Evaluate the area of the surface using double integral 
Surface of the cylinder $x^2+y^2=2x$ delimited by $z=0$ and
  $z=\sqrt{x^2+y^2}$.

I have to say that I found similar posts but none has helped me, some gave me wrong anwears. The surface integral I take always gave me $\pi$ in it. The real anwear is 8. I preffer polar coordinates, but it's ok without. 
 A: The surface $x^2+y^2=2x$ can be written in polar coordinates $(\rho,\phi)$ as
$$\rho=2\cos(\phi)$$
where $\phi \in[-\pi/2,\pi/2]$.

The vector $\vec r=\hat\rho \rho+\hat zz$ that locates a point on that surface can be written parametrically as 
$$\begin{align}
\vec r(\phi,z)&=\hat \rho(\cos(\phi))\rho(\phi)+\hat zz\\\\
&=\hat \rho(\phi)2\cos(\phi)+\hat zz
\end{align}$$

The vector surface element $\hat n\,dS$ of this surface is given by
$$\begin{align}
\hat n\,dS&=\left(\frac{\partial \vec r}{\partial \phi}\times\frac{\partial \vec r}{\partial z}\right)\,d\phi\,dz\\\\
&=\left(\left(-2\hat \rho\sin(\phi)+2\hat\phi \cos(\phi)\right)\times \hat z\right)\,d\phi\,dz\\\\
&=\left(2\hat \rho \cos(\phi)+2\hat \theta \sin(\phi)\right)\,d\phi\,dz
\end{align}$$
Hence, we have $dS=2\,d\phi\,dz$.

The surface is bounded by the plane $z=0$ and the $z=\sqrt{x^2+y^2}=2\cos(\phi)$.

Finally, we can write
$$\begin{align}
S&=\int_S (1)\,dS\\\\
&=\int_{-\pi/2}^\pi/2 \int_0^{2\cos(\phi)}(2)\,dz\,d\phi\\\\
&=4\int_{-\pi/2}^{\pi/2}\cos(\phi)\,d\phi\\\\
&=8
\end{align}$$
as was to be shown!
