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

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SECTION 15.10 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS 0 287<br />

13. R is a portion of an annular region (see the figure) that is easily describ<strong>ed</strong> in polar coordinates as<br />

R = { ( r , 8) 11 ::; r ::; ../2, 0 ::; (} ::; 7T / 2}. If we convert<strong>ed</strong> a double integral over R to polar coordinates the resulting region<br />

of integration is a rectangle (in the rO-plane), so we can create a transformation T here by letting u play the role of rand v the<br />

role of 0. Thus T is defin<strong>ed</strong> by x = u cos v , y = u sin v and T maps the rectangle S = { ( u, v) I 1 ::; u ::; ../2, 0 ::; v. ::; 7T / 2}<br />

in the uv-plane to R in the xy-plane.<br />

v<br />

y<br />

E.<br />

2<br />

s<br />

-T<br />

0<br />

II<br />

X<br />

2 1<br />

15 B(x, y) = 1 1 = 3 and x - 3y = (2u + v) - 3(u + 2v) = -u- 5v. To find the regionS in the uv-plane that<br />

· 8(u, v) 1 2 ·<br />

corresponds toR we first find the corresponding boundary under the given transformation. The line through {0, 0) and (2, 1) is<br />

y = ~x which is the image of u + 2v = ~(2u + v) => v = 0; the line through (2, 1) and (1, 2) is x + y = 3 wh ich is the<br />

image of (2u + v) + (u + 2v) = 3 => u + v = 1; the line through (0, 0) and (1, 2) is y = 2x which is the image of<br />

u + 2v = 2(2u + v) =><br />

u = 0. Thus S is the triangle 0 ::; v ::; 1 - u, 0 ~ u ::; 1 in the uv-plane and<br />

jj~ (x - 3y) dA = j 0<br />

1<br />

j 0<br />

1 -u ( -u - 5v) 131 dv du = -3 f 0<br />

1<br />

[uv + ~v 2 J::~ - u du .<br />

= - 3 j; (u - u 2 + ~{ 1 - u?) du = - 3[~u 2 - iu 3 - ~( 1- u?]~ = -3(~ - ~ + ~) = - 3<br />

17 B(x, y) = 12 0<br />

I = 6, x 2 = 4u 2 and the planar ellipse 9x 2 + 4y 2 ::; 36 is the image ~fthe disk u 2 + v 2 < 1. Thus<br />

. 8( u, v) 0 3 . -<br />

ffn x 2 dA =<br />

JJ (4u 2 )(6) dudv = f 0<br />

2<br />

"' f 0<br />

1<br />

(24r<br />

2<br />

cos 2 8)rdrd8 = 24J;.,.. cos 2 8d8 J 0<br />

1<br />

r 3 dr<br />

u2 +v2~ 1<br />

= 24[~ x +~sin 2x]~"' ar 4 ] ~ = 24(7r)(i) = 67!'<br />

19. 8(x, y) = , 1 2<br />

/v - ufv ' = .!., xy = u, y =xis the image of the parabola v 2 = u, y == 3x is the image of the parabola<br />

8(u, v) o 1 v<br />

j L x y dA = 1<br />

31: u ( ; ) dv du = 13<br />

v 2 = 3u, and the hyperbolas xy = 1, xy =a are the images of the lines u = 1 and u = 3 respectively. Thus<br />

21 . (a) :ix,<br />

a 0 0<br />

y, z~ = 0 b 0 = abc and since u = ~. v<br />

u,v,w a c<br />

0 0 c<br />

u (In v3u - ln vfu) du = J: u In J3 du = 4ln J3 = 2 ln 3.<br />

= J!.b' w = ~the solid enclos<strong>ed</strong> by the ellipsoid is the image of the<br />

JJJE dV = JJJ abcdudvdw = (abc)(<strong>vol</strong>umeof tlie ball) = ~1rabc<br />

u2+v2+w2 :S 1<br />

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