Convert a double integral into polar coordinates If $a$ is a positive number, compute
$$
\int_0^{2a}\int_{-\sqrt{2ay-y^2}}^0 \sqrt{x^2+y^2} \, dx \, dy.
$$
I wanted to use polar coordinates to compute this double integral. However, the lower $x$-limit
$$
x=-\sqrt{2ay-y^2}
$$
is equivalent to saying that $x$ satisfies
$$
x^2+(y-a)^2=a^2.
$$
I am still not sure how we can change this integral to polar coordinates...

Let $x=r\cos\theta,y=r\sin\theta$. Since $x\leq 0$, $\frac{\pi}2\leq \theta\leq\pi$. Since $x^2+y^2\leq 2ay$, $r^2\leq 2ar\sin\theta$ thus $r\leq 2a\sin\theta$. Hence 
\begin{align*}
\int_0^{2a}\int_{-\sqrt{2ay-y^2}}^0 \sqrt{x^2+y^2} \, dx \, dy&=\int\limits_{x^2+y^2\leq 2ay\\x\leq 0}\sqrt{x^2+y^2} \, dx \, dy\\&=\int_{\frac{\pi}2}^\pi \,d\theta\int_0^{2a\sin\theta}r^2\,dr\\&=\frac{8a^3}{3}\int_{\frac{\pi}2}^\pi(\sin\theta)^3\,d\theta.
\end{align*}
I believe that you can move on now.
