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

284 0 CHAPTER 15 MULTIPLE INTEGRALS
39. The region E of integration is the region above the cone z = .J x 2 + y 2 and below the sphere x 2 + y 2 + z 2 = 2 in the first
octant. Because E is in the first octant we have 0 ~ B ~ %. The cone has equation 1/> = "i (as in Example 4), so 0 ~ 1/> ~ .;f,
and 0 ~ p ~ ../2. So Ute integral becomes
f rr 14 0
.fa"'' 12 fov"i (p sin¢ cos B) (p sin c/> sin t1) p 2 sin¢ dp dB d¢ ·
= J~-r / 4 sin 3 1/>d¢ J 0
rr/ 2 sinB cost1dB .J 0
v"i p 4 dp = (Jorr/ 4. (1 - cos 2 1/>) sin 1/> dl/>) [~ sin 2 B]~ 12 [ip 5 ]:
- [1 cos 3 A .. - cos A..] rr/4.. 1. l ( '2) 5 - [~ - ~ - ( l - 1) ] . M - 4
fl- 5
- 3 '~' '~' o 2 5 V"' - 12 2 a 5 ~ 15
41 . The.region of integration is the solid sphere x 2 + y 2 + (z - 2? ~ 4 or equivalently
p 2 sin 2 c/> + (p cos lj> - 2) 2 = p 2 - 4pcos + 4 ~ 4 =>
p ~ 4cos¢, so 0 ~ 8 ~ 211', 0 ~ c/> ~ %• and
0:::; p ~ 4cos ¢.Also (x 2 + y 2 + z 2 ) 3 1 2 = (p 2 ) 3 1 2 = p 3 , so the integral becomes
J" 12 0
J:" J4. cos 0
z = rcot r/> 0
y
= 2:r [- (a2 - a2 sin2 c/Jo)3/2 - a3 sin3 ¢o cot¢o + a3]
;, ~7ra 3 [1 - ( cos 3 c/> 0 + sin 2 ¢ 0 cos 1/> 0 )] = lrra 3 (1 - cos ¢ 0 )
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