Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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282 0 CHAPTER 15 MULTIPLE INTEGRALS 21. In spherical coordinates, B is represented by {(p, 0, 4>) I 0 ~ p ~ 5, 0 ~ 0 ~ 27r, 0 ~ cf> ~ 1r}. Thus JJJ 8 (x 2 + y 2 + z 2 ) 2 dV = J; f 0 2 "' J:(p 2 ) 2 p 2 sin cf>dpdO dl/> = Io"' sin 1/>dl/> J~"' dB I~ p 6 dp = [- coscf>]~ [0]~" [ tp 7 ]~ = (2)(27r)(1s.~211) = 312 tJ07r ~ 140,249.7 23. In spherical coordinates, E is represented by {(p, 0, ¢) 12 ~ p ~ 3, 0 ~ 0 ~ 27r, 0 ~ cf> ~ 1f} and IIIs(:c 2 + y 2 ) dV = ]~"' g •r I 2 3 (p 2 sin 2 4>) p 2 sin cf>dpdO dcf> = Io" sin 3 cf>dcf> I~"' dO I: p 4 dp = I 0 "'(1- cos 2 1/>) sinl/>dl/> [ 0]~,. [ip 5 )~ = [-coscf>+ ~ cos 3 4>]~ (27r) : i(243 - 32) = (1 - ~ + 1 - U (27r) Cil) = 16~:" 25. In spherical coordinates, E is represented by {(p, 0, 4>) I 0 ~ p ~ 1, 0 ~ 8 ~ "i· 0 ~ 4> ~ ~}.Thus JJJE xe"' 2 + 112 +z 2 dV = J 0 "/ 2 J 0 Tr 12 I:(psin cf>cosO)eP 2 p 2 sin = Io"' 12 sin 2 1/>d¢ J; 12 cos(} d(} j 0 1 p 3 eP 2 dp .= fo" 12 H 1 - cos 2¢>) d¢ IoTr 12 cos(} d(} ( ~ p 2 eP2 J: -I: peP 2 dp) [integrate by parts with u = p 2 , dv = peP 2 dp] = a¢- i sin 2¢]~ /Z [sin(})~/ 2 [ ~p 2 eP 2 - ~eP 2 J: = (i- 0) (1- 0) (0 + ~) = i 27. The solid region is given byE= { (p, 0, cf>) I 0 ~ p ~ a, 0 ~ 0 ~ 21r, ~ ~ cf> ~ ~}and its volume is V = ffis dV = I:/ 6 3 J~ "' I; p 2 sin cf>dpd(}drjJ = I:/: sincf>dl/> I~ 11' dO I; p 2 dp . = [-coscf>J;~: [0) ~11' [~p 3 )~ = ( -~ + 4) (27r) (ta 3 ) = '\- 1 7ra 3 29. (a) Since p = 4 cos 4> implies p 2 = 4p cos 1/>, the equation is that of a sphere of radius 2 with center at (0, 0, 2). Thus (b) By the symmetry of the problem Mv= = M x= = 0. Then Hence (x,y,z) = (0,0, 2.1). M, 11 = I:" I; 13 I 0 4 co• 4> p 3 cos cf> sin cf> dp d¢ d(} = J~ " I; 13 cos cf> sin r/> ( 64 cos 4 rjJ) dcf> dO - f 2 "' 64 [-1 () -•] =71' 13 dO - f 2 71' .ll dO - 21 - Jo 6 cos '~' ¢=0 - Jo 2 - 7r @) 2012 Coni!"&• !.coming. All Rights Re>crvcd. Moy not be scanned, copied, or duplicated. or posted to a publicly accessible website, in "11ole or in par1.
SECTION 15.9 TRIPLE INTEGRALS IN SPHERICAL COORDINATES 0 283 31. (a) By the symmetry of the region, My: = 0 and Mx: = 0. Assuming constant density K, m = fffE K dV = K fffE dV = iJ< (from Example 4). Then · M .,y = Jjj~ z K dV = K J; ,. }~,. 14 j~o• ¢>(pcos ¢) p 2 sin r/J dp drpd8 = K J;" J0,. 14 sin ¢cos¢ [tP 4 ] ==~os¢ d¢ d8 =if< J:w f 0 "' 14 sin¢cos¢ (cos 4 ifJ) .d(j)d(} = iK J:" d(} J; 14 cos 5 {j}sin (j)d¢ = iK (8]~ " [ -i cos 6 t/>] ~ 14 = iK(27r)( -i) [ c~r - 1] = -fiK ( -~) = ;~ K .. (- __) (My: !vfu M.,u) ( 77rK/96) ( 7) Thusthecentro1d1s x,y,z = m'm'm = 0, 0, 1rf(jB = 0,0, 12. (b)' As in Exercise 23, x 2 + y 2 = p 2 sin 2 ¢ and I: ~ JJJE (x2 + y2) J( dV = K J;" fo"/4 focos¢ (p2 sin2 ifJ) p2 sin¢ dpd¢d8 = K J:.,.. fo"'/4 sin3 t/> [kPsJ::~os ¢ d¢d8 = iK J:" J; 14 sin 3 ¢cos"' ¢d¢d8 =if< .r:w d(} j~w/ 4 cos 5 ¢ (1 - cos 2 ¢) sin¢d¢ = tK [8]~"' [- t cos 6 ¢ + ~- cos 8 ¢]~ 14 = 1J
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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CHAPTER 16 REVIEW 0 339 TRUE-FALSE
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0 APPENDIX Appendix H Complex Numbe
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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