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

SECTION 15.10 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS 0 287
13. R is a portion of an annular region (see the figure) that is easily described in polar coordinates as
R = { ( r , 8) 11 ::; r ::; ../2, 0 ::; (} ::; 7T / 2}. If we converted a double integral over R to polar coordinates the resulting region
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
role of 0. Thus T is defined 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}
in the uv-plane to R in the xy-plane.
v
y
E.
2
s
-T
0
II
X
2 1
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
· 8(u, v) 1 2 ·
corresponds toR we first find the corresponding boundary under the given transformation. The line through {0, 0) and (2, 1) is
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
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
u + 2v = 2(2u + v) =>
u = 0. Thus S is the triangle 0 ::; v ::; 1 - u, 0 ~ u ::; 1 in the uv-plane and
jj~ (x - 3y) dA = j 0
1
j 0
1 -u ( -u - 5v) 131 dv du = -3 f 0
1
[uv + ~v 2 J::~ - u du .
= - 3 j; (u - u 2 + ~{ 1 - u?) du = - 3[~u 2 - iu 3 - ~( 1- u?]~ = -3(~ - ~ + ~) = - 3
17 B(x, y) = 12 0
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
. 8( u, v) 0 3 . -
ffn x 2 dA =
JJ (4u 2 )(6) dudv = f 0
2
"' f 0
1
(24r
2
cos 2 8)rdrd8 = 24J;.,.. cos 2 8d8 J 0
1
r 3 dr
u2 +v2~ 1
= 24[~ x +~sin 2x]~"' ar 4 ] ~ = 24(7r)(i) = 67!'
19. 8(x, y) = , 1 2
/v - ufv ' = .!., xy = u, y =xis the image of the parabola v 2 = u, y == 3x is the image of the parabola
8(u, v) o 1 v
j L x y dA = 1
31: u ( ; ) dv du = 13
v 2 = 3u, and the hyperbolas xy = 1, xy =a are the images of the lines u = 1 and u = 3 respectively. Thus
21 . (a) :ix,
a 0 0
y, z~ = 0 b 0 = abc and since u = ~. v
u,v,w a c
0 0 c
u (In v3u - ln vfu) du = J: u In J3 du = 4ln J3 = 2 ln 3.
= J!.b' w = ~the solid enclosed by the ellipsoid is the image of the
JJJE dV = JJJ abcdudvdw = (abc)(volumeof tlie ball) = ~1rabc
u2+v2+w2 :S 1
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