Calculus III, Spring 06
Grinshpan
CHANGE OF VARIABLES
EXAMPLE 1. Evaluate the integral
ZZ
x−y
cos
dxdy,
x+y
R
where R is the triangular region with vertices (0, 0), (1, 0), (0, 1).
SOLUTION. Here the region of integration is simple, but the function
f (x, y) = cos x−y
x+y is not. It seems reasonable to set
u = x − y,
v = x + y.
Then, as the point (x, y) varies in R, the point (u, v) varies in the
triangular region Q bounded by the lines u = v, v = 1, u = −v.
By the change of variables formula
ZZ
ZZ
x−y
dxdy =
cos(u/v)|J|dudv.
cos
x+y
Q
R
Since x = (u + v)/2 and y = (v − u)/2, we can compute the Jacobian factor:
J = xu yv − xv yu = 1/4 + 1/4 = 1/2.
So
ZZ
1
cos(u/v)|J|dudv =
2
Q
Z
1
=
2
Z
1Z v
cos(u/v)du dv
−v
0
1
0
u=v
v sin(u/v)
u=−v
Z
= sin 1
vdv
0
=
sin 1
.
2
1
1
dv
2
EXAMPLE 2. Let R be the region bounded by the lines y = x and
y = x + 1 and by the hyperbolas y = 1/x and y = 2/x. Evaluate the
double integral
ZZ
(x + y)dxdy.
R
SOLUTION. Here the function f (x, y) = x + y is easy to integrate, but the
region R is not so attractive.
Observe that the arcs y − x = 0, y − x = 1, xy = 1, xy = 2 bounding R are
easily expressed in terms of
u = y − x,
v = xy.
Thus, as the point (x, y) varies in R, the point (u, v) varies in the square
region Q:
0 ≤ u ≤ 1, 1 ≤ v ≤ 2.
By the change of variables formula
ZZ
ZZ
(x + y)dxdy =
(x + y)|J|dudv.
R
Q
To compute the Jacobian J we need xu , yu , xv , yv .
Writing ux = xy − x2 and subtracting v = xy we obtain x2 + ux − v = 0.
Using implicit differentiation we find that
−x
−1
1
x
=
, xv = −
=
.
xu = −
2x + u
x+y
2x + u
x+y
Since y − x = u we conclude that yu = xu + 1 and yv = xv . So
y
1
yu =
, yv =
.
x+y
x+y
It follows that J = xu yv − xv yu =
−1
x+y .
Consequently,
ZZ
ZZ
Z
(x + y)dxdy =
(x + y)|J|dudv =
R
Q
1
2Z 1
dudv = 1.
0