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Multiple Integrals, g.1. Double Integrals., Let A be a finite region of the xy-plane and let f (x, y) be a function of the, , dependent variables, , andy deined at every point in A. Divide the region A into, , narts of area dA1,.... dAn: Let (, , y+) be any point inside the rth elementary area, , Then the limit of the sum if it exists of 2 f y ) dA, when the number of, r=1, , b-divisions tends to infinity and thereby making area of each sub-division tend to, ro., is called he double ntegral of f (x, y) over the region A and is denoted by, , SSrfoy) dA., Thus, , SSr, , y) dA, , 1), , Lim, , ==, , n, , 0, , dA,, , 1, , 0, , The region "4* is called the region of integration. The term double integral, fers to the dimensionality of the region A. It can be noted here that this definition, orresponds to the definition, J f ) d r = Lim, , f) dx, , 2 f ) da, , n, , r=1, , .(2), , dx 0, r the definite integral of a single variable., , 9-2. Tripple Integrals., , The results obtained above for two dimensions can be extended to finite regions, R. Divide, three dimensions. Supposef (x,y,z) is a function defined in a closed region, into n sub-regions dr1, Ór2,, òrn. Let dv; be the volume of the jth region, , e region, If (iyj, 2j), , . . ., , is any, , point in this region then, Lim, , f D ) dvj., , n0i =, , f,y,2) dv, , it exists is denoted by, , is called the tripple integral off (x, y, z) over R. In this, onfine ourselves mainly to double integrals., , nd, , and the ordinates, , x, , =, , shall however, , integrals., , Evaluation of double, 9-3. Theorem., If the egion, , J2 ), , chapter we, , A is the area bounded by the curves y =f ), a, , and x, , =, , b then, , Sa) Sx,y) dy, 167), , d
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168, , Integrals Calculus, , where the, , integration, , y first and x, , is treated, , is camied with, as, , C, respect, , to, , X=b, , X*a, , y12, , constant., , roo. Let us divide the whole region, , Into elementary rectangles of dimensions, or and dy, , by, , drawing, , lines, , parallel, , coordinate axes. One such reccangle, , in, , the, , figure, , 9.1, , 8xoys, , the, , is shown, , S, , by the shaded region., , By=f, , Then from definition, -X, , X, =, , x, where (, , fig. 9.1, , Lim, , ), , f ) d r dy, , 0, , dy 0, , Y) is a point inside the rth rectangle and the summation extends to all such, , rectangles into which the region is divided., , Let us consider a vertical strip Po0'P'. We first sum up all the elementary, , rectangles into which this strip may be supposed to be divided. Thus we get the sum, of f,y,) dr dy over the strip POQ'P'. Let n be the total number of such strips., , Then sum of all such strips gives the sum (i)., Lim, , Therefore, d x 0 2fns) dr dy, dy0, Lim, , dx, , 1, , Where ( Ys) is a point inside the sth rectangle in the rth strip and m the, number, of rectangles in the th strip. Summation inside the bracket is to be performed first, keeping x, constant., Now we can write, , Lim m, d y 0 f G n s )dy =, where y, , and y2 are, , the extreme, , f,)dy, , values, , of y in the rth, bounded below and above by the curves., y=Ji), , 1i ), , and, , and, , 2, , region A, , y=J2() we can take, , y2=f2(), , Therefore from (il) we have, Lim, , dy-0, , strip., , 1, , Since the, , fy)dy J, , ), , f ) dy, , F () say., Introducing this result in (ii) we have, =, , dr, Lim, dy, , 0, , f y ) drdy=Lim, , .iv, , F ) dx
