Multivariate Calculus - Bruce E. Shapiro

134 LECTURE 16. DOUBLE INTEGRALS OVER RECTANGLES
then du = 2dx, because y is a constant. When x=1, u=1+y; When x=3, u=9+y.
Therefore,
∫ 3
x
(x 2 + y) 2 dx = 1 ∫ 9+y
du
2 u 2
2
1
1
2
= 1 2
1+y
∫ 9+y
1+y
u −2 du
= 1 2 (−u−1 ) ∣ 9+y
1+y
= 1 [ −1
2 9 + y + 1 ]
1 + y
Substituting this back into the original iterated integral, we find that
∫ 5 ∫ 3
∫
x
5
[
(x 2 + y) 2 dxdy = 1 −1
2 9 + y + 1 ]
dy
1 + y
= 1 2 [− ln(9 + y) + ln(1 + y)]|5 2
= 1 [(− ln 14 + ln 6) − (− ln 11 + ln 3)]
2
= 1 (6)(11)
ln
2 (14)(3) = 1 11
ln
2 7 = ln √ 11/7 ≈ 0.226
When we integrate over non-rectangular regions, the limits on the inner integral
will be functions that depend on the variable in the outer integral.
Example 16.9 Find the area of the triangle
R = {(x, y) : 0 ≤ x ≤ b, 0 ≤ y ≤ mx}
using double integrals and show that it gives the usual formula y = (base)(height)/2
Solution.
∫ b
A =
R dxdy =
0
∫ mx
0
dydx =
∫ b
0
mxdx = mb 2 /2
We can find the height of the triangle by oberserving that at x = b, y = mb, therefore
the height is h = mb, and hence
A = mb 2 /2 = (mb)(b)/2 = (height)(base)/2
We will consider integrals over non-rectangular regions in greater detail in the following
section.
Revised December 6, 2006. Math 250, Fall 2006