Thread: Need help with double integral & polar coords

1. Need help with double integral & polar coords

I've been stuck on this for a good while now, and it's really starting to irritate me. Any help I can get would be greatly appreciated.

I've been trying to evaluate this problem for a good while now. I've looked through countless websites which explain the process of converting everything into polar coordinates, but for some reason it's just not clicking.

The part I'm having the most problems with are the limits of integration. I just can't figure out how they're determined, and once I think I do have it figured out, things don't seem to work.

Any help I can get is greatly appreciated.

2. Originally Posted by Protoman
I've been stuck on this for a good while now, and it's really starting to irritate me. Any help I can get would be greatly appreciated.

I've been trying to evaluate this problem for a good while now. I've looked through countless websites which explain the process of converting everything into polar coordinates, but for some reason it's just not clicking.

The part I'm having the most problems with are the limits of integration. I just can't figure out how they're determined, and once I think I do have it figured out, things don't seem to work.

Any help I can get is greatly appreciated.
$\displaystyle x = r~cos(\theta)$
and
$\displaystyle y = r~sin(\theta)$

So the integrand becomes
$\displaystyle x^2y^2 = r^4~sin^2(\theta)~cos^2(\theta)$

The area element becomes
$\displaystyle dx~dy = r~dr~d \theta$

And the region of integration is the right half-circle of radius 2 centered on the origin. So the integral becomes:
$\displaystyle \int_0^2 \int_{-\pi}^{\pi}r^4~sin^2(\theta)~cos^2(\theta) \cdot r~d \theta~dr$

-Dan

3. Originally Posted by topsquark
And the region of integration is the right half-circle of radius 2 centered on the origin.

Thanks for the help...could you explain how you know this?

4. Originally Posted by Protoman
Thanks for the help...could you explain how you know this?
Have you tried drawing the region of integration? $\displaystyle x = \pm \sqrt{4 - x^2} \Rightarrow x^2 = 4 - y^2 \Rightarrow x^2 + y^2 = 2^2$. So for $\displaystyle 0 \leq y \leq 2$ you clearly have a semi-circle centred on the origin. This clearly suggests using polar coordinates .....