Multivariate Calculus - Bruce E. Shapiro

LECTURE 18. DOUBLE INTEGRALS IN POLAR COORDINATES 153
Transforming the integral to polar coordinates and using the fact that r 2 = x 2 + y 2
gives
I 2 =
=
=
∫ ∞ ∫ 2π
0
∫ ∞
0
∫ ∞
= 2π
0
e −r2 r
0
∫ ∞
e −r2 rdθdr
∫ 2π
0
e −r2 r(2π)dr
0
e −r2 rdr
dθdr
At this point we can make the substitution u = −r 2 .
r = 0, then u = 0 and when r = ∞, u = −∞
Then du = −2rdr, when
I 2 = 2π
= − 2π
∫ ∞
0
∫ −∞
0
e −r2 rdr
e u du
= −2πe u | −∞
0
= −2π(e −∞ − e 0 )
= −2π(0 − 1) = 2π
Therefore
∫ ∞
−∞
e −x2 dx = I = √ I 2 = √ 2π
Math 250, Fall 2006 Revised December 6, 2006.