Page 97 - Vector Analysis

§3.5 The Change of Variables Formula 93

Moreover, the curve xy ´ x + y = 0 corresponds to u = 0, while the lines x ´ y = 1 and
y = 0 correspond to v = 1 and u + v = 0, respectively; thus if E is the region enclosed by
u = 0, v = 1 and u + v = 0, then A = g(E).

vy

Eg A
u x

Figure 3.3: The image of E under g

Therefore,

żż ż

f (x, y)d(x, y) = f (x, y) d(x, y) = (f ˝ g)(u, v)|Jg(u, v)| d(u, v)

A g(E) E

ż 1ż 0 1 ż 1

= (u + v)2e´v2 dudv = 30 v3e´v2 dv

0 ´v

= 1 ż1 we´w dw = 1 ˇw=1 = 1(2 ´ )
6 ´ 6 (w + 1)e´wˇ ´6 e 1.
0
ˇw=0

Example 3.36 (Polar coordinates). In R2, when the domain over which the integral is taken

is a disk D, a particular type of change of variables is sometimes very useful for the purpose

of evaluating the integral. Let (x, y) = (x0 + r cos θ, y0 + r sin θ) ” ψ(r, θ), where (x0, y0) is
the center of D under consideration. If the radius of D is R, then D, up to removing a line

segment with length R, is the image of (0, R) ˆ (0, 2π) under ψ. Note that the Jacobian of

ψ is

ˇ B ψ1 B ψ1 ˇ ˇˇcos θ ´r sin θˇˇ
ˇ Bθ ˇ ˇ ˇ = r.
ˇ
Jψ (r, θ) = ˇ Br ˇ =
ˇ

ˇ B ψ2 B ψ2 ˇ ˇˇsin θ r cos θ ˇ
ˇ ˇ ˇ

ˇ Br Bθ ˇ

Therefore, if f : D Ñ R is Riemann integrable, then

żż ż

f (x, y) d(x, y) = f (x, y) d(x, y) = (f ˝ ψ)(r, θ)ˇˇJψ(r, θ)ˇˇ d(r, θ)

D ψ((0,R)ˆ(0,2π)) (0,R)ˆ(0,2π)

ż

= f (x0 + r cos θ, y0 + r sin θ) r d(r, θ) .

(0,R)ˆ(0,2π)