Answer
$\int_0^a\int_{-\sqrt{a^2-y^2}}^0x^2ydxdy=0$
Work Step by Step
For any positive real number $a$, $R=\{(x,y)|0\leq y\leq a,-\sqrt{a^2-y^2}\leq x\leq 0\}$ describes a left half of circle center at $(0,0)$ with radius $a$.
In polar coordinates, it can be expressed as $R=\{(r,\theta)|0\leq r\leq a,\pi/2\leq \theta\leq 3\pi/2\}$.
Convert the integral in polar coordinates and evaluate:
$\int_0^a\int_{-\sqrt{a^2-y^2}}^0x^2ydxdy=\int_0^a\int_{\pi/2}^{3\pi/2}(r\cos \theta)^2r\sin\theta \cdot rd\theta dr$
$=\int_0^a\int_{\pi/2}^{3\pi/2}r^4\cos^2\theta\sin\theta d\theta dr$
$=\int_0^a-\frac{r^4\cos^3\theta}{3}]_{\pi/2}^{3\pi/2}dr$
$=\int_0^a 0dr$
$=0$
