Multivariate Calculus - Bruce E. Shapiro

LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS 139
Example 17.4 Find the iterated integral
and sketch the set S.
∫ 1 ∫ y 2
0 0
2ye x dxdy
Solution We write the integral as
∫ 1 ∫ y 2
0
0
2ye x dxdy =
=
∫ 1
0
∫ 1
0
∫ y 2
2y e x dxdy =
0
2y[e y2 − e 0 ]dy =
∫ 1
0
∫ 1
0
2y
} {e x | y2
0 dy
2ye y2 dy −
The first integral on the right can be solve with the substitution
u = y 2 , du = 2dy
∫ 1
0
2ydy
when y = 0, u = 0, and when y = 1, u = 1. Thus y=0, u=0; when y=1, u=1. Thus
∫ 1
0
2ye y2 dy =
∫ 1
The second integral is straightforward:
0
e u du = e u | 1 0 = e1 − e 0 = e − 1
Therefore
∫ 1
0
2ydy = y 2∣ ∣ 1 0 = 1 − 0 = 1
Figure 17.6: The region in example 4.
∫ 1 ∫ y 2
0
0
2ye x dxdy =
∫ 1
0
2ye y2 dy −
∫ 1
= (e − 1) − 1 = e − 2
0
2ydy
We can draw the domain by observing that the outer integral – the one over y –
increase from zero to 1. For any fixed y within this region, x increases for 0 to y 2 .
As we move up along the y axis (the outer integral) we need to increase from x = 0
to x = y 2 , i.e., the curve of y = √ x.
Math 250, Fall 2006 Revised December 6, 2006.