# Lecture 17, November 2 • ∂ ·

```Lecture 17, November 2
• Formula for change of variables. When it comes to double integrals, one has
∫∫
∫∫
∂(x, y) du dv.
f (x, y) dx dy =
f (x(u, v), y(u, v)) &middot; ∂(u, v) Here, the additional factor inside the integral is the absolute value of the Jacobian
∂x ∂x ∂(x, y) ∂u ∂v ∂x ∂y ∂y ∂x
=
=
−
.
∂(u, v) ∂y ∂y ∂u ∂v ∂u ∂v
∂u ∂v
.....................................................................................
Example 1. Consider the region R in the xy-plane bounded by the lines
x + y = 1,
x + y = 2,
x − y = 0,
x − y = 1.
If we introduce the variables u = x − y and v = x + y, then we can write
∫∫
∫ 2∫ 1
x−y
u ∂(x, y) dx dy =
&middot;
du dv.
x+y
∂(u, v) 1
0 v
R
To compute the Jacobian in this case, we ﬁrst need to solve for x and y, namely
}
}
{
}
{
{
x = (u + v)/2
u + v = 2x
u=x−y
.
=⇒
=⇒
y = (v − u)/2
v − u = 2y
v =x+y
This allows us to diﬀerentiate x, y
∂x
∂(x, y) ∂u
=
∂(u, v) ∂y
∂u
with respect
∂x 1
∂v 2
=
∂y 1
− 2
∂v
to u, v and we now get
1 1
2 1 1
= + = .
1 4 4
2
2
Keeping this in mind, we can ﬁnally compute the given integral as
∫∫
∫ 2∫ 1
u
x−y
dx dy =
du dv
x+y
0 2v
1
R
[
]2
∫ 2 [ 2 ]1
∫ 2
u
ln v
ln 2
1
=
dv =
=
.
dv =
4v u=0
4 1
4
1
1 4v
```