Double integral bounds after polar transformation

[Cistra]

Homework Statement

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

Homework Equations

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!

 

[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.

 

[Mark44]

Homework Statement

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

Homework Equations

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)drd\theta, but I can't seem to understand the reasoning behind the bounds. Thanks for your help!

Here's why you have the bounds you show.
The region of integration for the iterated integral in rectangular coordinates was the first quadrant, 0 <= x <= infinity, 0 <= y <= infinity.
The region of integration won't change in switching to polar form, but its description usually does. Your first integration is with respect to r, so you want r to range from 0 to infinity. Your second integration is with respect to theta, so you want theta to range from 0 to pi/2. Your limits of integration are switched.

 

[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).

 

[Mark44]

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?

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.

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).