either side of the vertical line x 4 by a polygonal line without leaving the set, (bear in mind that the points on x 4 are not in the set)., A region is a domain in the complex plane with all, some, or none of its boundary points., Since an open connected set does not contain any boundary points, it is automatically a region., A region containing all its boundary points is said to be closed. The disk defined by Zz iZ 2, is an example of a closed region and is referred to as a closed disk. A region may be neither open, nor closed; the annular region defined by 1 Zz 5Z 3 contains only some of its boundary, points and so is neither open nor closed., , REMARKS, Often in mathematics the same word is used in entirely different contexts. Do not confuse the, concept of “domain” defined in this section with the concept of the “domain of a function.”, , 17.3 Sets in the Complex Plane, , |, , 829
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Exercises, , 17.3, , Answers to selected odd-numbered problems begin on page ANS-40., , In Problems 1–8, sketch the graph of the given equation., 1. Re(z) 5, , 2. Im(z) 2, , 17., , 3. Im(z 3i) 6, , 19., , 4. Im(z i) Re(z 4 3i), , 21., , 5. Zz 3i Z 2, , 6. Z2z 1 Z 4, , 23., , 7. Zz 4 3i Z 5, , 8. Zz 2 2i Z 2, 24., , In Problems 9–22, sketch the set of points in the complex plane, satisfying the given inequality. Determine whether the set is a, domain., 1, 11. Im(z) 3, 13. 2 Re(z 1), , 10. ZRe(z)Z, , 9. Re(z), , 26., , 5, , 14. 1 Im(z), , 4, , 25., , 2, , 12. Im(z i), , 17.4, , 1, 0, 16. Im(1/z), 2, 0 arg (z) 2p/3, 18. Zarg (z) Z, p/4, Zz iZ 1, 20. Zz iZ, 0, 2 Zz i Z 3, 22. 1 Zz 1 iZ, 2, Describe the set of points in the complex plane that satisfies, Zz 1 Z Zz iZ., Describe the set of points in the complex plane that satisfies, ZRe(z)Z ZzZ., Describe the set of points in the complex plane that satisfies, z2 z 2 2., Describe the set of points in the complex plane that satisfies, Zz iZ Zz iZ 1., , 15. Re(z 2), , 4, , Functions of a Complex Variable, , INTRODUCTION One of the most important concepts in mathematics is that of a function. You, may recall from previous courses that a function is a certain kind of correspondence between two, sets; more specifically: A function f from a set A to a set B is a rule of correspondence that assigns, to each element in A one and only one element in B. If b is the element in the set B assigned to the, element a in the set A by f, we say that b is the image of a and write b f (a). The set A is called, the domain of the function f (but is not necessarily a domain in the sense defined in Section 17.3)., The set of all images in B is called the range of the function. For example, suppose the set A is a set, of real numbers defined by 3 x q and the function is given by f (x) !x 2 3; then f (3) 0,, f (4) 1, f (8) !5, and so on. In other words, the range of f is the set given by 0 y q. Since, A is a set of real numbers, we say f is a function of a real variable x., , Functions of a Complex Variable When the domain A in the foregoing definition, of a function is a set of complex numbers z, we naturally say that f is a function of a complex, variable z or a complex function for short. The image w of a complex number z will be some, complex number u iv; that is,, w f (z) u(x, y) iv(x, y),, , (1), , where u and v are the real and imaginary parts of w and are real-valued functions. Inherent in the, mathematical statement (1) is the fact that we cannot draw a graph of a complex function w f (z), since a graph would require four axes in a four-dimensional coordinate system., Some examples of functions of a complex variable are, f (z) z2 4z,, y, , w = f (z), z, , domain of f, , v, , f (z) , range of f, w, , x, , u, , z, ,, z 1, 2, , f (z) z Re(z),, , z any complex number, z 2 i and z 2 i, z any complex number., , Each of these functions can be expressed in form (1). For example,, f (z) z2 4z (x iy)2 4(x iy) (x2 y2 4x) i(2xy 4y)., , (a) z-plane, , (b) w-plane, , FIGURE 17.4.1 Mapping from z-plane to, w-plane, , 830, , |, , Thus, u(x, y) x2 y2 4x, and v(x, y) 2xy 4y., Although we cannot draw a graph, a complex function w f (z) can be interpreted as a mapping, or transformation from the z-plane to the w-plane. See FIGURE 17.4.1., , CHAPTER 17 Functions of a Complex Variable
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v, , y, , EXAMPLE 1, u=1–, , x, , Find the image of the line Re(z) ⫽ 1 under the mapping f (z) ⫽ z2., , u, , SOLUTION For the function f (z) ⫽ z2 we have u(x, y) ⫽ x 2 ⫺ y 2 and v(x, y) ⫽ 2xy. Now,, Re(z) ⫽ x and so by substituting x ⫽ 1 into the functions u and v, we obtain u ⫽ 1 ⫺ y2 and, v ⫽ 2y. These are parametric equations of a curve in the w-plane. Substituting y ⫽ v/2 into, the first equation eliminates the parameter y to give u ⫽ 1 ⫺ v 2 /4. In other words, the image, of the line in FIGURE 17.4.2(a) is the parabola shown in Figure 17.4.2(b)., , x=1, (a) z-plane, , Image of a Vertical Line, , v2/4, , (b) w-plane, , FIGURE 17.4.2 Image of x ⫽ 1 is a, parabola, , We shall pursue the idea of f (z) as a mapping in greater detail in Chapter 20., It should be noted that a complex function is completely determined by the real-valued, functions u and v. This means a complex function w ⫽ f (z) can be defined by arbitrarily specifying, u(x, y) and v(x, y), even though u ⫹ iv may not be obtainable through the familiar operations on the, symbol z alone. For example, if u(x, y) ⫽ xy2 and v(x, y) ⫽ x2 ⫺ 4y3, then f (z) ⫽ xy2 ⫹ i(x2 ⫺ 4y3), is a function of a complex variable. To compute, say, f (3 ⫹ 2i), we substitute x ⫽ 3 and y ⫽ 2, into u and v to obtain f (3 ⫹ 2i) ⫽ 12 ⫺ 23i., , y, , i, , x, , –i, , FIGURE 17.4.3 f1(z) ⫽ z (normalized), , Complex Functions as Flows We also may interpret a complex function w ⫽ f (z), as a two-dimensional fluid flow by considering the complex number f (z) as a vector based, at the point z. The vector f (z) specifies the speed and direction of the flow at a given point z., FIGURES 17.4.3 and 17.4.4 show the flows corresponding to the complex functions f1(z) ⫽ z and, f2(z) ⫽ z 2, respectively., If x(t) ⫹ iy(t) is a parametric representation for the path of a particle in the flow, the tangent, vector T ⫽ x⬘(t) ⫹ iy⬘(t) must coincide with f (x(t) ⫹ iy(t)). When f (z) ⫽ u(x, y) ⫹ iv(x, y), it, follows that the path of the particle must satisfy the system of differential equations, dx, ⫽ u(x, y), dt, , y, , dy, ⫽ v(x, y)., dt, We call the family of solutions of this system the streamlines of the flow associated, with f (z)., x, , EXAMPLE 2, , Streamlines, , Find the streamlines of the flows associated with the complex functions (a) f1(z) ⫽ z and, (b) f2(z) ⫽ z2., SOLUTION (a) The streamlines corresponding to f1(z) ⫽ x ⫺ iy satisfy the system, FIGURE 17.4.4 f2(z) ⫽ z 2 (normalized), , dx, ⫽x, dt, dy, ⫽ ⫺y, dt, and so x(t) ⫽ c1et and y(t) ⫽ c2e⫺t. By multiplying these two parametric equations, we see, that the point x(t) ⫹ iy(t) lies on the hyperbola xy ⫽ c1c2., (b) To find the streamlines corresponding to f2(z) ⫽ (x2 ⫺ y2) ⫹ i 2xy, note that dx/dt ⫽ x2 ⫺ y2,, dy/dt ⫽ 2xy, and so, dy, 2xy, ⫽ 2, ., dx, x 2 y2, This homogeneous differential equation has the solution x2 ⫹ y2 ⫽ c2 y, which represents a, family of circles that have centers on the y-axis and pass through the origin., , Limits and Continuity The definition of a limit of a complex function f (z) as z S z0, has the same outward appearance as the limit in real variables., 17.4 Functions of a Complex Variable, , |, , 831
