3Integration in ℝ2 and ℝ2
IA Vector Calculus
3.2 Change of variables for an integral in R
2
Proposition.
Suppose we have a change of variables (
x, y
)
↔
(
u, v
) that is
smooth and invertible, with regions D, D
0
in one-to-one correspondence. Then
Z
D
f(x, y) dx dy =
Z
D
0
f(x(u, v), y(u, v))|J| du dv,
where
J =
∂(x, y)
∂(u, v)
=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
is the Jacobian. In other words,
dx dy = |J| du dv.
Proof.
Since we are writing (
x
(
u, v
)
, y
(
u, v
)), we are actually transforming from
(u, v) to (x, y) and not the other way round.
Suppose we start with an area
δA
0
=
δuδv
in the (
u, v
) plane. Then by
Taylors’ theorem, we have
δx = x(u + δu, v + δv) − x(u, v) ≈
∂x
∂u
δu +
∂x
∂v
δv.
We have a similar expression for δy and we obtain
δx
δy
≈
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
δu
δv
Recall from Vectors and Matrices that the determinant of the matrix is how
much it scales up an area. So the area formed by
δx
and
δy
is
|J|
times the area
formed by δu and δv. Hence
dx dy = |J| du dv.
Example. We transform from (x, y) to (ρ, ϕ) with
x = ρ cos ϕ
y = ρ sin ϕ
We have previously calculated that |J| = ρ. So
dA = ρ dρ dϕ.
Suppose we want to integrate a function over a quarter area D of radius R.
x
y
D
Let the function to be integrated be
f
=
exp
(
−
(
x
2
+
y
2
)
/
2) =
exp
(
−ρ
2
/
2). Then
Z
f dA =
Z
fρ dρ dϕ
=
Z
R
ρ=0
Z
π/2
ϕ=0
e
−ρ
2
/2
ρ dϕ
!
δρ
Note that in polar coordinates, we are integrating over a rectangle and the
function is separable. So this is equal to
=
h
−e
−ρ
2
/2
i
R
0
[ϕ]
π/2
0
=
π
2
1 − e
−R
2
/2
. (∗)
Note that the integral exists as R → ∞.
Now we take the case of x, y → ∞ and consider the original integral.
Z
D
f dA =
Z
∞
x=0
Z
∞
y=0
e
−(x
2
+y
2
)/2
dx dy
=
Z
∞
0
e
−x
2
/2
dx
Z
∞
0
e
−y
2
/2
dy
=
π
2
where the last line is from (*). So each of the two integrals must be
p
π/2, i.e.
Z
∞
0
e
−x
2
/2
dx =
r
π
2
.