Section 13.2: Iterated Integrals
It is usually difficult to evaluate double integrals as the limit of a Riemann sum. It is
much easier to evaluate double integrals as iterated integrals using partial integration.
Definition: Double integrals of the form
Z bZ d
f (x, y)dydx
a
d
Z
Z
or
b
f (x, y)dxdy
c
c
a
are called iterated integrals.
To evaluate the iterated integral
Z bZ d
Z b Z
f (x, y)dydx =
a
c
d
f (x, y)dy dx,
c
a
first integrate with respect to y and then with respect to x. Similarly, to evaluate
Z dZ b
Z d Z b
f (x, y)dxdy =
f (x, y)dx dy,
c
a
c
a
first integrate with respect to x and then with respect to y.
1
Z
4
2
0
Integrating with respect to x first,
Z 1Z 2
Z
4
2
(x − y )dxdy =
1
0
Z
1
1
0
"
2 #
1 5
x − xy 2 dy
5
0
1
Z 1
31
− y 2 dy
=
5
0
1
31
1 3 y− y =
5
3
0
88
=
.
15
Integrating with respect to y first,
Z 2Z 1
Z
4
2
(x − y )dydx =
1
2
(x − y )dxdy and
Example: Evaluate the iterated integrals
0
Z
1
1 #
1
x4 y − y 3 dx
3
1
0
Z 2
1
=
x4 −
dx
3
1
2
1 5 1 =
x − x 5
3
1
88
=
.
15
2
"
(x4 − y 2 )dydx.
Theorem: (Fubini’s Theorem)
If f is continuous on the rectangle R = [a, b] × [c, d], then
ZZ
Z bZ d
Z dZ b
f (x, y)dA =
f (x, y)dydx =
f (x, y)dxdy.
R
a
c
c
ZZ
a
xexy dA, where R = [0, 1] × [0, 1].
Example: Evaluate the double integral
By Fubini’s Theorem, integrating first with respect to x gives
ZZ
Z 1Z 1
xy
xexy dxdy = e − 2.
xe dA =
0
R
0
By Fubini’s Theorem, integrating first with respect to y gives
Z 1Z 1
ZZ
xy
xexy dydx
xe dA =
0
0
R
Z 1
=
exy |10 dx
Z0 1
(ex − 1)dx
=
0
= (ex − x)|10
= e − 2.
It is best to choose the order of integration that yields a simple integral.
Example: Find the volume of the solid S lying under the plane z = 2x + 4y + 2 and above
the rectangle R = [0, 1] × [1, 4].
By Fubini’s Theorem,
ZZ
V
=
(2x + 4y + 2)dA
Z
R
4Z 1
=
(2x + 4y + 2)dxdy
1
Z
0
4
=
Z1 4
=
h
1 i
x2 + 4xy + 2x 0 dy
(4y + 3)dy
4
= 2y 2 + 3y 1
= 39.
1