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ltiple Integrals, , 169, , F) dx =Ja, Hence, , from () and (v), , we, , JJ, General practice is, Remark, , e Curves, , x, , =, , .(v), , get, , 2), , i ) y ) dy| de, to omit, , the brackets, , on, , the, , right, , and write it, , JJ snaa -JJ Sihra.), ) y )de, de dy, dy, , 1. It can be shown in, the same way that if the, 81 O),* = 82 V) and y = c, y = d, then, , and, , SS, , 81, , Here the integration is first carried with, respect y, x treatingy as a constant i.e. we sum along a horizontal, , as, , Vii), , is bounded by, , de| dy, , () ), , -Sr», , region, , ..vi), , simply, , . vii), , if the brackets, , are, , omitted., , rip first., , Remark 2. If the region of integration is bounded, , yf0), , two curves as in firuge 9.2 even then we can consider, , bounded by the four lines, , y f0%), , y=h) , y =fa), X =a, , 9-4. Limits of, , and x = b., , Integration forSS t(x, y) dx dy., , We have seen above that the integral is evaluated, , fig. 9.2, , integrating with respect to y first treatingx as constant., , other words we first integrate in a vertical strip (strip parallel to y-axis). Limits of, 5 integration therefore are the values of y for the lowest and highest points of the, These values are in general function of x and are obtained from the equations, , the curves bounding the strip., , Second integrationis made with respect to x which gives strip-wise summation, m the first to the last strip. Hence the extreme left value of a for the region gives, lower limt and extreme right one as the upper limit. Examples given below make, , ar the task of deciding the limits and evaluating such integrals., dr dy, , Example 1. Evaluate, , Solution., , dr dy, (on integrating with respect to y first, treating * as cnstant)
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170, , Integrals Calculus, , putting x, , -, , 2, , sin, , 6, dt, , 4 sin G cos 0 do, , =, , asink62 cOs 6). 4sin cos Dd0, , 8sin, , 0, , cos?, , 0do, 2.2.1, , 2T (3), -, , Example, region, , in the, , 2. Evaluate, , SS, , (a+y)dr dy, , positive quadrant for which x *y, , Solution. X+y=l is, , the, , over, s, , Y, , l., ., , a straight line making, , intercepts unity from the positive direction of the axes., Hence the region of integration is the A OAB in this, case., , If we take any, , strip parallel, , to y-axis then it 15, side for which y=0, Dy, the line AB on the other side for which y = 1 -x. and, Limits, of r are clrealy 0to 11, , bounded by x-axis, , Hence, , on, , one, , JJ +) de dy = J +)ad, , ig. 9., , dr, , -9, , --3Example, , 3., , Evaluate, , =x andy =x, , Solution. Let, , us, , JSy( +) de dy, , flirst mark the region of, , the, , area, , betwed, , (Gorakhpur 2005, 0, integration which is bounded by, , which is a st. line, id, , hich is a parabola., , Obviously the region of integration is the region, , OPAQO., , over
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171, , s h u l t i p l eI n t e g r a l s, , Solving (i), x= 0, , and, , (i), , we, , x =, , or, , A1.1, , get, y, , 1,, , ives y=0 and y=l respectively. Thus, Whic, , is (1, 1)., divide the region into strips parallel to, and consider one such strip PQ. This strip, , point, , A, , the, heNow, ys, , d e d by the line y = x, , on one side and, , bou, , vr on the other side, ory the limits for will be y =x as lower, = t Thus, and y = Vr as upper limit. Then the limits, 1,., will be 0 to, =, , himit, , for x Hence the given integral, , -, , JJy( +), , de, , fig. 9.4, , dy, , V, , [e)--S)], -, , Example 4. Evaluate Sf rd9 dr over the, , circle, , area, , of the, , r asin 6, Solution. r=a sin 6 is a circle with its centre C on, , he line 6, , Initial line is tangent at the pole, , Circle is symmertical about the line 6 = ., , Hence, , Initial Line, fig. 9.5, , he value of the integal, 2, , C, , value of the integral of, the half circle, , 2, , n/2, , a sin, r d6 dr, sin, , 2 J, , d6