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Limit of a Function, , Definition 17.4.1, , Suppose the function f is defined in some neighborhood of z0, except possibly at z0 itself. Then, f is said to possess a limit at z0, written, lim f (z) L, , zSz0, , if, for each e, y, , v, , z0, , f(z), , z, , δ, , ε, L, , D, x, (a) δ -neighborhood, , u, (b) ε -neighborhood, , FIGURE 17.4.5 Geometric meaning of a, complex limit, , 0, there exists a d, , 0 such that Z f (z) LZ, , e whenever 0, , Zz z0 Z, , d., , In words, lim zSz0 f (z) L means that the points f (z) can be made arbitrarily close to the, point L if we choose the point z sufficiently close to, but not equal to, the point z0. As shown in, FIGURE 17.4.5, for each e-neighborhood of L (defined by Z f (z) LZ e) there is a d-neighborhood, of z0 (defined by Zz z0 Z d) so that the images of all points z z0 in this neighborhood lie in, the e-neighborhood of L., The fundamental difference between this definition and the limit concept in real variables, lies in the understanding of z S z0. For a function f of a single real variable x, lim xSx0 f (x) L, means f (x) approaches L as x approaches x0 either from the right of x0 or from the left of, x0 on the real number line. But since z and z0 are points in the complex plane, when we say, that lim zSz0 f (z) exists, we mean that f (z) approaches L as the point z approaches z0 from, any direction., The following theorem summarizes some properties of limits:, Limit of Sum, Product, Quotient, , Theorem 17.4.1, , Suppose lim zSz0 f (z) L1 and lim zSz0 g(z) L2. Then, (i) lim [ f (z) g(z)] L1 L2, zSz0, , (ii) lim f (z)g(z) L1L2, zSz0, , f (z), L1, ,, zSz0 g(z), L2, , (iii) lim, , L2 0., , Continuity at a Point, , Definition 17.4.2, , A function f is continuous at a point z0 if, lim f (z) f (z0)., , zSz0, , As a consequence of Theorem 17.4.1, it follows that if two functions f and g are continuous, at a point z0, then their sum and product are continuous at z0. The quotient of the two functions, is continuous at z0 provided g(z0) 0., A function f defined by, f (z) anzn an1zn1 p a2z2 a1z a0,, , an 0,, , (2), , where n is a nonnegative integer and the coefficients ai, i 0, 1, . . ., n, are complex constants,, is called a polynomial of degree n. Although we shall not prove it, the limit result lim z z0, zSz0, , indicates that the simple polynomial function f (z) z is continuous everywhere—that is, on, the entire z-plane. With this result in mind and with repeated applications of Theorem 17.4.1 (i), and (ii), it follows that a polynomial function (2) is continuous everywhere. A rational, function, f (z) , , g(z), ,, h(z), , where g and h are polynomial functions, is continuous except at those points at which h(z) is zero., 832, , |, , CHAPTER 17 Functions of a Complex Variable
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Derivative The derivative of a complex function is defined in terms of a limit. The symbol z, used in the following definition is the complex number x i y., Definition 17.4.3, , Derivative, , Suppose the complex function f is defined in a neighborhood of a point z0. The derivative of, f at z0 is, f 9(z0) lim, , DzS0, , f (z0 Dz) 2 f (z0), Dz, , (3), , provided this limit exists., If the limit in (3) exists, the function f is said to be differentiable at z0. The derivative of a, function w f (z) is also written dw/dz., As in real variables, differentiability implies continuity:, If f is differentiable at z0, then f is continuous at z0., Moreover, the rules of differentiation are the same as in the calculus of real variables. If f and g, are differentiable at a point z, and c is a complex constant, then, Constant Rules:, , d, d, c 0,, cf (z) c f 9(z), dz, dz, , (4), , Sum Rule:, , d, f f (z) g(z)g f 9(z) g9(z), dz, , (5), , Product Rule:, , d, f f (z)g(z)g f (z)g9(z) g(z) f 9(z), dz, , (6), , Quotient Rule:, Chain Rule:, , g(z) f 9(z) 2 f (z)g9(z), d f (z), c, d , dz g(z), fg(z)g 2, , (7), , d, f (g(z)) f 9(g(z))g9(z)., dz, , (8), , The usual Power Rule for differentiation of powers of z is also valid:, d n, z nz n 2 1 , n an integer., dz, EXAMPLE 3, , Using the Rules of Differentiation, , Differentiate (a) f (z) 3z4 5z3 2z and (b) f (z) , SOLUTION, , (9), , z2, ., 4z 1, , (a) Using the Power Rule (9) along with the Sum Rule (5), we obtain, f (z) 3 4z3 5 3z2 2 12z3 15z2 2., , (b) From the Quotient Rule (7),, f 9(z) , , (4z 1) 2z 2 z 2 4, 4z 2 2z, , ., (4z 1)2, (4z 1)2, , f (z0 Dz) 2 f (z0), DzS0, Dz, must approach the same complex number from any direction. Thus in the study of complex, variables, to require the differentiability of a function is a greater demand than in real variables., If a complex function is made up, such as f (z) x 4iy, there is a good chance that it is not, differentiable., In order for a complex function f to be differentiable at a point z0, lim, , 17.4 Functions of a Complex Variable, , |, , 833
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EXAMPLE 4, , A Function That Is Nowhere Differentiable, , Show that the function f (z) x 4iy is nowhere differentiable., SOLUTION With z x i y, we have, f (z z) f (z) (x x) 4i( y y) x 4iy x 4i y, and so, , lim, , DzS0, , f (z Dz) 2 f (z), Dx 4iDy, lim, ., Dz, DzS0 Dx iDy, , (10), , Now, if we let z S 0 along a line parallel to the x-axis, then y 0 and the value of (10) is 1., On the other hand, if we let z S 0 along a line parallel to the y-axis, then x 0 and the value, of (10) is seen to be 4. Therefore, f (z) x 4iy is not differentiable at any point z., , Analytic Functions While the requirement of differentiability is a stringent demand,, there is a class of functions that is of great importance whose members satisfy even more severe, requirements. These functions are called analytic functions., Definition 17.4.4, , Analyticity at a Point, , A complex function w f (z) is said to be analytic at a point z0 if f is differentiable at z0 and, at every point in some neighborhood of z0., A function f is analytic in a domain D if it is analytic at every point in D., The student should read Definition 17.4.4 carefully. Analyticity at a point is a neighborhood, property. Analyticity at a point is, therefore, not the same as differentiability at a point. It is left, as an exercise to show that the function f (z) Zz Z 2 is differentiable at