Double integral $\iint_D |x^3 y^3|\, \mathrm{d}x \mathrm{d}y$ Solve the following double integral
\begin{equation}
\iint_D |x^3 y^3|\, \mathrm{d}x \mathrm{d}y
\end{equation}
where $D: \{(x,y)\mid x^2+y^2\leq y \}$.
Some help please? Thank you very much.

Since
$$
D=\{(x,y)|\ x^2+y^2\le y\}=\left\{(x,y)|\ x^2+(y-\frac12)^2\le \frac14\right\},
$$
we have $y\in [0,1]$ for all $(x,y) \in D$.
Setting
$$
x=r\cos\theta,\ y=r\sin\theta+\frac12,\quad \theta \in [-\pi/2,3\pi/2],\ r \in [0,\frac12],
$$
we get
\begin{eqnarray}
\int_D|x^3y^3|\,dxdy&=&\int_Dy^3|x|^3\,dxdy=\int_{-\pi/2}^{3\pi/2}\int_0^{\frac12}r^4|\cos^3\theta|(r\sin\theta+\frac12)^3\,drd\theta\\
&=&\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\
&&-\int_0^{\frac12}\int_{\pi/2}^{3\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\
&=&\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\
&&-\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta-\frac12)^3\,drd\theta\\
&=&2\int_0^{\frac12}\int_0^{\pi/2}r^4\cos^3\theta\left[(r\sin\theta+\frac12)^3-(r\sin\theta-\frac12)^3\right]\,drd\theta\\
&=&2\int_0^{\frac12}\int_0^{\pi/2}r^4\cos^3\theta(3r^2\sin^2\theta+\frac14)\,drd\theta\\
&=&\int_0^1(\frac{3t^2}{7\cdot2^6}+\frac{1}{320})(1-t^2)\,dt\\
&=&\int_0^1\left[\frac{3}{7\cdot2^6}(t^2-t^4)+\frac{1}{320}(1-t^2)\right]\,dt\\
&=&\frac{3}{7\cdot2^6}(\frac13-\frac15)+\frac{1}{320}(1-\frac13)=\frac{1}{336}.
\end{eqnarray}
