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

270 D CHAPTER 15 MULTIPLE INTEGRALS
15.
HereT = {(x, y,z) I 0 $ x $ 1, 0 $ y $ 1 - x, O $ z $ 1- x - y}, so
Jffr x2 dV ::::: fot J;-a: fot -a:-y x2 dz dydx = fot fot-x x2(1-x- y) dy dx
= J; fot -a:(x2- ~3 - x2y) dy dx = fot [x2y.- xay - t X2y2J::~- a: dx
= J; [x 2 (1 - x)- x 3 (1 - x) - t x 2 (1- x?] dx
= j' 1 (lx 4 - x 3 + lx 2 ) dx = [...!. x 5 - l x 4 + l x 3 ) 1
0 2 2 10 4 . 6 0
1 1 1 1
- w - 4+ s = oo
17.
The projection of E on the yz-plane is the disk y 2 + z 2 $ 1. Using polar
coordinates y = r cos 8 and z = r sin 8, we get
f.ffe x dV = ffv [J4~2 +4:2 x d.x] dA = ~ ffv [4 2 -
(4y 2 + 4z 2 ) 2 ) dA
= 8 J;"" J:(1- r 4 ) 1·drd8 = 8 J~ 2 " dB J 0 \ r - r 5 ) dr
X
_ 8(211") [lr2 _ lr6] I _
16rr
- 2 6 0- 3
19. The plane 2x + y + z = 4 intersects' the xy-plane when
2x + y + 0 = 4 => y = 4 - 2x, so
4
E = {(x, y, z) I 0 $ x $ 2, 0 $ y $ 4- 2x, 0 $ z $ 4 - 2x - y} and
V = fo2 J~l -2 a: fo4-2x-v dz dy dx = fo2 fo4- 2:r. ( 4 - 2x - y) dy dx
r2 [4 2 1 2] y =4.- 2x dx
= Jo y - xy - 2 Y v= O
= J; [4(4 - 2x)- 2x(4 - 2x) - t(4 - 2x?] dx
= J; (2x 2 - 8x + 8) dx = [ ~x 3 - 4x 2 + 8x]~ ='¥
X
y
21. The plane y + z = 1 intersects the xy-plane in the line y = 1, so
E = { (x, y, z) I. - 1 $ x $ 1, x 2 $ y $ 1, 0 $ z $ 1 - y} and
V = JJJ ~ dV = .C 1
J:2 f 0
1 - 11 dz dy dx = f~ 1 f:.2 (1 - y) dy dx
1 [ 1 2] v=1
!
d }'1 (1 '.!
=
1 ") d
-1 y- 2Y y=x2 X = - 1 2 - X + 2X X
_ [ 1 1 3 + 1 5 ] 1 1 1 1 l 1 1 _ R
- 2X - .3 X lOX· - 1 = 2 - 3 + 10 + 2 - J + lO - 15
(- 1, 1, 0)
® 2012 Cenpge Learning. All Rights Rcsc:rvcd. M:ay noc be scnnncd, copied. or duplicalcd, or posted too publicly occcssible website, in whole or in p3tt.