z 0 but is differentiable, nowhere else. Hence, f (z) ZzZ 2 is nowhere analytic. In contrast, the simple polynomial f (z) z 2, is differentiable at every point z in the complex plane. Hence, f (z) z 2 is analytic everywhere., A function that is analytic at every point z is said to be an entire function. Polynomial functions, are differentiable at every point z and so are entire functions., , REMARKS, Recall from algebra that a number c is a zero of a polynomial function if and only if x c, is a factor of f (x). The same result holds in complex analysis. For example, since, f (z) z4 5z2 4 (z2 1)(z2 4), the zeros of f are i, i, 2i, and 2i. Hence,, f (z) (z i)(z i)(z 2i)(z 2i). Moreover, the quadratic formula is also valid. For, example, using this formula, we can write, f (z) z2 2z 2 (z (1 i))(z (1 i)), (z 1 i)(z 1 i)., See Problems 21 and 22 in Exercises 17.4., , Exercises, , 17.4, , Answers to selected odd-numbered problems begin on page ANS-40., , In Problems 1–6, find the image of the given line under the mapping f (z) z2., 1. y 2, 3. x 0, 5. y x, , 2. x 3, 4. y 0, 6. y x, , In Problems 7–14, express the given function in the form, f (z) u iv., 834, , |, , CHAPTER 17 Functions of a Complex Variable, , 7. f (z) 6z 5 9i, 8. f (z) 7z 9i z 3 2i, 9. f (z) z 2 3z 4i, , 10. f (z) 3z 2 2z, , 11. f (z) z3 4z, , 12. f (z) z4, , 13. f (z) z 1/z, , 14. f (z) , , z, z1
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In Problems 15–18, evaluate the given function at the indicated, points., 2, , 3, , 2, , 15. f (z) 2x y i(xy 2x 1), , (a) 2i, (b) 2 i, (c), 16. f (z) (x 1 1/x) i(4x2 2y2 4), (a) 1 i, (b) 2 i, (c), 17. f (z) 4z i z Re(z), (a) 4 6i, (b) 5 12i, (c), 18. f (z) ex cos y iex sin y, (a) pi/4, (b) 1 pi, (c), , 33. f (z) , , 1 4i, , z, z, , z 2 2 2z 2, zS1 i, z 2 2 2i, , zS1, , xy21, z21, , In Problems 25 and 26, use (3) to obtain the indicated derivative, of the given function., 25. f (z) z2, f (z) 2z, 26. f (z) 1/z, f (z) 1/z2, , In Problems 27–34, use (4)–(8) to find the derivative f (z) for, the given function., 27. f (z) 4z3 (3 i)z2 5z 4, , z, z 2 3i, , 36. f (z) , , 2i, z 2 2z 5iz, 2, , z3 z, z 2 4 3i, 38. f (z) 2, 2, z 4, z 2 6z 25, 39. Show that the function f (z) z is nowhere differentiable., 40. The function f (z) |z|2 is continuous throughout the entire, complex plane. Show, however, that f is differentiable only at, the point z 0. [Hint: Use (3) and consider two cases: z 0, and z 0. In the second case let z approach zero along a, line parallel to the x-axis and then let z approach zero along, a line parallel to the y-axis.], 37. f (z) , , lim, , 24. lim, , 5z 2 2 z, z3 1, , 35. f (z) , , In Problems 23 and 24, show that the given limit does not exist., zS0, , 34. f (z) , , 3 pi/3, , lim, , 23. lim, , 3z 2 4 8i, 2z i, , In Problems 35–38, give the points at which the given function, will not be analytic., , zSi, , 22., , 32. f (z) (2z 1/z)6, , 2 7i, , 19. lim (4z3 5z2 4z 1 5i), , 5z 2 2 2z 2, zS1 2 i, z1, 4, z 21, 21. lim, zSi z 2 i, , 30. f (z) (z5 3iz3)(z4 iz3 2z2 6iz), 31. f (z) (z2 4i)3, , 5 3i, , In Problems 19–22, the given limit exists. Find its value., , 20., , 29. f (z) (2z 1)(z2 4z 8i), , In Problems 41–44, find the streamlines of the flow associated, with the given complex function., 41. f (z) 2z, 43. f (z) 1/ z, , 42. f (z) iz, 44. f (z) x2 iy2, , In Problems 45 and 46, use a graphics calculator or computer, to obtain the image of the given parabola under the mapping, f (z) z2., 45. y 12 x2, , 46. y (x 1)2, , 28. f (z) 5z4 iz3 (8 i)z2 6i, , 17.5, , Cauchy–Riemann Equations, , INTRODUCTION In the preceding section we saw that a function f of a complex variable z is, , analytic at a point z when f is differentiable at z and differentiable at every point in some neighborhood of z. This requirement is more stringent than simply differentiability at a point because, a complex function can be differentiable at a point z but yet be differentiable nowhere else. A, function f is analytic in a domain D if f is differentiable at all points in D. We shall now develop, a test for analyticity of a complex function f (z) u(x, y) iv(x, y)., , A Necessary Condition for Analyticity In the next theorem we see that if a function, f (z) u(x, y) iv(x, y) is differentiable at a point z, then the functions u and v must satisfy a pair, of equations that relate their first-order partial derivatives. This result is a necessary condition, for analyticity., Theorem 17.5.1, , Cauchy–Riemann Equations, , Suppose f (z) u(x, y) iv(x, y) is differentiable at a point z x iy. Then at z the first-order, partial derivatives of u and v exist and satisfy the Cauchy–Riemann equations, 0u, 0v, , 0x, 0y, , and, , 0u, 0v, ., 0y, 0x, , 17.5 Cauchy–Riemann Equations, , (1), , |, , 835
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PROOF: Since f (z) exists, we know that, f 9(z) lim, , DzS0, , f (z Dz) 2 f (z), ., Dz, , (2), , By writing f (z) u(x, y) iv(x, y) and z x i y, we get from (2), f 9(z) lim, , DzS0, , u(x Dx, y Dy) iv(x Dx, y Dy) 2 u(x, y) 2 iv(x, y), ., Dx iDy, , (3), , Since this limit exists, z can approach zero from any convenient direction. In particular, if, z S 0 horizontally, then z x and so (3) becomes, f 9(z) lim, , DxS0, , u(x Dx, y) 2 u(x, y), v(x Dx, y) 2 v(x, y), i lim, ., Dx, DxS0, Dx, , (4), , Since f (z) exists, the two limits in (4) exist. But by definition the limits in (4) are the first partial, derivatives of u and v with respect to x. Thus, we have shown that, f 9(z) , , 0u, 0v, i ., 0x, 0x, , (5), , Now if we let z S 0 vertically, then z i y and (3) becomes, f 9(z) lim, , DyS0, , u(x, y Dy) 2 u(x, y), v(x, y Dy) 2 v(x, y), i lim, ,, iDy, DyS0, iDy, , (6), , which is the same as, f 9(z) i, , 0u, 0v, ., 0y, 0y, , (7), , Equating the real and imaginary parts of (5) and (7) yields the pair of equations in (1)., If a complex function f (z) u(x, y) iv(x, y) is analytic throughout a domain D, then the real, functions u and v must satisfy the Cauchy–Riemann equations (1) at every point in D., EXAMPLE 1, , Using the Cauchy–Riemann Equations, , The polynomial f (z) z2 z is analytic for all z and f (z) x2 y2 x i(2xy y). Thus,, u(x, y) x2 y2 x and v(x, y) 2xy y. For any point (x, y), we see that the Cauchy–, Riemann equations are satisfied:, 0u, 0v, 2x 1 , 0x, 0y, EXAMPLE 2, , and, , 0u, 0v, 2y ., 0y, 0x, , Using the Cauchy–Riemann Equations, , Show that the function f (z) (2x2 y) i( y2 x) is not analytic at any point., SOLUTION We identify u(x, y) 2x2 y and v(x, y) y2 x. Now from, 0u, 4x, 0x, , and, , 0v, 2y, 0y, , 0u, 1x, 0y, , and, , 0v, 1, 0x, , we see that 0u/0y 0v/0x but that the equality 0u/0x 0v/0y is satisfied only on the line, y 2x. However, for any point z on the line, there is no neighborhood or open disk about z in, which f is differentiable. We conclude that f is nowhere analytic., 836, , |, , CHAPTER 17 Functions of a Complex Variable
