15.2 Double Integrals over General Regions
In section 15.1, we learned how to integrate z = f (x, y) over a rectangular region R = [a, b] × [c, d].
ˆ
x
f (x, y)dA = ∫
R
b
a
∫c
d
f (x, y)dydx = ∫
d
c
b
∫a f (x, y)dxdy.
ˆ When the integrand can be written as f (x, y) = g(x)h(y), the
special case of Fubini applies, and, if R is a rectangle, then
x
f (x, y)dA = ∫
R
Example 1: Evaluate
b
d
b
d
∫c g(x)h(y)dydx = (∫a g(x)dx)(∫c h(y)dy).
a
s
R
e2x+y dA, where R = [0, 1] × [4, 5].
What if the region of integration, D, is not rectangular?
Example 2: Consider the region D bounded by y = x2 and y = x, as shown below. Evaluate
1
x
∫0 ∫x2 4xy dydx.
1
Suppose z = f (x, y) is continuous over a region D in the xy-plane, as shown below.
There are two types of plane regions:
Type I: A plane region D is said to be of Type I if D lies between the graphs of two continuous functions
of x, that is D = {(x, y)|a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}.
If f is continuous on a Type I region D = {(x, y)|a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}, then
ZZ
f (x, y) dA =
D
Z b Z g2 (x)
a
f (x, y) dydx
g1 (x)
Type II: A plane region D is said to be of Type II if D lies between the graphs of two continuous functions
of y, that is D = {(x, y)|c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}.
If f is continuous on a Type II region D = {(x, y)|c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}, then
ZZ
f (x, y) dA =
D
Z d Z h2 (y)
c
2
h1 (y)
f (x, y) dxdy
Example 3: Sketch the region D bounded by x = y 2 and x = 4. Using a set of inequalities, describe D
as both a Type I and Type II region.
Example 4: Sketch the triangular region D with vertices (0, 0), (0, 4) and (2, 4). Evaluate
x
D
3
(x2 +y)dA.
Exercise 1: Sketch the region of integration and evaluate
x
D
xey dA, where D is the region bounded
D
by y = 0, y = x2 and x = 2.
Exercise 2: Evaluate
x
√
y
dA, where D is the region bounded by y = 0, y = x, and x = 4.
2
1+x
4
Example 5: Evaluate ∫
1
0
1
2
∫x sin(y )dydx.
Example 6: Sketch the region of integration and change the order of integration for ∫
1
2−y
∫
−2 y 2
5
f (x, y)dxdy.
Recall: If f (x, y) ≥ 0 and f is continuous on the region D, then the
volumexV of the solid that lies above D and under the surface f (x, y)
is V =
f (x, y)dA.
D
Example 7: Find the volume of the solid bounded by the three coordinate planes and the plane
2x + 3y + 4z = 12.
NOTE: If we integrate the constant function f (x, y) = 1 over a region
D, we get the area of D, that is:
A(D) =
x
1 dA
D
6