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M u l t i p l eI n t e g r a l s, , 173, , ('a sin 0, r d6 dr., , Evaluate, , Evaluate J JJ de, EvaluateS2, , (KVP 2013), , dr over the arca of the circler =a cos 6., , Evaluate J +y)* dr dy over the area bounded by cllipse, .5,. Change of order of integration,, , $9-5. Ch, , Let us consider the integral, , (Gorakhpur 92, 9), , 8nU,, , GS) drdy, , This integral is a summation of strips parallel to y-axis. If we change the order, teoration, then the summation will be that of strips parallel to r-axis, and so, , Sre 2dding strips we must add up all elements in a strips parallel to the x-axis., , bet, , Following procedure is adopted tor the change of order of integration., , Mark the region ofintegration by drawing the curves, , y f i )y =f2 ) , x =a andx = b., ) Now divide the region of integration into different parts (if necessary) by, , drawing lines parallel tox-axs through the points where strips parallel tox-axis change, their character. These points may be points of intersection of any two of, = f ) y=fa ) , x = a and r, , = b falling in the region of integration., , i) Draw elementary strips parallel to x-axis in any one part of the region., The value of x in terms of y at the extremities of the elementary strips give limits of, for that part of the region., , Civ) For other parts of the region we proceed similarly and add all such integrals., (MGKY 2013), Following examples will clarify the method., , Example 5. Change the order of integration, , in, , f ) dr dy., , x =0, B, , Solution. The region of integration is bounded, byy=0,yx,x =0 and x = a. Thus 0AB is the region, , Xa, , of integration (figure 9.6), Consider an elementary strip parallel to the axis, pf x. Values of x at the extremities of this strip are, =y andx = a. These will be the lower and upper limits, of x, Also y varies from O to B, hence its limits are 0, , 0, , a., , xy0, , A, , Fig. 9.6, , Therefore, , s)ddy, f%y) dy d, , 7), , Example 6, Change the order of integration, , 4)Pnd hence find its value., , in, , i dy, (Gorakhpur 2004, MGKV 2014,, , Purvanchal, , 2014)
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174, The, boundedSolution., by the lines, , Integrals Calculus, , the lines yy ;region, infinite, boundary, =x,x, urst, ary., , of, , =, , Thus it, , 0, , integration is, (y-axis) and an\, , is, as shown in the upper half of the, figure 9.7., us divide, the region into, LO, Xaxis and, strips parallel, one such, The, extremities of thisconsider, lie on x 0 onstrip., strip, and, one, side, =x, , quadrant,, et, , x=0, yX, , =, , ony, limits ofx, , or x, , the other, side: Hence the, are from 0 to, y., Limits of y are, clearly 0 to co., =y, , on, , Fig. 9.7, , Therefore, , dr, , Example, , 7., , Change the, , Ca cos a, , order, , -), , *tan a, , 1, of integration in the double integral, , fy) de dy., , Solution. Here the, of, bounded by the line region integration, y=x tan a the circle, =a-x* orx+y =a the line, x, is, , is y-axis and the linex, to y-axis., , a cos a, , B, , 0, , which, which is parallel x=0, =, , =, , (Gorakhpur 88; Purvanchal 90), , Thus the region of, integration is clearly, OAB as shown in, 9.8, figure, The circle r+ =a and, the line, =x, , y, , the, , y=xtan, , tana, , intersect at A. Solving, we find, coordinates of A as (a cos a, athem, sin a), Now we can see from the, figure that the, to, , X=a cos, fig. 9.8, , strips parallel x-axis change their character at A. Thus, the region of, divided into two parts namely OLA and LAB., integration is, Now in the region LAB any, strip parallel tor-axis is bounded, side and the circle, by y-axis on one, , V, , 2+y2= 42, x = V(a-, , or, , the other side. Hence the limits of are 0 to, x, v(a- y') and the limits of y, OL to OB i.e. a sin a to a., on, , Again in the region OAL, , and the line y, , =, , x, , tan a, , or x, , any, , =y cot, , a, , are, , strip parallel, on, , to x-axis lies on, y-axis on one side, the oher side. Therefore the, limits, , be 0 to y cot a and the limits of y are clearly 0 to a sin a., , of x will, bea
