Section 13.3: Double Integrals over General Regions
In Section 13.2, we learned how to integrate functions of two variables over rectangular
regions. In this section, we will consider integration over more general plane regions.
Definition: There are two types of plane regions.
• A region D in the xy-plane is said to be of type I if it lies between the graphs of two
continuous functions of x. That is,
D = {(x, y)|a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}
where g1 and g2 are continuous on [a, b].
• A region D in the xy-plane is said to be of type II if it lies between the graphs of two
continuous functions of y. That is,
D = {(x, y)|c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}
where h1 and h2 are continuous on [c, d].
Theorem: Suppose that f is a continuous function over some region D.
1. If D is a type I region, then
Z bZ
ZZ
g2 (x)
f (x, y)dydx.
f (x, y)dA =
a
D
g1 (x)
2. If D is a type II region, then
ZZ
Z
d
Z
h2 (y)
f (x, y)dA =
D
f (x, y)dxdy.
c
h1 (y)
Example: Evaluate the double integral
ZZ
x cos ydA,
D
where D is the region bounded by y = 0, y = x2 , and x = 2.
The region D is a type I region which can be described as
D = {(x, y)|0 ≤ x ≤ 2, 0 ≤ y ≤ x2 }.
Applying the previous theorem gives
ZZ
2
Z
x2
Z
x cos ydA =
x cos ydydx
D
0
Z
0
2
=
Z
0
2
x2
x sin y dx
0
x sin(x2 )dx
0
2
1
2 = − cos(x )
2
0
1
(1 − cos(4)).
=
2
=
Example: Find the volume of the solid that lies under the elliptic paraboloid z = 3x2 + y 2
and above the region D bounded by y = x and x = y 2 − y.
To find the points of intersection of these curves, let
y2 − y
y 2 − 2y
y(y − 2)
y
=
=
=
=
y
0
0
0, 2.
The region D is a type II region which can be described as
D = {(x, y)|0 ≤ y ≤ 2, y 2 − y ≤ x ≤ y}.
Applying the previous theorem gives
ZZ
Z
2
2
(3x + y )dA =
D
2
Z
y
(3x2 + y 2 )dxdy
0
y
Z 2
=
(x3 + xy 2 )
dy
y 2 −y
Z
0
2
y 2 −y
(y 6 + y 4 − 2y 3 )dy
0
2
1 7 1 5 1 4 =
y + y − y 7
5
2 0
128 32
=
+
−2
7
5
794
=
.
35
=
Example: Evaluate the iterated integral
Z
0
1
Z
3
2
ex dxdy.
3y
R 2
If we try to evaluate the integral in its present form, then we must first evaluate ex dx,
which can only be accomplished using infinite series. So we must change the order of integration. The region of integration can be expressed as
D = {(x, y)|0 ≤ y ≤ 1, 3y ≤ x ≤ 3}.
By sketching this region, we find that an alternative representation of D is
D = {(x, y)|0 ≤ x ≤ 3, 0 ≤ y ≤
Thus, the double integral can be evaluated as
Z 1Z 3
Z
x2
e dxdy =
0
3y
3
0
Z
x/3
Z
x
}.
3
2
ex dydx
0
3
1
2
xex dx
3 0
Z
1 9 u
e du
=
6 0
9
1 u =
e
6 0
1 9
=
(e − 1).
6
=
Theorem: (Properties of Double Integrals)
Suppose that f and g are integrable functions over some region D. Then,
1.
ZZ
ZZ
ZZ
[f (x, y) ± g(x, y)]dA =
D
2.
f (x, y)dA ±
D
ZZ
ZZ
cf (x, y)dA = c
f (x, y)dA
D
D
3. If f (x, y) ≤ g(x, y) for all (x, y) ∈ D, then
ZZ
ZZ
f (x, y)dA ≤
g(x, y)dA.
D
g(x, y)dA
D
D
4. If D = D1 ∪ D2 , where D1 and D2 do not overlap, then
ZZ
ZZ
ZZ
f (x, y)dA =
f (x, y)dA +
D
D1
f (x, y)dA.
D2
5. The area of the region D is given by
ZZ
Area =
1dA.
D