Section 15.3 Double Integrals over General Regions
Ruipeng Shen
April 7
1
Basic Definition
Let us consider how to integrate a function not just over rectangles but also over regions of more
general shape, as illustrated in left part of figure 1. We suppose that D is a bounded region,
which means that D can be enclosed in a rectangular region R as shown in the right part of the
figure. Thus we can define a new function F with domain R by
f (x, y), if (x, y) ∈ D;
F (x, y) =
.
(1)
0,
if (x, y) ∈ R \ D
z
z
z=f(x,y)
z=F(x,y)
y
y
D
x
R
x
Figure 1: Double integral over a general region
If F is integrable over R, then we define the double integral of f over D by
ZZ
ZZ
f (x, y) dA =
F (x, y) dA.
D
R
In the case where f (x, y) ≥ 0, we can still interpret
that lies above D and under the surface z = f (x, y).
1
RR
D
f (x, y) dA as the volume of the solid
2
Integral over a Region of Type I
A plane region D 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], as shown in the figure below. In order to evaluate the
y
y=g2(x)
D
y=g1(x)
x
b
a
Figure 2: Region of Type I
RR
double integral D f (x, y) dA, we choose a rectangle R = [a, b] × [c, d] that contains D, as in
figure 3, and we let F be the function given by (1). Then, by Fubini’s theorem we have
ZZ
ZZ
b
Z
f (x, y) dA =
d
Z
F (x, y) dA =
D
R
y
Z
b
Z
g2 (x)
F (x, y) dy dx =
a
c
f (x, y) dy dx.
a
g1 (x)
y=d
y=g2(x)
x=a
x=b
y=g1(x)
y=c
x
Figure 3: Integral over Region of Type I
2
Proposition 1. If f is continuous on a type I region D such that
D = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)},
ZZ
Z b Z g2 (x)
f (x, y) dy dx.
then
f (x, y) dA =
g1 (x)
a
D
Similarly if f is continuous on a type II region D0 such that
D0 = {(x, y) : c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)},
Z d Z h2 (y)
f (x, y) dx dy.
f (x, y) dA =
ZZ
then
D0
h1 (y)
c
ZZ
(x + 2y) dA where D is the region bounded by the parabola y = 2x2
Example 2. Evaluate
D
and y = 1 + x2 .
y
(-1,2)
(1,2)
y=1+x2
D
y=2x2
x
Figure 4: Region bounded by parabola y = 2x2 and y = 1 + x2
Solution
This is a type I region given by
D = {(x, y) : −1 ≤ x ≤ 1, 2x2 ≤ y ≤ 1 + x2 }.
Thus we have
ZZ
Z
1
Z
1+x2
(x + 2y) dA =
1
(x + 2y) dy dx =
2x2
−1
1
D
Z
Z
=
−1
1+x2
xy + y 2 y=2x2 dx
x(1 + x2 ) + (1 + x2 )2 − x · 2x2 − (2x2 )2 dx
−1
1
Z
=
(1 + x + 2x2 − x3 − 3x4 ) dx
−1
1
2x3
x4
3x5
32
x2
+
−
−
=
.
= x+
2
3
4
5 −1
15
3
Z
1
Z
1
Example 3. Evaluate the iterated integral
y
sin(y 2 ) dy dx.
x
0
y=1
(1,1)
D
y=x
x
Figure 5: Region for integral in example 3
Solution Let D be the triangle with vertices (0, 0), (1, 1) and (0, 1), as shown in figure 5. This
region is of both Type I and Type II. Therefore we have
Z 1Z 1
ZZ
Z 1Z y
2
2
sin(y ) dy dx =
sin(y ) dA =
sin(y 2 ) dx dy
0
x
Z
=
0
D
1
0
0
1
1 − cos 1
1
2
2
.
y sin(y ) dy = − cos(y ) =
2
2
0
4