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175, M u l i p l eI n t e g r a l s, , Hence we have, , fa cos a(a?, , fa,y), , y, , de dy =., , cot a, , fr,y) dy dr, , tan a, , f , y ) dy de., a sin a, , Example, , 8. Change, , the order of integration, , in the, , integral, , (a, , (Gorakhpur 99), The, , Solution., , region, , circle), y, 242 a x =0 (a, the lines x, , Darabola), , (aj, , the region, , =, , and, , of integration is bounded, , V(ar), , =, , 0 and, , of integration is OPACO,, , or, , x, , =, , fig., , y2, a., , Thus, , P yax, , 9.9., , (a,a), , to x-axis change their, The strips parallel, the circle. Let, characterat the highestCpointCof, to the circle. Then the, , C, , at, , tangent, LM be the, is obviously divided into three, integration, of, region, CMA and LPM., OLC,, namely, parts, of the circle, we have, From the equation, x - ax+y =, , x={a, , out of these, , two, , V(a*-x*), , =, , y, , x-0, , ar, , =, , by, , ww, , 0, , fig., , xa, M, , A, 9.9, , v(a-4)}, , values{a- v(a-4y)}, , is lesser than, , l a + V(a-4y)}, The lesser, the region CMA., Now any, , X, , on one, , value corresponds, , strip, , in the, , to the, , on, , OLC whereas the greater, , OLC is bounded, , region, , side and the circle, , region, , the other, , -, , V(a-4y}., , Since the, , we, , (a/2, a/2), , C is, , point, , parabola, , the, , side for which, , a, , value x= {a, , by, , a, , on, , the circle, , region CMA, the strip lies, the other. So the limits of x are from{a, on, , +, , on, , =, , to, , ax or, , shall take the lesser, , hence the limits, , are from 0 to a/2., in the, , y, , one, , one, , V(a-4y)}, , for y, , side and the line, to, , a, , and the limits, , of y are obviously 0 to, lies, Again in the region LPM the strip, x a on, , be, , 2 to a, , as, , the other. So the limits, the, , point, , P is, , for x, , (a, a). Hence, , on, , are, , 2, , the parabola, to a, , on one, , side and the, , whereas the limits for y ill, , the given integral
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Integrals Calculus, , 176, , (ita-a-45}, , vdy dt, , S 9-6. Change of order in polar coordinate., Process is similar to that adopted in Cartesian Coordinates. The only difference, , nce, , is that in order to evaluate anintegral of the type, , the summation is completed by dividing the region into triangular strips (unlike the, rectangular strips in Cartesian Coordinates). Then we make strip-wise summation to, cover the whole area for which the value 6 will be suitably chosen. When we change, the order of integration we get an integral of the type, , de, JJ, se.)a, Here the integration is first performed with respect to 6,r being kept constant, , which means that the summation is first made inside a circular strip. Now the value, of r is chosen so as to cover the whole region. Following example is given as an, , illustration of the method., , Example, , 9., , Change, , the, , order, of integration in, n/2 (2a cos, f,0) do dr., , (Gorakhpur 83), , Solution. The region of integration is bounded, by r = 0 (origin), r = 2a cos, , (a, 0),, , 6=0, , which, , is, , (a circle with centre, , the, , initial, , line, , and, , =-axis). Thus the region of inte- gration is the, semi-circle OQAO as shown in the figure 9.10., To change the order of integration consider, the elementary circular arc PQ ( varying and, remainingconstant). Thus r is bounded by the initial, line one side for which 6 = 0 and the circle on the, , P, fig. 9.10, , other side for which, , 6, , cos, , which is obtained from the equation of the circle. Since the diameter of the, 2a hence the limits for r will be 0 to 2a. Hence the given, , 2a cos(/a), , integral, , circles, , fo,0) drde, , S 9-7. Change of variables., (Transfomnation) In the evaluation of multiple integral sometimes it 1s, convenient to change the variables. The process of changing the variables in a, mulup, integral is called the transformation of multiple integrals. Suppose we want to, , the multiple integral, , transtor