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Important., , By themselves, the Cauchy–Riemann equations are not sufficient to ensure analyticity., However, when we add the condition of continuity to u and v and the four partial derivatives,, the Cauchy–Riemann equations can be shown to imply analyticity. The proof is long and complicated and so we state only the result., Theorem 17.5.2, , Criterion for Analyticity, , Suppose the real-valued functions u(x, y) and v(x, y) are continuous and have continuous, first-order partial derivatives in a domain D. If u and v satisfy the Cauchy–Riemann equations, at all points of D, then the complex function f (z) u(x, y) iv(x, y) is analytic in D., EXAMPLE 3, , Using Theorem 17.5.2, , For the function f (z) , , y, x, 2i 2, we have, 2, x y, x y2, 2, , y2 2 x 2, 0u, 0v, 2, , 2 2, 0x, 0y, (x y ), , and, , 2xy, 0u, 0v, 2, ., 2 2, 0y, 0x, (x y ), , In other words, the Cauchy–Riemann equations are satisfied except at the point where, x2 y2 0; that is, at z 0. We conclude from Theorem 17.5.2 that f is analytic in any, domain not containing the point z 0., The results in (5) and (7) were obtained under the basic assumption that f was differentiable, at the point z. In other words, (5) and (7) give us a formula for computing f (z):, f 9(z) , , 0u, 0v, 0v, 0u, i, , 2i ., 0x, 0x, 0y, 0y, , (8), , For example, we know that f (z) z2 is differentiable for all z. With u(x, y) x2 y2, 0u/0x 2x,, v(x, y) 2xy, and 0v/0x 2y, we see that, f (z) 2x i2y 2(x iy) 2z., Recall that analyticity implies differentiability but not vice versa. Theorem 17.5.2 has an analogue, that gives sufficient conditions for differentiability:, If the real-valued functions u(x, y) and v(x, y) are continuous and have continuous firstorder partial derivatives in a neighborhood of z, and if u and v satisfy the Cauchy–Riemann, equations at the point z, then the complex function f (z) u(x, y) iv(x, y) is differentiable, at z and f (z) is given by (8)., The function f (z) x2 y2 i is nowhere analytic. With the identifications u(x, y) x2 and, v(x, y) y2, we see from, 0u, 0v, 0u, 0v, 2x,, 2y, and, 0,, 0, 0x, 0y, 0y, 0x, that the Cauchy–Riemann equations are satisfied only when y x. But since the functions u,, 0u/0x, 0u/0y, v, 0v/0x, and 0v/0y are continuous at every point, it follows that f is differentiable on, the line y x and on that line (8) gives the derivative f (z) 2x 2y., , Harmonic Functions We saw in Chapter 13 that Laplace’s equation 02u/0x2 02u/0y2 0, , occurs in certain problems involving steady-state temperatures. This partial differential equation, also plays an important role in many areas of applied mathematics. Indeed, as we now see, the, real and imaginary parts of an analytic function cannot be chosen arbitrarily, since both u and v, must satisfy Laplace’s equation. It is this link between analytic functions and Laplace’s equation, that makes complex variables so essential in the serious study of applied mathematics., , Definition 17.5.1, , Harmonic Functions, , A real-valued function f(x, y) that has continuous second-order partial derivatives in a domain D, and satisfies Laplace’s equation is said to be harmonic in D., , 17.5 Cauchy–Riemann Equations, , |, , 837
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A Source of Harmonic Functions, , Theorem 17.5.3, , Suppose f (z) u(x, y) iv(x, y) is analytic in a domain D. Then the functions u(x, y) and, v(x, y) are harmonic functions., We will see in Chapter 18 that, an analytic function possesses, derivatives of all orders., , PROOF: In this proof we shall assume that u and v have continuous second-order partial, derivatives. Since f is analytic, the Cauchy–Riemann equations are satisfied. Differentiating both, sides of 0u/0x 0v/0y with respect to x and differentiating both sides of 0u/0y 0v/0x with respect, to y then give, 0 2u, 0 2v, , 2, 0x 0y, 0x, , and, , 0 2u, 0 2v, , ., 2, 0y 0x, 0y, , With the assumption of continuity, the mixed partials are equal. Hence, adding these two equations gives, 0 2u, 0 2u, 2 0., 2, 0x, 0y, This shows that u(x, y) is harmonic., Now differentiating both sides of 0u/0x 0v/0y with respect to y and differentiating both sides, of 0u/0y 0v/0x with respect to x and subtracting yield, 0 2v, 0 2v, 2 0., 2, 0x, 0y, , Harmonic Conjugate Functions If f (z) u(x, y) iv(x, y) is analytic in a domain D,, then u and v are harmonic in D. Now suppose u(x, y) is a given function that is harmonic in D., It is then sometimes possible to find another function v(x, y) that is harmonic in D so that, u(x, y) iv(x, y) is an analytic function in D. The function v is called a harmonic conjugate, function of u., Harmonic Function/Harmonic Conjugate Function, , EXAMPLE 4, , (a) Verify that the function u(x, y) x3 3xy2 5y is harmonic in the entire complex plane., (b) Find the harmonic conjugate function of u., SOLUTION, , (a) From the partial derivatives, 0u, 3x 2 2 3y 2,, 0x, , 0 2u, 6x,, 0x 2, , 0u, 6xy 2 5,, 0y, , 0 2u, 6x, 0y 2, , we see that u satisfies Laplace’s equation:, 0 2u, 0 2u, , 6x 2 6x 0., 0x 2, 0y 2, (b) Since the harmonic conjugate function v must satisfy the Cauchy–Riemann equations,, we must have, 0v, 0u, , 3x 2 2 3y 2, 0y, 0x, , and, , 0v, 0u, 6xy 5., 0x, 0y, , (9), , Partial integration of the first equation in (9) with respect to y gives v(x, y) 3x2y y3 h(x)., From this we get, 0v, 6xy h9(x)., 0x, Substituting this result into the second equation in (9) gives h (x) 5, and so h(x) 5x C., Therefore, the harmonic conjugate function of u is v(x, y) 3x2y y3 5x C. The analytic, function is f (z) x3 3xy2 5y i(3x2y y3 5x C )., 838, , |, , CHAPTER 17 Functions of a Complex Variable
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REMARKS, Suppose u and v are the harmonic functions that comprise the real and imaginary parts of an, analytic function f (z). The level curves u(x, y) c1 and v(x, y) c2 defined by these functions, form two orthogonal families of curves. See Problem 32 in Exercises 17.5. For example, the, level curves generated by the simple analytic function f (z) z x iy are x c1 and y c2., The family of vertical lines defined by x c1 is clearly orthogonal to the family of horizontal, lines defined by y c2. In electrostatics, if u(x, y) c1 defines the equipotential curves, then, the other, and orthogonal, family v (x, y) c2 defines the lines of force., , Exercises, , 17.5, , Answers to selected odd-numbered problems begin on page ANS-40., , In Problems 1 and 2, the given function is analytic for all z., Show that the Cauchy–Riemann equations are satisfied at every, point., 1. f (z) z, , 3, , 2, , 2. f (z) 3z 5z 6i, , In Problems 3–8, show that the given function is not analytic at, any point., 3. f (z) Re(z), , 4. f (z) y ix, , 5. f (z) 4z 6z 3, 7. f (z) x 2 y 2, , 6. f (z) z 2, , 8. f (x) , , In Problems 9–14, use Theorem 17.5.2 to show that the given, function is analytic in an appropriate domain., 9. f (z) ex cos y iex sin y, 