Multivariate Calculus - Bruce E. Shapiro

130 LECTURE 16. DOUBLE INTEGRALS OVER RECTANGLES
Figure 16.3: A partition of the rectangle R = [0, 6] × [0, 4] into size equally-sized
squares can be used to calculate ∫ R
f(x, y)dA as described in example 16.1.
Theorem 16.2 Suppose that R = U + V where U, V , and R are rectangles such
that U and V do not overlap (except on their boundary) then
f(x, y)dA = f(x, y)dA + f(x, y)dA
R
U
in other words, the integral over the union of non-overlapping rectangles is the sum
of the integrals.
Theorem 16.3 Iterated Integrals. If R = [a, b] × [c, d] is a rectangle in the xy
plane and f(x, y) is integrable on R, then
∫ b
(∫ d
) ∫ d
(∫ b
)
f(x, y, )dA = f(x, y)dy dx = f(x, y)dx dy
R
a
c
The parenthesis are ordinarily omitted in the double integral, hence the order of the
differentials dx and dy is crucial,
f(x, y, )dA =
R
∫ b ∫ d
a
c
f(x, y)dydx =
V
c
a
∫ d ∫ b
c
a
f(x, y)dxdy
Note: If R is not a rectangle, then you can not reverse the order of integration
as we did in theorem 16.3.
Example 16.2 Repeat example 1 as an iterated integral.
Solution. The corresponding iterated integral is
∫ 6
(∫ 4
)
R f(x, y)dA = (x + y)dy dx
=
0
∫ 6
0
0
(∫ 4
xdy +
0
∫ 4
0
)
ydy dx
In the first integral inside the parenthesis, x is a constant because the integration is
over y, so we can bring that constant out of the integral:
∫ 6
( ∫ 4 ∫ 4
)
R f(x, y)dA = x dy + ydy dx
0
Revised December 6, 2006. Math 250, Fall 2006
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