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

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288 0 CHAPTER 15 MULTIPLE INTEGRALS<br />

2 2 2<br />

(b) If we approximate the surface. of the earth by the ellipsoid 6<br />

;., 82<br />

+ 6<br />

i 782<br />

+ 6<br />

; 562<br />

_ = 1, then we can estimate<br />

the <strong>vol</strong>ume of the earth by finding the <strong>vol</strong>ume of the solid E enclos<strong>ed</strong> by the ellipsoid. From part (a), this is<br />

fffE dV = ~11"(6378)(6378)(6356) ~ 1.083 X 10 12 km 3 •<br />

(c) The moment ofintertia about the z-axis is I% = JJJE (x 2 + y 2 ) p(x, y, z) dV, where E is the solid enclos<strong>ed</strong> by<br />

xz yz zz 18(x y z) I<br />

2 + b 2<br />

+ 2 = 1. As in part (a), we use the transformation x = au, y = bv, z = cw, so 8<br />

( ' ' ) = abc and<br />

a . c u,v,w<br />

I , = JJJE (x 2 + y 2 ) kdV = f.(.[ k(a 2 u 2 + b 2·u 2 )(abc) dudv dw<br />

u2+ v2+w2 ~ 1<br />

= abck f 0<br />

" f 0<br />

2 " J 0<br />

1 ( a<br />

2 p<br />

2<br />

sin 2 ¢ cos 2 () + b 2 p 2 sin 2 ¢ sin 2 8) p 2 sin ¢ dp d8 drjJ<br />

= abck [a 2 f 0 ,.. f~,.. f 0<br />

1 (p 2 sin<br />

2<br />

rP cos 2 B) p 2 sin 1/> dp d() dlj> + b 2 f 0 " f~,.. f 0<br />

1 (p<br />

2<br />

sin 2 rjJ sin 2 B) p 2 sin rP dp dB dr!>]<br />

=. a 3 bck J 0<br />

" sin 3 rjJ drjJ f 0<br />

2<br />

" cos 2 B dB J; p 4 dp + ab 3 ck fo.,.. sin 3 rjJdrjJ f~" sin 2 B dB J; p 4 dp<br />

= a 3 bck [ ~ cos 3 ¢-cos I/>] ~ [ ~B + i sin2 B]~" [iP 5 ] ~ + ab 3 ck [t cos 3 rjJ- cosr/J]~ [~B- i sin 28] ~" [tP 5 ]~<br />

= a 3 bck (t) (1r) ( t) + ab 3 ck (1) (1r) (i) = rt7r (a 2 + b 2 ) abck<br />

23. Letting u = x - 2y and v = 3x - y, we have x = f.(2v- u) and y = f(v- 3u). Then 8 8 ((x, y)) = ~ - 1 / 5 215 1 = .!.<br />

0 0<br />

u,v -3/5 1/5 5<br />

and R is the image of the rectangle enclos<strong>ed</strong> by the lines u = 0, u = 4, v = 1, and v = 8. Thus<br />

8(xy) 1.:...1/2 1/ 21 1<br />

25. Letting u = y- x, v = y + x, we nave y = !(u + v), x = Hv - u). Then (u: v) = i = 8 1 2 112<br />

- and R is the<br />

2<br />

image of the trapezoidal region with vertices ( - 1, 1), ( -2, 2), {2, 2), and {1, 1). Thus<br />

y -x<br />

!h 121" u I 11 11 2 [ u]u=v 112<br />

R Y +X '1 -v V 1 V u= -v 1<br />

cos--dA-= cos - - - 2<br />

du dv = - 2<br />

v sin- dv = - 2<br />

2v sin{1) dv = ~sin 1<br />

.<br />

27. Letu=x+yandv=-x+y.Thenu+v = 2y => v=Hu+v) andu - v=2x => X=~(u-v).<br />

8(x y) 11/2 -1/ 2! 1<br />

- 8<br />

( ') = =- 2<br />

.Now[u[= [x+y[< [x[+[y[ - l

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