Line integrals - Surface area Here is my task:
Calculate surface area of $2(x^{2}+y^{2})^{2}=xy$ between surface $x^{2}+y^{2}=z$ and $z=0$.
Here is my attempt to solve this problem.
Firstly, I transformed line $2(x^{2}+y^{2})^{2}=xy$ to polar form, putting $x=\rho\cos \phi$ and $y=\rho\sin \phi$. I got $\rho=\frac{1}{2}\sqrt{\sin 2\phi}$. It looks like this:
http://s13.postimg.org/wvwox80s7/math.png
To use formula surface_area=$\int z(x,y)ds$ (I don't know how to write line integral symbol in latex so I used symbol for "ordinary" integral instead), we must find first $ds$, which by definitions equals $$ds=\sqrt{(\frac{\mathrm{d} x}{\mathrm{d} \phi})^{2}+(\frac{\mathrm{d} y}{\mathrm{d} \phi})^{2}}d\phi$$
We have:
$x=\rho\cos \phi=\frac{1}{2}\sqrt{\sin 2\phi}\cos\phi$, $y=\rho\sin \phi=\frac{1}{2}\sqrt{\sin 2\phi}\sin\phi$. After differentiating $x$ and $y$ with respect to $\phi$ and putting it in expression for $ds$, we get (if I didn't make mistake somewhere in calculations) $ds=\frac{1}{2\sqrt{\sin 2\phi}}d\phi$. Now we can put everything in our expression for surface area:
$\int z(x,y)ds$=$4\int_{0}^{\pi/4}\left [ (\frac{1}{2}\sqrt{\sin 2\phi}\cos\phi)^{2} + (\frac{1}{2}\sqrt{\sin 2\phi}\sin\phi)^{2} \right ]\frac{1}{2\sqrt{\sin 2\phi}}d\phi=4\int_{0}^{\pi/4}\frac{1}{4}\sin2\phi(\cos^{2}\phi+\sin^{2}\phi)\frac{1}{2\sqrt{\sin 2\phi}}d\phi=\frac{1}{2}\int_{0}^{\pi/4}\frac{\sin{2\phi}}{\sqrt{\sin{2\phi}}}d\phi$
I have no idea how to solve this integral. Any suggestion?
 A: PARAMETERIZATION OF SURFACE
We will use $\phi$ and $z$ to parameterize the surface.  The vector that locates a point on the surface is denoted by
$$\vec r(\phi,z)=\hat \rho(\phi)\rho(\phi)+\hat zz$$ 
where $\hat \rho$ and $\hat z$ are radial and axial unit vectors, respectively.
We have $\rho(\phi)=\frac12 \sqrt{\sin (2\phi)}$ for $0\le \phi \le \pi/2$ and $\pi\le \phi \le 3\pi/2$.  ______________________________________________________________________
SURFACE LIMITS
We will be integrating over $\phi$ and $z$.  Symmetry considerations show that the surface integral over the entire domain is simply four times that over $0\le\phi\le\pi/4$.
We are give that $z$ is bounded below by $z=0$ and above by $z=\rho^2(\phi)=\frac12 \sin(2\phi)$.
The Jacobian $J$ for the surface integral is given by
$$\begin{align}
J&=\left|\frac{\partial \vec r(\phi,z)}{\partial \phi}\times\frac{\partial \vec r(\phi,z)}{\partial z}\right|\\\\
&=\left|(\hat \rho(\phi)\rho(\phi)-\hat \phi(\phi)\rho'(\phi))\times(\hat z)\right|\\\\
&\sqrt{\rho^2(\phi)+\rho'^2(\phi)}
\end{align}$$
where $\rho'(\phi)$ is the derivative of $\rho(\phi)$ with respect to $\phi$ and is 
$$\rho'(\phi)=\frac12 \frac{\cos(2\phi)}{\sqrt{\sin(2\phi)}}$$
Therefore, we have 
$$J=\frac12\frac{1}{\sqrt{\sin(2\phi)}}$$

SURFARCE INTEGRAL
The surface area $S$ is given by
$$\begin{align}
S&=4\int_{0}^{\pi/4}\int_{0}^{\sin(2\phi)/4}\frac12\frac{1}{\sqrt{\sin(2\phi)}}dzd\phi\\\\
&=\frac12\int_{0}^{\pi/4}\sqrt{\sin(2\phi)}d\phi\\\\
&=\frac14\int_0^{\pi/2}\sqrt{\sin(\phi)}d\phi \tag 1\\\\
\end{align}$$
Enforcing the substitution $u=\cos^2\phi$, then $du=-2\cos \phi\,\sin \phi\,d\phi$.  We also have $\sin \phi = (1-u)^{1/2}$.  Thus, $(1)$ becomes
$$\begin{align}
S&=\frac18\int_0^1 u^{-1/2}(1-u)^{-1/4}du\\\\
&=\frac18\,B(1/2,3/4)\\\\
&=\frac18\,\frac{\Gamma(1/2)\,\Gamma(3/4)}{\Gamma(5/4)}\\\\
\end{align}$$
Now, using the Duplication Formula 
$$\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\Gamma(2z)$$
and the functional equation 
$$\Gamma(z+1)=z\Gamma(z)$$
reveals that $\Gamma(5/4)=2^{-1/2}\Gamma(3/2)/\Gamma(3/4)=2^{-3/2}\sqrt{\pi}/\Gamma(3/4)$.  Putting it all together yields
$$S=\bbox[5px,border:2px solid #C0A000]{\frac14\,\frac{\Gamma^2(3/4)}{\sqrt{2\pi}}}$$
