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

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270 D CHAPTER 15 MULTIPLE INTEGRALS<br />

15.<br />

HereT = {(x, y,z) I 0 $ x $ 1, 0 $ y $ 1 - x, O $ z $ 1- x - y}, so<br />

Jffr x2 dV ::::: fot J;-a: fot -a:-y x2 dz dydx = fot fot-x x2(1-x- y) dy dx<br />

= J; fot -a:(x2- ~3 - x2y) dy dx = fot [x2y.- xay - t X2y2J::~- a: dx<br />

= J; [x 2 (1 - x)- x 3 (1 - x) - t x 2 (1- x?] dx<br />

= j' 1 (lx 4 - x 3 + lx 2 ) dx = [...!. x 5 - l x 4 + l x 3 ) 1<br />

0 2 2 10 4 . 6 0<br />

1 1 1 1<br />

- w - 4+ s = oo<br />

17.<br />

The projection of E on the yz-plane is the disk y 2 + z 2 $ 1. Using polar<br />

coordinates y = r cos 8 and z = r sin 8, we get<br />

f.ffe x dV = ffv [J4~2 +4:2 x d.x] dA = ~ ffv [4 2 -<br />

(4y 2 + 4z 2 ) 2 ) dA<br />

= 8 J;"" J:(1- r 4 ) 1·drd8 = 8 J~ 2 " dB J 0 \ r - r 5 ) dr<br />

X<br />

_ 8(211") [lr2 _ lr6] I _<br />

16rr<br />

- 2 6 0- 3<br />

19. The plane 2x + y + z = 4 intersects' the xy-plane when<br />

2x + y + 0 = 4 => y = 4 - 2x, so<br />

4<br />

E = {(x, y, z) I 0 $ x $ 2, 0 $ y $ 4- 2x, 0 $ z $ 4 - 2x - y} and<br />

V = fo2 J~l -2 a: fo4-2x-v dz dy dx = fo2 fo4- 2:r. ( 4 - 2x - y) dy dx<br />

r2 [4 2 1 2] y =4.- 2x dx<br />

= Jo y - xy - 2 Y v= O<br />

= J; [4(4 - 2x)- 2x(4 - 2x) - t(4 - 2x?] dx<br />

= J; (2x 2 - 8x + 8) dx = [ ~x 3 - 4x 2 + 8x]~ ='¥<br />

X<br />

y<br />

21. The plane y + z = 1 intersects the xy-plane in the line y = 1, so<br />

E = { (x, y, z) I. - 1 $ x $ 1, x 2 $ y $ 1, 0 $ z $ 1 - y} and<br />

V = JJJ ~ dV = .C 1<br />

J:2 f 0<br />

1 - 11 dz dy dx = f~ 1 f:.2 (1 - y) dy dx<br />

1 [ 1 2] v=1<br />

!<br />

d }'1 (1 '.!<br />

=<br />

1 ") d<br />

-1 y- 2Y y=x2 X = - 1 2 - X + 2X X<br />

_ [ 1 1 3 + 1 5 ] 1 1 1 1 l 1 1 _ R<br />

- 2X - .3 X lOX· - 1 = 2 - 3 + 10 + 2 - J + lO - 15<br />

(- 1, 1, 0)<br />

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