# Using polar coordinates to solve a double integral

• Apr 3rd 2011, 10:38 AM
bazingasmile
Using polar coordinates to solve a double integral
Hi,
I am confused on how to go from rectangular components to polar ones. I'm going to show you the problem I'm trying to solve and the steps I took to solve it. I would really be glad if anyone could take a look and tell me if I'm doing something wrong.
Thanks for viewing,
Bazinga

The question is:
Use polar coordinates to solve the following integral.

$\int^{3}_{-3} \int^{\sqrt{9-y^2}}_{-\sqrt{9-y^2}} e^{-x^2-y^2} \mathrm{d}x\mathrm{d}y$

I thought that $\sqrt{9-y^2}$ and $-\sqrt{9-y^2}$ formed two semi-circles (the positive one in the 1st and 4th quadrant, the other one in the 2nd and 3rd quadrant), which formed a circle.

Then the double integral in polar coordinates would be:

$\int^{2\pi}_{0} \int^{3}_{0} e^{-r^2}r \mathrm{d}r\mathrm{d}\theta$

Is that right?
• Apr 3rd 2011, 11:07 AM
FernandoRevilla
Right.
• Apr 3rd 2011, 11:08 AM
bazingasmile
Thank you very much! I'll mark this post as solved as it could help somebody else in the future.