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

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
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