10. f (z) x sin x cosh y i(y cos x sinh y), 2, , 2, , f (z) x 2 y 2 2xyi; x-axis, f (z) 3x 2y 2 6x 2y 2i; coordinate axes, f (z) x 3 3xy 2 x i( y 3 3x 2y y); coordinate axes, f (z) x 2 x y i( y 2 5y x); y x 2, Use (8) to find the derivative of the function in Problem 9., Use (8) to find the derivative of the function in Problem 11., , 23. u(x, y) x, , 24. u(x, y) 2x 2xy, 2, , 25. u(x, y) x y, , 2, , 26. u(x, y) 4xy3 4x3y x, 27. u(x, y) loge(x2 y2), 28. u(x, y) ex (x cos y y sin y), , 2, , 11. f (z) e x 2 y cos 2xy ie x 2 y sin 2xy, 12. f (z) 4x2 5x 4y2 9 i(8xy 5y 1), , 29. Sketch the level curves u(x, y) c1 and v(x, y) c2 of the, , y, x21, 13. f (z) , 2i, 2, 2, (x 2 1) y, (x 2 1)2 y 2, 3, 2, x xy x, x 2y y 3 2 y, 14. f (x) , , i, x 2 y2, x 2 y2, , In Problems 15 and 16, find real constants a, b, c, and d so that, the given function is analytic., 15. f (z) 3x y 5 i(ax by 3), 16. f (z) x2 axy by2 i(cx2 dxy y2), , 17.6, , 17., 18., 19., 20., 21., 22., , In Problems 23–28, verify that the given function u is harmonic., Find v, the harmonic conjugate function of u. Form the corresponding analytic function f (z) u iv., , y, x, i 2, x 2 y2, x y2, , 2, , In Problems 17–20, show that the given function is not analytic, at any point but is differentiable along the indicated curve(s)., , analytic function f (z) z2., 30. Consider the function f (z) 1/z. Describe the level curves., 31. Consider the function f (z) z 1/z. Describe the level curve, v(x, y) 0., 32. Suppose u and v are the harmonic functions forming the real, and imaginary parts of an analytic function. Show that the, level curves u(x, y) c1 and v(x, y) c2 are orthogonal. [Hint:, Consider the gradient of u and the gradient of v. Ignore the, case where a gradient vector is the zero vector.], , Exponential and Logarithmic Functions, , INTRODUCTION In this and the next section, we shall examine the exponential, logarithmic,, , trigonometric, and hyperbolic functions of a complex variable z. Although the definitions of, these complex functions are motivated by their real variable analogues, the properties of these, complex functions will yield some surprises., , Exponential Function Recall that in real variables the exponential function f (x) ex, , has the properties, , f (x) f (x), , and, , f (x1 x2) f (x1)f (x2)., , 17.6 Exponential and Logarithmic Functions, , (1), |, , 839
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We certainly want the definition of the complex function f (z) ez, where z x iy, to reduce, ex for y 0 and to possess the same properties as in (1)., We have already used an exponential function with a pure imaginary exponent. Euler’s, formula,, eiy cos y i sin y,, , y a real number,, , (2), , played an important role in Section 3.3. We can formally establish the result in (2) by using the, Maclaurin series for ex and replacing x by iy and rearranging terms:, q (iy)k, (iy)2, (iy)3, (iy)4, e iy a, 1 iy , , , p, 2!, 3!, 4!, k 0 k!, , Maclaurin series for, cos y and sin y., , a1 2, , y2, y4, y6, y3, y5, y7, , 2, p b i ay 2, , 2, pb, 2!, 4!, 6!, 3!, 5!, 7!, , cos y i sin y., , For z x iy, it is natural to expect that, e xiy e xe iy, e xiy ex (cos y i sin y)., , and so by (2),, , Inspired by this formal result, we make the following definition., Exponential Function, , Definition 17.6.1, , ez e xiy ex (cos y i sin y)., , (3), , The exponential function ez is also denoted by the symbol exp z. Note that (3) reduces to ex when, y 0., , Complex Value of the Exponential Function, , EXAMPLE 1, , Evaluate e1.74.2i., SOLUTION With the identifications x 1.7 and y 4.2 and the aid of a calculator, we have,, to four rounded decimal places,, e1.7 cos 4.2 2.6837, , e1.7 sin 4.2 4.7710., , and, , It follows from (3) that e1.74.2i 2.6837 4.7710i., The real and imaginary parts of e z, u(x, y) e x cos y and v(x, y) e x sin y, are continuous, and have continuous first partial derivatives at every point z of the complex plane. Moreover, the, Cauchy–Riemann equations are satisfied at all points of the complex plane:, 0u, 0v, e x cos y , 0x, 0y, , and, , 0u, 0v, e x sin y ., 0y, 0x, , It follows from Theorem 17.5.2 that f (z) e z is analytic for all z; in other words, f is an entire, function., , Properties We shall now demonstrate that e z possesses the two desired properties given, , in (1). First, the derivative of f is given by (5) of Section 17.5:, , f (z) e x cos y i(e x sin y) e x (cos y i sin y) f (z)., As desired, we have established that, d z, e e z., dz, 840, , |, , CHAPTER 17 Functions of a Complex Variable
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Second, if z1 ⫽ x1 ⫹ iy1 and z2 ⫽ x2 ⫹ iy2, then by multiplication of complex numbers and the, addition formulas of trigonometry, we obtain, , y, z + 4π i, 3π i, , f (z1) f (z2) ⫽ e x1 (cos y1 ⫹ i sin y1) e x 2 (cos y2 ⫹ i sin y2), , z + 2π i, , ⫽ e x1 ⫹ x2 f( cos y1 cos y2 2 sin y1 sin y2) ⫹ i ( sin y1 cos y2 ⫹ cos y1 sin y2)g, , πi, z, x, z – 2πi, , ⫽ e x1 ⫹ x2 f( cos( y1 ⫹ y2) ⫹ i sin( y1 ⫹ y2)g ⫽ f (z1 ⫹ z2)., e z1e z2 ⫽ e z1 ⫹ z2., , In other words,, , –π i, , (4), , It is left as an exercise to prove that, e z1, ⫽ e z1 2 z2., e z2, , –3π i, , FIGURE 17.6.1 Values of f (z) ⫽ e z at the, four points are the same, , Periodicity Unlike the real function ex, the complex function f (z) ⫽ ez is periodic with, , y, , πi, , x, , –πi, , the complex period 2pi. Since e2pi ⫽ cos 2p ⫹ i sin 2p ⫽ 1 and, in view of (4), ez⫹2pi ⫽ eze2pi ⫽ ez, for all z, it follows that f (z ⫹ 2pi) ⫽ f (z). Because of this complex periodicity, all possible, functional values of f (z) ⫽ ez are assumed in any infinite horizontal strip of width 2p. Thus,, if we divide the complex plane into horizontal strips defined by (2n ⫺ 1)p ⬍ y ⱕ (2n ⫹ 1)p,, n ⫽ 0, ⫾1, ⫾2, . . ., then, as shown in FIGURE 17.6.1, for any point z in the strip ⫺p ⬍ y ⱕ p, the, values f (z), f (z ⫹ 2pi), f (z ⫺ 2pi), f (z ⫹ 4pi), and so on, are the same. The strip ⫺p ⬍ y ⱕ p, is called the fundamental region for the exponential function f (z) ⫽ ez. The corresponding flow, over the fundamental region is shown in FIGURE 17.6.2., , Polar Form of a Complex Number In Section 17.2, we saw that the complex number z, could be written in polar form as z ⫽ r (cos u ⫹ i sin u). Since eiu ⫽ cos u ⫹ i sin u, we can now, write the polar form of a complex number as, z ⫽ reiu., , FIGURE 17.6.2 Flow over the fundamental, region, , For example, in polar form z ⫽ 1 ⫹ i is z ⫽ "2e pi>4 ., , Circuits In applying mathematics, mathematicians and engineers often approach the same, problem in completely different ways. Consider, for example, the solution of Example 10 in, Section 3.8. In this example we used strictly real analysis to find the steady-state current ip(t) in, an LRC-series circuit described by the differential equation, L, , d 2q, dq, 1, ⫹R, ⫹ q ⫽ E0 sin gt., dt, C, dt 2, , Electrical engineers often solve circuit problems such as this using complex analysis. To, illustrate, let us first denote the imaginary unit !⫺1 by the symbol j to avoid confusion with, the current i. Since current i is related to charge q by i ⫽ dq/dt, the differential equation is the, same as, L, , 1, di, ⫹ Ri ⫹ q ⫽ E0 sin gt., dt, C, , Moreover, the impressed voltage E0 sin gt can be replaced by Im(E0e jgt ), where Im means the, “imaginary part of.” Because of this last form, the method of undetermined coefficients suggests, that we assume a solution in the form of a constant multiple of complex exponential—that is,, ip(t) ⫽ Im(Ae jgt). We substitute this expression into the last differential equation, use the fact that, q is an antiderivative of i, and equate coefficients of e jgt:, a jLg ⫹ R ⫹, , 1, b A ⫽ E0 gives A ⫽, jCg, , E0, 1, R ⫹ j aLg 2, b, Cg, , ., , 17.6 Exponential and Logarithmic Functions, , |, , 841
