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

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SECTION 15.7 TRIPLE INTEGRALS 0 269<br />

Converting to polar coordinates x = r cos fJ; z = r sin fJ we have<br />

15.7 Triple Integrals<br />

If<br />

, r 2 ri r3 J2 2 d d dx ri r3 [ 1 2 2] v=2 d d · ri r3 3 2<br />

1. JB<br />

d d<br />

xy~ dV = Jo Jo _ 1 xyz y z = Jo Jo 2xy z v=- 1 z x = Jo Jo 2 x z z x<br />

= r1 (lxz3]==3 dx = fl TI xdx = 27 x2] 1 = TI<br />

Jo 2 ==O Jo 2 4 o 4<br />

2 " 2 ru- · . r2 r· 2 [ 2 ] x=y- z r2 r= 2 [( 2 ]<br />

3. J 0 J; Jo "(2 x - y)cJ:J: dydz=Jo J; x - xy x=O dydz =Jo Jo y- z) - (y-z)y dydz<br />

2<br />

2<br />

= J; J 0 " (z - yz) dydz = J~ 2 [yz 2 - h 2 z ] ~ =~ 2 2<br />

dz = J 0<br />

(z 4 - %z 5 ) dz<br />

5 ! 2 r2= r tn x - y d dxd = !2 r2= [- -v] v=ln:r; d d = !2 r2z (- - ln x + 0) dxd<br />

. 1 Jo Jo x e y z 1 Jo x e v=O x z 1 Jo xe xe z<br />

= Jt J:= (-i + x) dxdz = g [-x + %x 2 J::~= dz<br />

· = f 1<br />

2 (- 2z + 2z<br />

2 ) dz = [- z 2 + ~ z 3 ] ~ = - 4 + 1 36<br />

+ 1 - ~ = ~<br />

7. J 0<br />

.,. 12 J~ J 0 "' cos(x+y+z)dzdxdy = j~rr/ 2 g [sin(x+y +z)J::~ dxdy<br />

= f 0<br />

.,. 12 J~ (sin(2x + y) - sin(x + y)J dx dy<br />

= j~,./ 2 [- % cos(2x + y) + cos(x + y)J::~ dy<br />

= f 0<br />

,. 12 [- ~ cos 3y + cos 2y + ~ cos y - cos y] dy<br />

1. . 3 1 . 2 1 . ] .,./2 1 1 1<br />

= [-6 SID y + 2 S Ln y - 2 sm y 0 = G. - 2 = -S<br />

9. JJJE ydV = J; fa"' J;::: ydzdydx = J; J; [yz]~== ~~ dydx = f 0<br />

3<br />

f0"' 2y 2 dydx<br />

- r3 (1y3] y = x dx - r3 1x3 dx - lx4] 3 - 81 = ll<br />

- Jo 3 11=o - Jo 3 - G o - 6 2<br />

[z · .! tan -l ~] x = z dz dy<br />

} } E X + Z 1 71 0 X + Z 1 71 Z Z x =O<br />

11. / " ( ( - 2<br />

- z-<br />

2 dV = 14141= ~ dx dz dy = 1414<br />

4 4<br />

= f 1 f 11<br />

(tan- 1 (1)-tan- 1 4 4<br />

(o)] dzdy = f 1 I:(~- o) dz' dy = ~ f 1<br />

[zJ::: dy<br />

= ~ f 1<br />

4<br />

(4-y)dv= ~ [4v - h<br />

2<br />

]~ · = ~ (16-8 - 4+~) = 9 ;<br />

13. Here E = {(x,y,z) I 0 :::; x:::; 1, 0 :::; y:::; -fo, O :::; z :::; 1 +x +y}, so<br />

JJJE 6xydV = f 0<br />

1<br />

fov'X J;+.,+u 6xydzdydx = J~ fov'X [6xy zJ : : ~+"'+ 11 dydx<br />

= f~ fov'X 6xy(1 + x + y) dy dx = / 0<br />

1<br />

(3xy 2 + 3x 2 y 2 + 2x~ 3 ) ~=~ dx<br />

= t (3x2 + 3x3 + 2x5f2) ·dx = [x3 + ~x·l + :!x7f 2] 1 = 65<br />

0 4 7 0 28<br />

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