Page 96 - Vector Analysis

92 CHAPTER 3. Multiple Integrals

Let (u, v) = (x, x ´ 2y). Then (x, y) = g(u, v) = ( u ´ v ) thus
u, ;
2

ˇ 1 0 ˇ 1
ˇ ˇ

Jg (u, v) = ˇ1 ´ 1 ˇ = ´ 2 .
ˇ 2 ˇ
ˇ
ˇ2

Define E as the triangle with vertices (0, 0), (4, 0), (4, 4). Then A = g(E).

vy

g

E A

u x

Figure 3.2: The image of E under g

Therefore,

żż 1ż ( )

f (x, y) d(x, y) = f (x, y) d(x, y) = f g(u, v) d(u, v)
2E
A g(E)

1ż4żu ? 1 ż 4 [ 2 3 2 5 ]ˇv=u
= (u ´ v) v dvdu = uv 2 ´ v 2 ˇ du
40 0 40 3 5 ˇv=0

1 ż 4 (2 2) 5 1 2 7 ˇu=4 256
= ´ u 2 du = ˆ u 2 ˇ = .
40 3 5 15 7 ˇu=0 105

Example 3.35. Let A be the region in the first quadrant of the plane bounded by the
curves xy ´ x + y = 0 and x ´ y = 1, and f : A Ñ R be given by

f (x, y) = x2y2(x + y)e´(x´y)2 .

ż

We would like to evaluate the integral f (x, y) d(x, y).

A

Let (u, v) = (xy ´ x + y, x ´ y). Unlike the previous two examples we do not want
to solve for (x, y) in terms of (u, v) but still assume that (x, y) = g(u, v). By the inverse

function theorem,

Jg (u, ˇ = ( B (u, v) )´1 = ˇˇy ´ 1 x + 1ˇˇ´1 = 1 = ´x 1 .
v)ˇ ˇ 1 1´ +
B (x, y) ´1 ˇ ´y + x´ 1 y
ˇ(u,v)=g´1(x,y) ˇ ˇ