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The quantity Z R j(Lg 1/Cg) is called the complex impedance of the circuit. Note that the, modulus of the complex impedance, ZZZ "R 2 (Lg 2 1>Cg)2 , was denoted in Example 10, of Section 3.8 by the letter Z and called the impedance., Now, in polar form the complex impedance is, Lg 2, Z ZZ Ze ju, , tan u , , where, , 1, Cg, , R, , ., , Hence, A E0 /Z E0 /(ZZZe ju), and so the steady-state current can be written as, ip(t) Im, , E0 ju jgt, e e ., ZZZ, , The reader is encouraged to verify that this last expression is the same as (35) in Section 3.8., , Logarithmic Function The logarithm of a complex number z x iy, z 0, is defined, as the inverse of the exponential function—that is,, w ln z, , z ew., , if, , (5), , In (5) we note that ln z is not defined for z 0, since there is no value of w for which ew 0. To, find the real and imaginary parts of ln z , we write w u iv and use (3) and (5):, x iy euiv eu (cos v i sin v) eu cos v ieu sin v., The last equality implies x eu cos v and y eu sin v. We can solve these two equations for u, and v. First, by squaring and adding the equations, we find, e2u x2 y2 r 2 ZzZ 2, , and so, , u loge ZzZ,, , where loge ZzZ denotes the real natural logarithm of the modulus of z. Second, to solve for v, we, divide the two equations to obtain, y, tan v ., x, This last equation means that v is an argument of z; that is, v u arg z. But since there is, no unique argument of a given complex number z x iy, if u is an argument of z, then so is, u 2np, n 0, 1, 2, . . . ., Definition 17.6.2, , Logarithm of a Complex Number, , For z 0, and u arg z,, ln z loge|z| i(u 2np),, , n 0,, , 1,, , 2, . . . ., , (6), , As is clearly indicated in (6), there are infinitely many values of the logarithm of a, complex number z. This should not be any great surprise since the exponential function, is periodic., In real calculus, logarithms of negative numbers are not defined. As the next example will, show, this is not the case in complex calculus., , Complex Values of the Logarithmic Function, , EXAMPLE 2, , Find the values of (a) ln(2), (b) ln i, and (c) ln(1 i)., SOLUTION, , (a) With u arg(2) p and loge|2| 0.6932, we have from (6), ln(2) 0.6932 i(p 2np)., , 842, , |, , CHAPTER 17 Functions of a Complex Variable
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(b) With u arg(i) p/2 and loge|i| loge 1 0, we have from (6), ln i ia, , p, 2npb., 2, , In other words, ln i pi/2, 3pi/2, 5pi/2, 7pi/2, and so on., (c) With u arg(1 i) 5p/4 and loge|1 i| loge "2 0.3466, we have from (6), ln(1 i) 0.3466 i a, , EXAMPLE 3, , 5p, 2npb ., 4, , Solving an Exponential Equation, , Find all values of z such that ez !3 i., , SOLUTION From (5), with the symbol w replaced by z, we have z ln( !3 i). Now, | !3 i | 2 and tan u 1/ !3 imply that arg( !3 i) p/6, and so (6) gives, z log e2 ia, , p, 2npb, 6, , or z 0.6931 ia, , p, 2npb., 6, , Principal Value It is interesting to note that as a consequence of (6), the logarithm, of a positive real number has many values. For example, in real calculus, loge5 has only one, value: loge 5 1.6094, whereas in complex calculus, ln 5 1.6094 2npi. The value of ln 5, corresponding to n 0 is the same as the real logarithm loge 5 and is called the principal value, of ln 5. Recall that in Section 17.2 we stipulated that the principal argument of a complex number,, written Arg z, lies in the interval (p, p]. In general, we define the principal value of ln z as, that complex logarithm corresponding to n 0 and u Arg z. To emphasize the principal value, of the logarithm, we shall adopt the notation Ln z. In other words,, Ln z loge|z| i Arg z., , (7), , Since Arg z is unique, there is only one value of Ln z for each z 0., EXAMPLE 4, , Principal Values, , The principal values of the logarithms in Example 2 are as follows:, (a) Since Arg(2) p, we need only set n 0 in the result given in part (a) of Example 2:, Ln(2) 0.6932 pi., (b) Similarly, since Arg(i) p/2, we set n 0 in the result in part (b) of Example 2 to, obtain, Ln i , , p, i., 2, , (c) In part (c) of Example 2, arg(1 i) 5p/4 is not the principal argument of z 1 i., The argument of z that lies in the interval (p, p] is Arg(1 i) 3p/4. Hence, it follows, from (7) that, Ln(1 i) 0.3466 , , 3p, i., 4, , Up to this point we have avoided the use of the word function for the obvious reason that, ln z defined in (6) is not a function in the strictest interpretation of that word. Nonetheless, it is, customary to write f (z) ln z and to refer to f (z) ln z by the seemingly contradictory phrase, multiple-valued function. Although we shall not pursue the details, (6) can be interpreted as an, infinite collection of logarithmic functions (standard meaning of the word). Each function in the, collection is called a branch of ln z. The function f (z) Ln z is then called the principal branch, of ln z, or the principal logarithmic function. To minimize the confusion, we shall hereafter, simply use the words logarithmic function when referring to either f (z) ln z or f (z) Ln z ., 17.6 Exponential and Logarithmic Functions, , |, , 843
