# Double Integral

1. Dec 31, 2011

### JG89

1. The problem statement, all variables and given/known data

Integrate $f(x,y) = e^{-(x^2 + y^2)}$ over the set $A = \{ (x,y): x > 0, y > 0, x^2 + y^2 < a \}$

2. Relevant equations

3. The attempt at a solution

The polar coordinate transformation $g(r,\theta) = (r cos\theta, r sin\theta)$ is a diffeomorphism from A to the set $B = \{(r,\theta): 0 < r < a, 0 < \theta < \pi / 2 \}$, so I can use the change of variables theorem.

So $\int_A e^{-(x^2 + y^2)} = \int_B re^{-r^2} = \int_0^a \int_0^{\pi/2} re^{-r^2} d \theta dr = (\pi / 2) \int_0^a re^{-r^2} dr$.

Let $u = -r^2$. Then $-du/2 = rdr$, so we have $(\pi / 2) \int_0^a re^{-r^2} dr = (- \pi / 4) \int_0^{-a^2} e^u du = (- \pi / 4)(e^{-a^2} - 1)$.

I get the same answer if I reverse the order of integration. Is this answer correct?

2. Dec 31, 2011

### SammyS

Staff Emeritus
It all looks fine to me.

3. Jan 1, 2012

### DavidAlan

All good buddy. Why even mention diffeomorphisms? This be calculus, there be dragons there.

4. Jan 1, 2012

### JG89

Because as far as I know, the change of variables theorem requires g to be a C^r diffeomorphism.