Multivariate Calculus - Bruce E. Shapiro

152 LECTURE 18. DOUBLE INTEGRALS IN POLAR COORDINATES
We can solve this final integral by making the substitution u = 4 − r 2 , hence du =
−2rdr. When r = 0, u = 4 and when r = 2, u = 0. Therefore
S
√
4 − x 2 − y 2 dA = − π 2
= − π 4
∫ 0
4
∫ 0
4
u 1/2 (du/2)
u 1/2 du
= − π ∣
2 ∣∣∣
0
4 3 u3/2
= π 6
(4 3/2)
= 4π 3
4
Example 18.7 Find the value of
∫ ∞
−∞
e −x2 dx
Solution. There is no possible substitution that will allow us to find a closed form
solution to the indefinite integral
∫
e −x2 dx
However, by using polar coordinates we can find the definite integral ∫ ∞
−∞ e−x2 dx,
as follows. First, we define the quantity
I =
∫ ∞
−∞
e −x2 dx
which is the definite integral we are trying to find. Then
I 2 =
=
=
=
(∫ ∞
−∞
∫ ∞
−∞
∫ ∞ ∫ ∞
) (∫ ∞
e −x2 dx
e −x2 dx
−∞ −∞
∫ ∞ ∫ ∞
−∞
−∞
∫ ∞
−∞
e −x2 dx
−∞
e −y2 dy
e −x2 e −y2 dxdy
e −(x2 +y 2) dxdy
This is an integral over the entire xy plane; in polar coordinates this is the set
S = {(r, θ) : 0 ≤ r ≤ ∞, 0 ≤ θ ≤ 2π}
Revised December 6, 2006. Math 250, Fall 2006
)