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Some familiar properties of the logarithmic function hold in the complex case:, ln(z1z2) ⫽ ln z1 ⫹ ln z2, , z1, ln a b ⫽ ln z1 ⫺ ln z2., z2, , and, , (8), , Equations (8) and (9) are to be interpreted in the sense that if values are assigned to two of the, terms, then a correct value is assigned to the third term., , Properties of Logarithms, , EXAMPLE 5, , Suppose z1 ⫽ 1 and z2 ⫽ ⫺1. Then if we take ln z1 ⫽ 2pi and ln z2 ⫽ pi, we get, ln(z1z2) ⫽ ln(⫺1) ⫽ ln z1 ⫹ ln z2 ⫽ 2pi ⫹ pi ⫽ 3pi, z1, ln a b ⫽ ln(⫺1) ⫽ ln z1 ⫺ ln z2 ⫽ 2pi ⫺ pi ⫽ pi., z2, , Just as (7) of Section 17.2 was not valid when arg z was replaced with Arg z , so too (8) is, not true, in general, when ln z is replaced by Ln z. See Problems 45 and 46 in Exercises 17.6., , y, , Analyticity The logarithmic function f (z) ⫽ Ln z is not continuous at z ⫽ 0 since f (0) is, not defined. Moreover, f (z) ⫽ Ln z is discontinuous at all points of the negative real axis. This, is because the imaginary part of the function, v ⫽ Arg z, is discontinuous only at these points., To see this, suppose x0 is a point on the negative real axis. As z S x0 from the upper half-plane,, Arg z S p, whereas if z S x0 from the lower half-plane, then Arg z S ⫺p. This means that, f (z) ⫽ Ln z is not analytic on the nonpositive real axis. However, f (z) ⫽ Ln z is analytic throughout, the domain D consisting of all the points in the complex plane except those on the nonpositive, real axis. It is convenient to think of D as the complex plane from which the nonpositive real, axis has been cut out. Since f (z) ⫽ Ln z is the principal branch of ln z, the nonpositive real axis, is referred to as a branch cut for the function. See FIGURE 17.6.3. It is left as exercises to show, that the Cauchy–Riemann equations are satisfied throughout this cut plane and that the derivative of Ln z is given by, , branch, cut, x, , FIGURE 17.6.3 Branch cut for Ln z, y, , d, 1, Ln z ⫽, z, dz, for all z in D., , i, , FIGURE 17.6.4 shows w ⫽ Ln z as a flow. Note that the vector field is not continuous along the, , x, –i, , (9), , branch cut., , Complex Powers Inspired by the identity xa ⫽ ea ln x in real variables, we can define com-, , plex powers of a complex number. If a is a complex number and z ⫽ x ⫹ iy, then za is defined by, za ⫽ ea ln z,, , z ⫽ 0., , (10), , a, , FIGURE 17.6.4 w ⫽ Ln z as a flow, , In general, z is multiple-valued since ln z is multiple-valued. However, in the special case when, a ⫽ n, n ⫽ 0, ⫾1, ⫾2, . . ., (10) is single-valued since there is only one value for z2, z3, z⫺1, and, so on. To see that this is so, suppose a ⫽ 2 and z ⫽ reiu, where u is any argument of z. Then, e2 ln z ⫽ e 2 (loger ⫹ iu) ⫽ e 2 loger ⫹ 2iu ⫽ e 2 loger e2iu ⫽ r 2 eiueiu ⫽ (reiu )(reiu ) ⫽ z2., If we use Ln z in place of ln z, then (10) gives the principal value of z a., , Complex Power, , EXAMPLE 6, 2i, , Find the value of i ., SOLUTION With z ⫽ i, arg z ⫽ p/2, and a ⫽ 2i , it follows from (10) that, i2i ⫽ e2i[loge1⫹i(p/2⫹2np)] ⫽ e⫺(1⫹4n)p, where n ⫽ 0, ⫾1, ⫾2, . . . . Inspection of the equation shows that i2i is real for every value of n., Since p/2 is the principal argument of z ⫽ i, we obtain the principal value of i2i for n ⫽ 0. To, four rounded decimal places, this principal value is i 2i ⫽ e⫺p ⫽ 0.0432., 844, , |, , CHAPTER 17 Functions of a Complex Variable
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Exercises, , 17.6, , Answers to selected odd-numbered problems begin on page ANS-41., , In Problems 1–10, express e z in the form a ib., p, p, 1. z , i, 2. z i, 6, 3, p, p, 3. z 1 , i, 4. z 2 2, i, 4, 2, 3p, 5. z p pi, 6. z p , i, 2, 7. z 1.5 2i, 8. z 0.3 0.5i, 9. z 5i, 10. z 0.23 i, , In Problems 35–38, find all values of z satisfying the given equation., 38. e2z ez 1 0, , 40. 3i/p, (1 i), , 42. (1 "3i)3i, , 41. (1 i), , e 2 3pi, 12. 3 pi>2, e, , In Problems 43 and 44, find the principal value of the given, quantity. Express answers in the form a ib., 43. (1)(2i/p), , 44. (1 i)2i, , 45. If z1 i and z2 1 i, verify that, , iz, , 14. f (z) e, , z2, , 16. f (z) e1/z, , 15. f (z) e, , 37. ez1 ie2, , 39. (i)4i, , Ln(z1z2) Ln z1 Ln z2., , In Problems 13–16, use Definition 17.6.1 to express the given, function in the form f (z) u iv., 13. f (z) e, , 36. e1/z 1, , In Problems 39–42, find all values of the given quantity., , In Problems 11 and 12, express the given number in the form a ib., 11. e15pi/4 e1pi/3, , 35. ez 4i, , 46. Find two complex numbers z1 and z2 such that, , 2z, , Ln(z1/z2) Ln z1 Ln z2., 47. Determine whether the given statement is true., , In Problems 17–20, verify the given result., e z1, 17. |ez| ex, 18. z2 e z1 2 z2, e, 19. ezpi ezpi, 20. (ez)n enz, n an integer, 21. Show that f (z) e z is nowhere analytic., 2, 22. (a) Use the result in2 Problem 15 to show that f (z) e z is an, entire function., 2, (b) Verify that u(x, y) Re(e z ) is a harmonic function., In Problems 23–28, express ln z in the form a ib., 23. z 5, 24. z ei, 25. z 2 2i, 26. z 1 i, 27. z "2 "6i, 28. z "3 i, , In Problems 29–34, express Ln z in the form a ib., 29. z 6 6i, 30. z e3, 31. z 12 5i, 32. z 3 4i, 5, 33. z (1 "3i), 34. z (1 i)4, , 17.7, , (a) Ln(1 i)2 2 Ln(1 i), (b) Ln i 3 3 Ln i, (c) ln i 3 3 ln i, 48. The laws of exponents hold for complex numbers a and b:, zazb zab,, , za, zab,, zb, , (za)n zna,, , n an integer., , However, the last law is not valid if n is a complex number., Verify that (i i)2 i 2i, but (i 2)i i 2i., 49. For complex numbers z satisfying Re(z), 0, show that (7), can be written as, Ln z , , y, 1, loge(x2 y2) i tan1 ., x, 2, , 50. The function given in Problem 49 is analytic., , (a) Verify that u(x, y) loge(x2 y2) is a harmonic function., (b) Verify that v(x, y) tan1( y/x) is a harmonic function., , Trigonometric and Hyperbolic Functions, , INTRODUCTION In this section we define the complex trigonometric and hyperbolic func-, , tions. Analogous to the complex functions ez and Ln z defined in the previous section, these, functions will agree with their real counterparts for real values of z. In addition, we will show, that the complex trigonometric and hyperbolic functions have the same derivatives and satisfy, many of the same identities as the real trigonometric and hyperbolic functions., , Trigonometric Functions If x is a real variable, then Euler’s formula gives, eix cos x i sin x, , and, , eix cos x i sin x., , By subtracting and then adding these equations, we see that the real functions sin x and cos x can, be expressed as a combination of exponential functions:, sin x , , e ix 2 eix, ,, 2i, , cos x , , e ix eix, ., 2, , 17.7 Trigonometric and Hyperbolic Functions, , (1), , |, , 845
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Using (1) as a model, we now define the sine and cosine of a complex variable:, Definition 17.7.1, , Trigonometric Sine and Cosine, , For any complex number z x iy,, sin z , , e iz 2 eiz, 2i, , and, , cos z , , e iz eiz, ., 2, , (2), , As in trigonometry, we define four additional trigonometric functions in terms of sin z, and cos z:, sin z, 1, 1, 1, , cot z , , csc z , , sec z , ., cos z, cos, z, tan z, sin z, , tan z , , (3), , When y 0, each function in (2) and (3) reduces to its real counterpart., , Analyticity Since the exponential functions eiz and eiz are entire functions, it follows, , that sin z and cos z are entire functions. Now, as we shall see shortly, sin z 0 only for the real, numbers z np, n an integer, and cos z 0 only for the real numbers z (2n 1)p/2, n an, integer. Thus, tan z and sec z are analytic except at the points z (2n 1)p/2, and cot z and, csc z are analytic except at the points z np., , Derivatives Since (d/dz)ez ez, it follows from the Chain Rule that (d/dz)eiz ieiz and, , (d /dz)eiz ieiz. Hence,, , d, d e iz 2 eiz, e iz eiz, sin z , , cos z., dz, dz, 2i, 2, In fact, it is