Multivariate Calculus - Bruce E. Shapiro

LECTURE 16. DOUBLE INTEGRALS OVER RECTANGLES 129
6. The volume of box i is V i = f(u i , v i )∆A i .
7. The total volume under the surface is
n∑
V i =
i=1
n∑
f(u i , v i )∆A i
i=1
We then obtain the double integral by taking the limit as the number of squares
becomes large and their individual sizes approach zero.
Definition 16.1 Let f(x, y) : R ⊂ R 2 ↦→ R be a function of two variables defined
on a rectangle R. Then the Double Integral of f over R is defined as
R
f(x, y)dA =
lim
n→∞,∆A i →0
n∑
f(u i , v i )∆A i
i=1
if this limit exists. The summation on the right is called a Riemann Sum. If the
integral exits, the function f(x, y) is said to be integrable over R.
Theorem 16.1 If f(x, y) is bounded on a closed rectangle R, and is continuous
everywhere except for possibly only finitely many smooth curves in R, then f(x, y)
is integrable on R. Furthermore, if f(x, y) is continuous everywhere on R then
f(x, y) is integrable on R.
Example 16.1 Approximate
using a Riemann Sum for
R
f(x, y)dA
f(x, y) = x + y
and R is the rectangle [0, 6] × [0, 4], with a partition that breaks R into squares by
the lines x = 2, x = 4, and y = 2, and choosing points at the center of each square
to define the Riemann sum.
Solution. See figure 16.3. Choose points (u i , v i ) in the center of each square, at
(1, 1), (3, 1), (5, 1), (1, 3), (3, 3), (5, 3), as shown. Since each ∆A i = 4, the Riemann
Sum is then
R
f(x, y)dA ≈ ∑ n
f(u i, v i )∆A i
i=1
= 4 (f(1, 1) + f(3, 1) + f(5, 1) + f(1, 3) + f(3, 3) + f(5, 3))
= 4 (2 + 4 + 6 + 4 + 6 + 8) = 120
Math 250, Fall 2006 Revised December 6, 2006.