By transforming to polar coordinates, prove $\int_0^{a/\sqrt{2}}\int_x^\sqrt{a^2-x^2}(x^2+y^2)^{1/2}dydx=\frac{\pi a^3}{12}$.
I know that I have to change the function using Jacobian discriminant. Once that is done I will be integrating the new function which will end in $drd\theta$ instead of $dydx$. The thing I can not figure out is how to change the limits from the Cartesian coordinates to polar coordinates. Can someone please tell me how they can be found so I can solve the rest of the problem.
Sketch the first quadrant of $x^2+y^2\le a^2$, then shade the integration region. You'll see $0\le x\le\frac{a}{\sqrt{2}},\,x\le y\le\sqrt{a^2-x^2}$ iff $0\le r\le a,\,\frac{\pi}{4}\le\theta\le\frac{\pi}{2}$. So the integral is $\int_{\pi/4}^{\pi/2}d\theta\int_0^ar^2dr$, which agrees with the required result.