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

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276 0 CHAPTER 15 MULTIPLE INTEGRALS<br />

1 L2 L2 L2 La<br />

£ 3·222 = 8<br />

55. (a) The triple integral will attain its maximum when the integrand 1 - x 2 - 2y 2 - 3z 2 is positive in the region E and negative<br />

everywhere else. For if E contains some region F where the integrand is negative, the integral could be increas<strong>ed</strong> by<br />

excluding F from E, and if E fails to contain some part G of the region where the integrand is positive, the integral could<br />

be increas<strong>ed</strong> by including Gin E. So we require that x 2 + 2y 2 + 3z 2 $ 1. This describes the region bound<strong>ed</strong> by the<br />

ellipsoid x 2 + 2y 2 + 3z 2 = 1.<br />

(b) The maximwn value of JjJE (1 - x 2 - 2y 2 -<br />

3z 2 ) dV occurs when E is the solid region bound<strong>ed</strong> by the ellipsoid<br />

x 2 + 2y 2 + 3z 2 = 1. The projection of E on the xy-plane is the plaf1ar region bound<strong>ed</strong> by the ellipse x 2 + 2y 2 = i, so<br />

and<br />

using a CAS.<br />

15.8 Triple Integrals in Cylindrical Coordinates<br />

.1. (a) From Equations I, x = r cos 9 = 4 cos i = 4 · ~ = 2,<br />

y '= rsinfJ = 4sin ~ = 4 · ~ = 2.J3, z = ~ 2,<br />

3<br />

so the point is<br />

X<br />

I<br />

- 2 •<br />

I<br />

l (4, 3. -2)<br />

Y · (2, 2.J3, -2) in rectangular coordinates ..<br />

(b)<br />

(2.-¥. 1)<br />

x = 2cos(-~) = 0, y = 2sin( - ~l) = - 2,<br />

and z == 1, so the point is (0, - 2, 1) in rectangular coordinates.<br />

X<br />

© 2012 Ceng3gc Learning. All Rights Rcscn·cd~ M:l)' not be SC>UuJcd, copi<strong>ed</strong>. or duplicat<strong>ed</strong>, or po5tcd loa publicly accessible wcb!litc, in whole or in part.

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