Solução_Calculo_Stewart_6e

F.
15.
SECTION 16.9TX.10
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ET SECTION 15.9 ¤ 257
∂(x, y)
∂(u, v) = 1/v −u/v 2
0 1 = 1 v , xy = u, y = x is the image of the parabola v2 = u, y =3x is the image of the parabola
v 2 =3u, and the hyperbolas xy =1, xy =3are the images of the lines u =1and u =3respectively. Thus
R
xy dA =
3
√ 3u
1
√ u
u 1
v
dv du =
3
1
u ln √ 3u − ln √
u du = 3
u ln √ 3 du =4ln √ 3=2ln3.
1
17. (a)
a 0 0
∂(x, y, z)
∂(u, v, w) = 0 b 0
= abc and since u = x a , v = y b , w = z the solid enclosed by the ellipsoid is the image of the
c
0 0 c
ball u 2 + v 2 + w 2 ≤ 1. So
E dV =
u 2 +v 2 +w 2 ≤ 1
(b) If we approximate the surface of the earth by the ellipsoid
abc du dv dw =(abc)(volume of the ball) = 4 3 πabc
x 2
6378 + y2
2 6378 + z2
=1, then we can estimate
2 63562 the volume of the earth by finding the volume of the solid E enclosed by the ellipsoid. From part (a), this is
E dV = 4 3 π(6378)(6378)(6356) ≈ 1.083 × 1012 km 3 .
19. Letting u = x − 2y and v =3x − y,wehavex = 1 (2v − u) and y = 1 y)
(v − 3u). Then∂(x,
5 5
∂(u, v) = −1/5 2/5
−3/5 1/5 = 1 5
and R is the image of the rectangle enclosed by the lines u =0, u =4, v =1,andv =8. Thus
R
x − 2y
4
3x − y dA =
0
8
1
u
v
1 5 dv du = 1 5
4
0
udu
8
1
1
v dv = 1 1 u2 4
5 2 0
8
ln |v| = 8 ln 8.
1 5
21. Letting u = y − x, v = y + x,wehavey = 1 (u + v), x = 1 y)
(v − u). Then∂(x,
2 2
∂(u, v) = −1/2 1/2
1/2 1/2 = −1 and R is the
2
image of the trapezoidal region with vertices (−1, 1), (−2, 2), (2, 2),and(1, 1). Thus
cos y − x 2
v
y + x dA = cos u v − 1 2 du dv = 1 2
v sin u u = v
dv = 1 2 v u = −v 2
R
1
−v
1
2
1
2v sin(1) dv = 3 2 sin 1
23. Let u = x + y and v = −x + y. Thenu + v =2y ⇒ y = 1 (u + v) and u − v =2x ⇒ x = 1 (u − v).
2 2
∂(x, y)
∂(u, v) = 1/2 −1/2
1/2 1/2 = 1 .Now|u| = |x + y| ≤ |x| + |y| ≤ 1 ⇒ −1 ≤ u ≤ 1,and
2
|v| = |−x + y| ≤ |x| + |y| ≤ 1 ⇒ −1 ≤ v ≤ 1. R is the image of the square
region with vertices (1, 1), (1, −1), (−1, −1),and(−1, 1).
So
R ex+y dA = 1 1
1
2 −1 −1 eu du dv = 1 2 e
u 1 1
−1 v = e − −1 e−1 .