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177, , f u l i p l eI n t e g r a l s, , where, of variables, 1, v,, another system =, x, ? (4, v) and y= y (u, v), say, would be completed in the, , a n o ., , hetransformationthe, hen )To determine, , following three, , yubstita, , eometrical, , considerations., , To determine the, =, know that dr dy J diu dv,, , in), , new, , steps., , This is done, , function f (r, y) in terms of u,, new value, eliminations. Let F (u, v) be its, cfitutions and, an, new limits. This is also, i ) The assignment of, v., , algebraic, , by algebraic, process and, , integration. From Differential Calculus., is defined, the Jacobian of transformation and, , element of, , where, , is, , Ou, , J=, , av, Thus we have, , SSsy) dr dy= JJ Fu, v)Jdu a, , Example 10., , Evaluate the doubleintegral, , (a-*2+y) dr dy, , changing to polar coordinates., from cartesian, Solution. Laws of transformation, , to, , polar, , is, , x =rcos 6, y =rsin 6., , + y =# (cos20+ sin20) =2, or, , J, , or, , 06, sin, , Cos 6, , - rsin 6, , r cos 6, , dx dy = r dO dr., , Now the region of integration is bounded by y = 0 (*-axis), y = v(a2- ? ) (a, , cle with centre at oigin), x andx =a,, circle of radius a. In order to cover this, =0, , d, , that the limits of r should be from 0 to, , a, , clearly the positive quadrant of, by means of polar coordinates we, , This is, area, , and those of 6 from 0, , to., , ve, , -, , +hdedy =J, , Prdd, , d0=102-, , Hence, , we
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Integrals Culculus, , 178, , Example 11. Evatuate the integra, , dr dy, , r+y), by changing to polar coordinates., Solution., , Region, , of, , integration is bounded, , the lines, , by, , (Purvanchul 90), , y = 0, y =x, * = 0, , and, , x =, , 1., , Y |AXISS, , Thus the region of integration is the region, , OAB (fig. 9.11) where B is the point (1, 1)., , 1(1.1), , Equation of the line AB is, =, , 1, , r cos 6= 1, , or, , r= sec6., Therefore in any, , OAB is divided r varies from 0 to, and then for all such strips 6 varies from 0, , which the, , sec, , triangular strip OPQ into, , A, , area, , y 0, , fig. 9.11, , to, , Thus the integral when transformed to polars, , 'sec 6(r cos 0) rde dr, cos3 0 de, , =, , 4, , =log(sec0 + tan0), , n/4, , -ose tan, =log(V+ 1)., S9-8. Area by double integral., Element of area in Cartesian Coordinates is dr dy and r d6 dr in polar, , coordinates. Hence the area is obtained by evaluating the double integral JJ drdin cartesian coordinates or, , rd dr in case of polar coordinates, integratior, , extending over the area under consideration. Thus the area bounded by the curve, y =f1 ) , y =f2 t),x =a and x = b is given by the double integral, dr dy, , S 9-9. Volume under, , a, , Surface., , Let A be the region in the y-plane and z = f(x, y) be a surface. We hav, , find the volume between this surface and the region A.
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79, fullipleI n t e g r a l s, , or to get this volume let us consider a small rectangle of arca dx dy in, In ore, , Now construct a vertical prism with dr dy as base bounded at the top, , h er e g i o nA., surface., , the, , given, , The volume of this prism is therefore z dx dy. Now the required, of similar prisms constructed over all the elementary rectangles, , olume 1s c o m p o s, , py, , the g i v e n r e g i o n, , Hence the required, b e, , JS z, , =, , dr dy where the limits of x and y, , are, , yet, , assigncd., , If the, , thcn, , volume, , region, , required, , the, , is bounded by the curves, of integration, y =fi ( ) , y =f2 (o), x = a and x =, , b,, , volume, , z dx dy, , Si ), Now, , Surface., , We, , and y from the equation of the, evaluate the integral to get the volume., as above, consider the area dy dz on yz plane and construct prism, value of, , substituting the, can, , Note. If we,, drawing lines parallel tor, y, , axis the, , f integration., , z, , in terms of, , x, , required volume, , ffxdy dz, , =, , on, dz dr on xz plane and erecting the prism, fSy dr dz under, the required volume, , Similarly