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

292 0 CHAPTER 15 MULTIPLE INTEGRALS
17. y
r..rx , . 11 1 2 y-..rx
1 D 1 + x2 dA = o 1 o 1 : x2 dy dx = o 1 + x2 [ h ) y:o dx
! .
r y 11
1
X 1
= l --dx = [! In(1 + x 2 )) = ! In 2
2 0 1 + x2 ·I o 4
0 X
19.
fj ,
f 2J,B-y~
0 ydA = J o 112 ydxdy
= fo2 y[x] ::~~v2 dy = J; y(8 ~ y2- y2) dy
= f 0
2
(8y - 2y
3) dy = [4y 2 - tY 4 ]~ = 8
21 .
y
/L (xz + y2)3/ 2 dA = 17r/313 (r2)3/2r drd0
= 11r/3 dO 13 r4dr = [0)~/3 [ir:;)~
1f 3 5
811f
=35=5
23. fffe xy dV = .1: j~x J;+v xy dz dy dx = .{ 0
3
.fox xy [ z J::~+v dy dx = J; ./~"' xy(x + y) dy dx
= ]~ 3 J;(x 2 y + xy 2 ) dydx = .r; [tx 2 y 2 + ~x1lJ;;:~ dx = J; (tx 4 + tx 4 ) dx
6 / '3 4 d [ 1 5] 3 81
=
40 5
6, 0 X X = GX 0 = 2 = . '
25. Iff v 2 z 2 d\l=f 1 J~ ]' 1 -v 2 - z
2
y2 z 2 dxdzdv=t r v'~-~~ 2 y 2 z 2 (1-v 2 - z 2 ) dz dv
E - 1 -~ 0 - 1 · -Vl -y2
= J;1r J; (r 2 cos 2 O)(r 2 sin 2 O)(l - 1·2)rdrd()
= J 0
2 Tr Jd ;t sin 2 2(J(r 5 - r 7 ) drdO
r2rr 1 (1 40) [ 1 6 1 8 ] r = 1 d() 1 [0 1 . 49] 2,- 2Tr "
= Jo S - COS 6r - ii 1 ' r=O = TIJ2 - 4 SID 0 = 192 = 96
= ll f " sin 3 () dO = lll [- cos 0 + i- cos 3 OJ " = !11
s Jo 5 3 . o 15
29 V _ r 2 r4 ( 2 2) d d _ 1 .2 [ 2 4 3J u=4 d . _ r2 ( 3 2 8 ) d . _
. - J o JJ X + 4y y X - 0 X y + 3 y u= l , X - Jo X + 4 :r. - 176
31:
V = f 2 j' 11 r< 2 - ul/ 2 dzdxdy = rz rv (1-ly) dxdy
Jo o .lo .Jo .lo 2
y
X