Double Integration With Polar Coordinates

[sarahxox]

Here is the integral:
\(\displaystyle \[\int_{0}^{\sqrt{2}} \int_{x}^{\sqrt{4-x^{2}}} \sin (x^{2}+y^{2}) dydx\)
I am suppose to convert it to polar coordinates. I think this is correct. This is what I got:
$\displaystyle \int_{0}^{\pi/4} \int_{x}^{\sqrt{2}} r\sin (r^{2})drd\Theta$
But then when I go to integrate it to dr i get:
$\displaystyle \int_{0}^{\pi/4} \frac{\cos(r^{2})}{2}d\Theta$
evaluted from 0 to 4, but then there is no point of the outside integral. So I am lost. Any help would be greatly appreciated.
Thanks!

 

[galactus]

The $\displaystyle \sqrt{4-x^{2}}$ suggests a circle of radius 2.

$\displaystyle \int_{0}^{\frac{\pi}{4}}\int_{0}^{2}rsin(r^{2})drd{\theta}$