Chapter 13: Multiple Integrals
Section 13.1: Double Integrals
Consider a function f of two variables defined on a closed rectangle
R = [a, b] × [c, d] = {(x, y) ∈ R2 |a ≤ x ≤ b, c ≤ y ≤ d}.
Partition the intervals [a, b] and [c, d] as
a = x0 < x1 < · · · < xi−1 < xi < · · · < xm = b,
c = y0 < y1 < · · · < yj−1 < yj < · · · < yn = d.
This forms a partition, P , of the rectangle R into mn subrectangles
Rij = [xi−1 , xi ] × [yj−1 , yj ].
Let ∆xi = xi − xi−1 and ∆yj = yj − yj−1 . Then the area of the rectangle Rij is
∆Aij = ∆xi ∆yj .
Choose a representative point (x∗ij , yij∗ ) in each subrectangle Rij and form the double Riemann
sum
m X
n
X
f (x∗ij , yij∗ )∆Aij .
i=1 j=1
Let ||P || denote the norm of the partition P , which is the length of the longest diagonal of
all subrectangles Rij .
Definition: The double integral of f over R is
ZZ
f (x, y)dA = lim
R
||P ||→0
m X
n
X
f (x∗ij , yij∗ )∆Aij ,
i=1 j=1
provided this limit exists. If the limit exists, then f is called integrable.
Theorem: (Volume Using Double Integrals)
If f (x, y) ≥ 0 and f is continuous on the rectangle R, then the volume of the solid region
that lies above R and below the surface z = f (x, y) is given by
ZZ
V =
f (x, y)dA.
R
Example: Estimate the volume of the solid region that lies below the surface z = xy and
above the rectangle R = [6, 12] × [4, 8].
(a) Use a Riemann sum with m = 3, n = 2, and choose the representative points to be the
upper right corners of the subrectangles.
The rectangle R can be partitioned as shown in Figure 1. The representative points
are (8, 6), (8, 8), (10, 6), (10, 8), (12, 6), (12, 8), and the area of each subrectangle is 4.
Thus, the volume of the region can be approximated as
V
≈ 4[f (8, 6) + f (8, 8) + f (10, 6) + f (10, 8) + f (12, 6) + f (12, 8)]
= 4[8(6) + 8(8) + 10(6) + 10(8) + 12(6) + 12(8)]
= 1680.
8
8,8
10,8
𝑅12
6
𝑅22
8,6
6
𝑅32
10,6
𝑅11
4
12,8
12,6
𝑅21
8
𝑅31
10
12
Figure 1: Partition of the rectangle R = [6, 12] × [4, 8] into mn = 6 subrectangles with
representative points taken as the upper right corners.
(b) Use the Midpoint Rule to estimate the volume.
In this case, the representative points are (7, 5), (7, 7), (9, 5), (9, 7), (11, 5), and (11, 7).
Again the area of each subrectangle is 4. Thus, the volume of the region can be
approximated as
V ≈ 4[7(5) + 7(7) + 9(5) + 9(7) + 11(5) + 11(7)] = 1296.
ZZ
(4 − 2y)dA, where R = [0, 1]×[0, 1] by identifying
Example: Evaluate the double integral
R
it as the volume of a solid.
Let z = 4 − 2y. The value of the double integral is the volume of the solid region shown in
Figure 2. Thus,
ZZ
1
V =
(4 − 2y)dA = (1 × 1 × 2) + (1 × 1 × 2) = 3.
2
R
𝑧
2
2
1
𝑥
𝑦
1
Figure 2: Sketch of the solid region under the plane z = 4 − 2y and above the rectangle
R = [0, 1] × [0, 1].