<s>[INST] Compute the area of a surface, encountering a strange integral compute the area of this yellow surface, which is actually a paraboloid:
$x^2+y^2=2az$(yellow one), cutted by $(x^2+y^2)^2=2a^2xy$(blue one)
To compute the part be surrounded in the blue surface, use polor coordinates:
\begin{cases}
x=r\cos\theta\\
y=r\sin\theta\\
z=\frac{r^2}{2a}\\
\end{cases}
use Gauss efficient to compute the area:
\begin{cases}
E=(\frac{\partial x}{\partial r})^2+(\frac{\partial y}{\partial r})^2+(\frac{\partial z}{\partial r})^2&=1+\frac{r}{a}\\ 
F=\frac{\partial x}{\partial r}\frac{\partial x}{\partial\theta}+\frac{\partial y}{\partial r}\frac{\partial y}{\partial\theta}+\frac{\partial z}{\partial r}\frac{\partial z}{\partial\theta}&=0\\
G=(\frac{\partial x}{\partial\theta})^2+(\frac{\partial y}{\partial\theta})^2+(\frac{\partial z}{\partial\theta})^2&=r^2\\ 
\end{cases}
$$S=\iint\limits_D\sqrt{EG-F^2}drd\theta$$
where D becomes $\left\{ (r,\theta )|\theta \in \left[ 0,\frac{\pi}{2} \right] \cup \left[ \pi ,\frac{3}{2}\pi \right] ,\mathrm{r}\in \left[ 0,\mathrm{a}\sqrt{\sin 2\theta} \right] \right\}$ since the blue surface can be written as $r^2=a^2\sin2\theta$.
I can't find anything wrong until now, but this intergal is extreme complex, and Mathematica give me a non elementary solution. However, the standard solution given by the text book to this question is$\frac{20-3 \pi}{9} a^{2}$
How can i get that answer? Or what's wrong with my method?
 [/INST] I am coming from assumption, that to obtain desired answer you should have cylinder $(x^2+y^2)^2=2a^2xy$ in place of brought in title $x^2+y^2=2a^2xy$ (blue one).
Then we will have $\sqrt{1+z'_x{^2}+z'_y{^2}} = \frac{1}{a}\sqrt{a^2+x^2+y^2}$. After writing double integral it is enough to use usual polar coordinates, so we will have
$$S=\frac{4}{a}\int\limits_{0}^{\frac{\pi}{4}}\,d\phi \int\limits_{0}^{a\sqrt{\sin 2\phi}}r\sqrt{a^2+r^2}\,dr = \frac{4}{3} a^2 \left( \int\limits_{0}^{\frac{\pi}{4}}(\sin \phi +\cos \phi)^3\,d \phi - \frac{\pi}{4} \right ) =\\
=\frac{4}{3} a^2  \left( 2\sqrt{2} \int\limits_{0}^{\frac{\pi}{4}} \sin^3 \left( \phi+\frac{\pi}{4} \right )\,d\phi - \frac{\pi}{4}\right ) =\\
=\frac{4}{3} a^2 \left(2\sqrt{2} \int\limits_{0}^{\frac{\pi}{4}} \left( \cos^2 \left( \phi+\frac{\pi}{4} \right )-1  \right )\,d \left( \cos \left( \phi+\frac{\pi}{4} \right )-1  \right ) - \frac{\pi}{4}\right ) = \frac{a^2}{9}(20-3\pi)$$
</s>