13.1 Double Integrals over Rectangles
In this chapter, we study the integral of multi-variable functions. First, we consider a
function f of two variables dened on a closed rectangle
R = [a, b] × [c, d] = { (x, y) ∈ R2 | a ≤ x ≤ b, c ≤ y ≤ d }
and f (x, y) ≥ 0 over R.
Problem
f?
. What is the volume of the solid that lies above R and under the graph of
The rst step is to divide the rectangle into subrectangles. We accomplish this by
dividing the interval [a, b] into m subintervals
a = x0 < x1 < ... < xm−1 < xm = b
of equal width ∆x = (b − a)/m and dividing [c, d] into n subintervals
c = y0 < y1 < ... < yn−1 < yn = d
of equal width ∆y = (d − c)/n. By drawing lines parallel to the coordinate axes
through these partition points we form the subrectangles
Rij = [xi−1 , xi ] × [yj−1 , yj ] = { (x, y) ∈ R2 | xi−1 ≤ x ≤ xi , yj−1 ≤ y ≤ yj }
each with area ∆A = ∆x∆y .
Next we choose a sample point (x∗ij , yij∗ ) in each Rij . Over each Rij we construct a
box whose height is given by f (x∗ij , yij∗ ) and volume is f (x∗ij , yij∗ )∆A.
Then the volume of the solid that lies above R and below the surface z = f (x, y) is
given by the double integral which is dened as
¨
f (x, y)dA := lim
n
m X
X
m,n→∞
R
f (x∗ij , yij∗ )∆A
i=1 j=1
. f is called integrable if the limit in the denition exists.
2. Another notation for the double integral is
Note 1
Note
¨
¨
f (x, y)dA =
R
f (x, y)dxdy
R
Example 1. Evaluate the integral
¨
4 dA
R
by identifying it as a volume of a solid, where R = [−1, 0] × [−3, 3].
Example 2. If R = { (x, y) ∈ R2 | − 1 ≤ x ≤ 1, −2 ≤ y ≤ 2 }, evaluate the integral
¨ p
R
1 − x2 dA