Multivariate Calculus - Bruce E. Shapiro

150 LECTURE 18. DOUBLE INTEGRALS IN POLAR COORDINATES
C rdrdθ = ∫ 2π
=
0
∫ 2π
= a2
2
∫ a
0
0
∫ 2π
0
= a2
2 (2π)
= πa 2
rdrdθ
∣
1 ∣∣∣
a
2 r2 dθ
Example 18.5 Find the volume V of the solid under the function
and over the region
f(x, y) = e −x2 −y 2
0
dθ
S = {(r, θ) : 5 ≤ r ≤ 7, 0 ≤ θ ≤ π/2}
Solution. This region is both r-simple and theta-simple, so it is not difficult to
“set-up” the integral:
V =
R
f(r, θ)rdrdθ =
∫ π/2 ∫ 7
0 5
f(r, θ)rdrdθ
To convert the function f(x, y) to a function in polar coordinates we observe that
since
x 2 + y 2 = r 2
then
f(x, y) = e −(x2 +y 2) = e −r2
Thus the integral becomes
V =
∫ π/2 ∫ 7
0 5
In the inner integral, make the substitution
u = −r 2
e −r2 rdrdθ
so that
du = −2dr
Revised December 6, 2006. Math 250, Fall 2006