I have a problem that is related to double integral and I think that it can be solved by using polar coordinates. However, I am not sure about my answer and I need your answer to verify my own.
Calculate $$ \iint_{D} \sqrt{x^2+y^2} dxdy $$ in which D: $x^2 + y^2 \leq 2y$ and $|x| \leq y$
Since $|x|\le y$, $\theta\in[\pi/4,\,3\pi/4]$. Since $r^2\le 2y=2r\sin\theta$, $r\le2\sin\theta$. The integral is$$\begin{align}\int_{\pi/4}^{3\pi/4}d\theta\int_0^{2\sin\theta}r^2dr&=\frac83\int_{\pi/4}^{3\pi/4}\sin\theta(1-\cos^2\theta)d\theta\\&=\frac89[\cos^3\theta-3\cos\theta]_{\pi/4}^{3\pi/4}\\&=\frac{20}{9}\sqrt{2}.\end{align}$$