# Double integral bounds after polar transformation

1. Dec 9, 2008

### Cistra

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

Basically I'm just trying to convert a double integral into polar coordinates, but when I do it I get confused with my bounds.

2. Relevant equations

3. The attempt at a solution

$$4\int_0^{\infty}\int_0^{\infty}e^{-(u^2+v^2)}u^{2x-1}v^{2y-1}dudv$$

(x and y are just numbers, not variables). Then I transform the integral given polar coordinates $$u=r\cos\theta, v=r\sin\theta$$ and I get

$$4\int_a^b\int_c^d e^{-r^2}(r\cos\theta)^{2x-1}(r\sin\theta)^{2y-1}rdrd\theta=4\int_a^b\int_c^d e^{-r^2}r^{2(x+y)-1}\cos^{2x-1}\theta\sin^{2y-1}\theta drd\theta$$.

I know that $$a=0,b=\infty,c=0,d=\frac{\pi}{2}$$ so that the integral bounds are $$\int_0^{\infty}\int_0^{\pi/2}f(r,\theta)rdrd\theta$$, but I can't seem to understand the reasoning behind the bounds. Thanks for your help!

Last edited: Dec 9, 2008
2. Dec 9, 2008

### mutton

In the uv-plane, the integral is over the first quadrant. Graph this region and describe it in terms of polar coordinates:

Since the region extends infinitely, the radius r goes from 0 to infinity. Since the angle $$\theta$$ is 0 on the positive u-axis and $$\pi/2$$ on the positive v-axis, $$\theta$$ goes from 0 to $$\pi/2$$.

So your bounds don't match up with their respective variables.

3. Dec 9, 2008

4. Dec 9, 2008

### Cistra

Hrm, okay thank you. Are you saying that it's supposed to be $$\int_0^{\pi/2}\int_0^{\infty}f(r,\theta)rdrd\theta$$ rather than what I've put? Because I'm splitting this up into the product of integrals rather than evaluating the double integral for itself, so I get

$$\int_0^{\infty}e^{-r^2}r^{2(x+y)-1}dr\cdot\int_0^{\pi/2}\cos^{2x-1}\theta\sin^{2y-1}\theta d\theta$$ in order to prove that $$\Gamma(x)\Gamma(y)=\Gamma(x+y)B(x,y)$$.

5. Dec 10, 2008

### Staff: Mentor

Yes, because that's from the order in dr dtheta.
If you want to integrate in the reverse order, I think that will work, too, because of the uncoupled nature of your f(r, theta), which can be written as g(r) * h(theta). That will, of course, switch the limits of integration.