by considering the, lines parallel to y-axis, his base by drawing, area, , integration., Hue limits of, , Example, , P= 4ax and its, , 12. Find, , by, , double, , =, , integration the, , area, , bounded, , latus rectum., , Solution. The, , curve, , under due limits, , by the parabola, , required, , is symmetrical about x-axis, hence, , area, , dx dy, , 2, , dx, , 2, , J V(Aax) dr, , =, , 4, , va/2|, , Jo, , =Va a2 =2, Example 13., , a, , Find, , cos ), , and outside the circle, Solution. r =a is a circle with, , (1 +, , r=a, , integration the, , double, , by, , (1, , +, , r, , =, , cos 6), , inside, , the, , cardiod, , a., , centre as, , the cardiod ABOCA thus the required, hows by the shaded region in fig 9.12., , area, , pole, , and radius, , area is as, , a, , where, , as, , B, , In order to evaluate this area let us integrate, a6 dr on the shaded area. This area can be divided, , to triangular strips., , Consider, , one, , such strip OPQ., , this strip 6 remains constant and r varies t, a, , (on, , the, , circle), , to r = 4 (1+ cos6), , on, , the, , ardiod. For all the strips into which the area can Dc, Vided 6 varies from - n/2 to it/2., , y =a, , ci, , ya(1+cos8), fig. 9.12
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181, , M u l t i p l eI n t e g r a l s, , 35ka t, , 24 3 2 1, , 16, n l e 15. Evalhuate by double integral the volume of the region enclosed by, the, plane, , x = 0, y =0, z = 0, and x +y +z = a, , Soution., , Here, , a vertical column is, , bounded, , by thc, , . The latter plane cufs the y-plane in the line a, , planes z = 0 and, , - x - y=0. So the area, , h i c h the volume stands is the region in xy-plany bounded by the lines, , above which, , y, Hence, , the, , =, , 0, y=a-x,X =0,x = a, , volume, , 2de, , =, , dy, , (a x-y) dr dy, , -f--, , dx, , aa-)-x(a -x) -, , ", , 2 a - 2 - a +) d, , EXERCISE 9.2, , Change the order, , of integration in the following integrals., , TT", fr,y) dr dy, 1, fa fa/x*, 2., , 3., , 4., , Sa/2 x, , -, , J2a, Give, , a, , Vdr dy, la, Vdr dy, , sketch of the, , integration, 5., , Show that, , region of integration and, , (Gorakhpur 86, 95, 2003), change the order of, , V de dy, , (2vax, , Change the order, , of, , dr dy=, , sa 2N, , integration in, , Change the order of integration in J, , /44, , dy dr, , T- 5, 2ax, , 2ax, , Vde dy, , Yd dy, , (Purvanchal 2003; I12; Gorakhpur 87)
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182, , 8., , Integrals Calculus, xe xy dr dy by changing the order of integration, , Evaluate, , (Gorakhpur 92, 94, 99, 2006, 07; Purvanchal 9, , 9., , Evaluate, , 94), (Purvanchal 89), , J, , y dr dy, Hint: Change the order ofintegration], , Evaluate the following intogral: by changing to polar coordinates., , x dr dy, , 10., , 2+y2, dx dy, , 11., 12., , + y coordinates, , Transform to, , polar, , and integrate, , fVa-1+x2+ y2, , dr dy, , the integral being extended over all positive values of x and y, , subject to xr +y2 s 1., 13., 14., 15., , Find by double integration the area of a circle of radius a., , Find the whole arca of the curve, (2a -y), Find by double integration the area of one loop of the curve, , a'x=, , 2= a, 16., , 17., , cos 29., , Find the mass of a circular plate of diameter a, whose density at any, , point is k times the distance from a fixed point on the circumference., Find the mass of area, =y. Ifp, , betweeny2 =x andx?, , Hint Mass, 18., 19., , (Purvanchal 89), , +yhdray, , Find the volume of the sphere + y2 +2 = a?., Find the volume in the positive octant of the, ellipsoid, =, , 20., , =k+y, , Find the volume of the cylinder, , 1., , +y - ax = 0 bounded by the, , planes z=0 and z=x., 21., , Transform the integral, , -* +, , d dy by changing to, , polar coordinated and hence evaluate it., 22., , 23., , Evaluate, Evaluate, , (Gorakhpur 2008), , -Fzd, , +y2, , by changing to polar form, , (Gorakhpur 2002), , x dr dy, , +, , Gorakhpur 200