readily shown that the forms of the derivatives of the complex trigonometric functions are the same as the real functions. We summarize the results:, d, sin z cos z, dz, , d, cos z sin z, dz, , d, tan z sec 2z, dz, , d, cot z csc 2z, dz, , d, sec z sec z tan z, dz, , d, csc z csc z cot z., dz, , (4), , Identities The familiar trigonometric identities are also the same in the complex case:, sin(z) sin z, , cos(z) cos z, , cos2z sin2z 1, sin(z1, , z2) sin z1 cos z2, , cos(z1, , z2) cos z1 cos z2 sin z1 sin z2, , sin 2z 2 sin z cos z, , cos z1 sin z2, , cos 2z cos2z sin2z., , Zeros To find the zeros of sin z and cos z we need to express both functions in the form, u iv. Before proceeding, recall from calculus that if y is real, then the hyperbolic sine and, hyperbolic cosine are defined in terms of the real exponential functions ey and ey:, sinh y , , e y 2 ey, 2, , and, , cosh y , , e y ey, ., 2, , (5), , Now from Definition 17.7.1 and Euler’s formula we find, after simplifying,, sin z , , e i(x iy) 2 ei(x iy), e y ey, e y 2 ey, sin x a, b i cos x a, b., 2i, 2, 2, , Thus from (5) we have, , sin z sin x cosh y i cos x sinh y., 846, , |, , CHAPTER 17 Functions of a Complex Variable, , (6)
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It is left as an exercise to show that, cos z cos x cosh y i sin x sinh y., , (7), , From (6), (7), and cosh2y 1 sinh2y, we find, |sin z|2 sin2 x sinh2y, 2, , 2, , (8), , 2, , |cos z| cos x sinh y., , and, , (9), , Now a complex number z is zero if and only if |z|2 0. Thus, if sin z 0, then from (8) we must, have sin2 x sinh2y 0. This implies that sin x 0 and sinh y 0, and so x np and y 0., Thus the only zeros of sin z are the real numbers z np 0i np, n 0, 1, 2, . . . . Similarly,, it follows from (9) that cos z 0 only when z (2n 1)p/2, n 0, 1, 2, . . . ., EXAMPLE 1, , Complex Value of the Sine Function, , From (6) we have, with the aid of a calculator,, sin(2 i) sin 2 cosh 1 i cos 2 sinh 1 1.4031 0.4891i., In ordinary trigonometry we are accustomed to the fact that |sin x| 1 and |cos x| 1. Inspection, of (8) and (9) shows that these inequalities do not hold for the complex sine and cosine, since, sinh y can range from q to q. In other words, it is perfectly feasible to have solutions for, equations such as cos z 10., EXAMPLE 2, , Solving a Trigonometric Equation, , Solve the equation cos z 10., SOLUTION From (2), cos z 10 is equivalent to (eiz eiz)/2 10. Multiplying the last, equation by eiz then gives the quadratic equation in eiz:, e2iz 20e iz 1 0., From the quadratic formula we find eiz 10 3 !11. Thus, for n 0, 1, 2, … , we, have iz loge(10, 3 !11) 2npi. Dividing by i and utilizing loge(10 3 !11) , loge(10 3 "11), we can express the solutions of the given equation as z 2np, i loge(10 3 "11)., , Hyperbolic Functions We define the complex hyperbolic sine and cosine in a manner, analogous to the real definitions given in (5)., Definition 17.7.2, , Hyperbolic Sine and Cosine, , For any complex number z x iy,, sinh z , , e z 2 ez, 2, , and, , cosh z , , e z ez, ., 2, , (10), , The hyperbolic tangent, cotangent, secant, and cosecant functions are defined in terms of, sinh z and cosh z:, tanh z , , sinh z, 1, 1, 1, , coth z , , sech z , , csch z , ., cosh z, tanh z, cosh z, sinh z, , (11), , The hyperbolic sine and cosine are entire functions, and the functions defined in (11) are analytic except at points where the denominators are zero. It is also easy to see from (10) that, d, sinh z cosh z, dz, , and, , d, cosh z sinh z ., dz, , (12), , It is interesting to observe that, in contrast to real calculus, the trigonometric and hyperbolic functions are related in complex calculus. If we replace z by iz everywhere in (10) and compare the results, with (2), we see that sinh(iz) i sin z and cosh(iz) cos z. These equations enable us to express, 17.7 Trigonometric and Hyperbolic Functions, , |, , 847
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sin z and cos z in terms of sinh(iz) and cosh(iz), respectively. Similarly, by replacing z by iz in (2) we, can express, in turn, sinh z and cosh z in terms of sin(iz) and cos(iz). We summarize the results:, sin z i sinh(iz),, sinh z i sin(iz),, , cos z cosh(iz), , (13), , cosh z cos(iz)., , (14), , Zeros The relationships given in (14) enable us to derive identities for the hyperbolic functions utilizing results for the trigonometric functions. For example, to express sinh z in the form, u iv we write sinh z i sin(iz) in the form sinh z i sin(y ix) and use (6):, sinh z i [sin(y) cosh x i cos(y) sinh x]., Since sin(y) sin y and cos(y) cos y, the foregoing expression simplifies to, Similarly,, , sinh z sinh x cos y i cosh x sin y., , (15), , cosh z cosh x cos y i sinh x sin y., , (16), , It also follows directly from (14) that the zeros of sinh z and cosh z are pure imaginary and are,, respectively,, pi, z npi and z (2n 1) ,, n 0, 1, 2, . . . ., 2, , Periodicity Since sin x and cos x are 2p-periodic, we can easily demonstrate that sin z, and cos z are also periodic with the same real period 2p. For example, from (6), note that, sin(z 2p) sin(x 2p iy), sin(x 2p) cosh y i cos(x 2p) sinh y, sin x cosh y i cos x sinh y;, that is, sin(z 2p) sin z. In exactly the same manner, it follows from (7) that cos(z 2p) cos z., In addition, the hyperbolic functions sinh z and cosh z have the imaginary period 2pi. This last, result follows from either Definition 17.7.2 and the fact that ez is periodic with period 2pi, or, from (15) and (16) and replacing z by z 2pi., , Exercises, , 17.7, , Answers to selected odd-numbered problems begin on page ANS-41., , In Problems 1–12, express the given quantity in the form a ib., 1. cos(3i), , 2. sin(2i), , p, 3. sina ib, 4, , 4. cos(2 4i), , p, 3ib, 2, 8. csc(1 i), 3p, 10. sinha, ib, 2, , 5. tan(i), , 6. cota, , 7. sec(p i), 9. cosh(pi), 11. sinha1 , , p, ib, 3, , 12. cosh(2 3i), , In Problems 13 and 14, verify the given result., , 17. sinh z i, 19. cos z sin z, , 18. sinh z 1, 20. cos z i sin z, , In Problems 21 and 22, use the definition of equality of complex, numbers to find all values of z satisfying the given equation., 21. cos z cosh 2, , 22. sin z i sinh 2, , 23. Prove that cos z cos x cosh y i sin x sinh y., 24. Prove that sinh z sinh x cos y i cosh x sin y., 25. Prove that cosh z cosh x cos y i sinh x sin y., 26. Prove that |sinh z|2 sin2y sinh2 x., 27. Prove that |cosh z|2 cos2y sinh2 x., 28. Prove that cos2z sin2z 1., , p, 5, i ln 2b , 2, 4, p, 3, 14. cosa i ln 2b i, 2, 4, , 29. Prove that cosh2z sinh2z 1., , In Problems 15–20, find all values of z satisfying the given, equation., , 31. Prove that tanh z is periodic with period pi., , 13. sina, , 15. sin z 2, , 848, , |, , 30. Show that tan z u iv, where, , u, , 16. cos z 3i, , CHAPTER 17 Functions of a Complex Variable, , sinh 2y, sin 2x, and v , ., cos 2x cosh 2y, cos 2x cosh 2y, , 32. Prove that (a) sin z sin z and (b) cos z cos z.
