Find coordinates of area's center of mass. So we have area $D:{(x^{2}+y^{2})^{2}=2a^{2}xy, x>0, y>0}$. $x= r \cdot cos(\varphi)$ and $y= r \cdot sin(\varphi)$
I want to find the center of mass by formula $y_{c} = \frac{M_{ox}}{m}$, where $M_{ox}=\iint_{D}(\gamma \cdot y \cdot dxdy)$ gamma is density and $m=\iint_{D}(\gamma \cdot dxdy)$.
First of all I decided to use polar coordinates. So $m=\frac{a^{2}}{2}$. and I have problem finding $M_{ox}=\int_{0}^{\pi/2}d(\varphi)\int_{0}^{a \cdot \sqrt{sin(2\varphi)}}r^{2} \cdot sin(\varphi)d(\varphi)$
Help me please to find that integral. Or maybe you have better solutions for this task.

\begin{align}
M_x
&= \int_0^{\pi/2}d\theta\int_0^{a\sqrt{\sin2\theta}}\gamma r^2\sin\theta\,dr \\
&= \gamma\int_0^{\pi/2}d\theta \dfrac13r^3\sin\theta\Big|_0^{a\sqrt{\sin2\theta}} \\
&= \gamma a^3\dfrac{2\sqrt{2}}{3}\int_0^{\pi/2}\sin^{5/2}\theta\cos^{3/2}\theta d\theta \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}2\int_0^{\pi/2}\sin^{2\frac74-1}\theta\cos^{2\frac54-1}\theta d\theta \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}\beta\left(\dfrac74,\dfrac54\right) \\
&= \gamma a^3\dfrac{\sqrt{2}}{3}\dfrac{\Gamma\left(\dfrac74\right)\Gamma\left(\dfrac54\right)}{\Gamma(3)} \\
&= \gamma a^3\dfrac{\sqrt{2}}{6}\dfrac34\Gamma\left(1-\dfrac14\right)\Gamma\left(1+\dfrac14\right) \\
&= \gamma a^3\dfrac{\sqrt{2}}{6}\dfrac34\dfrac{\pi\dfrac14}{\sin\pi\dfrac14} \\
&= \color{blue}{\dfrac{1}{16}\gamma a^3\pi}
\end{align}
where $\beta(x,y)$ is Beta function and $\Gamma\left(1-x\right)\Gamma\left(1+x\right)=\dfrac{\pi x}{\sin\pi x}$ is Euler's reflection formula.
