Surface area inside cylinder Find the surface area of the part $\sigma$: $x^2+y^2+z^2=4$ that lies inside the cylinder $x^2+y^2=2y$
So, the surface is a sphere of $R=2$. It looks there should be double integral to calculate the surface, but how, which way?
 A: Your cylinder is given by:
$$x^2 + (y-1)^2 = 1$$
The sphere fully overlaps the region in the plane given by the cylinder, so the region in the plane occupied by the surface is the same as the cylinder. The intersection of the cylinder and the sphere happens over
$$D:\{(x,y):\, -1\le x\le 1 \,\,\text{ and }\,\, 1-\sqrt{1-x^2} \le y\le 1+\sqrt{1-x^2}\, \}$$
You want 
$$2\iint_D \sqrt{f_x^2+f_y^2+1}\, dA$$
where $f(x,y) = \sqrt{4-x^2-y^2}$.
A: The surface in question is symmetric with respect to reflection in the $xy$-plane. Therefore the area is
$$
A=2\iint_D \sqrt{f_x^2+f_y^2+1}\, dxdy=4\iint_D\frac {dxdy}{\sqrt {4-x^2-y^2}}\tag1
$$ 
where we used $f(x,y) = \sqrt{4-x^2-y^2}$. $D$ is the crosssection of the cylinder with the $xy$-plane, which is a circle with radius 1 centered at the point $(0,1)$. 
To compute the integral $(1)$ it is suggestive to use polar coordinates $x=r\cos\phi,y=r\sin\phi$ and observe that the equation for the circle $D$ ($x^2+y^2=2y$) in polar coordinates reads:
$$
r=2\sin\phi,\quad(0\le\phi\le\pi).
$$
Therefore:
$$\begin{align}
A&=4\iint_D\frac {rdrd\phi}{\sqrt {4-r^2}}\\
&=4\int_0^{\pi}d\phi\int_0^{2\sin\phi}\frac {rdr}{\sqrt {4-r^2}}\\
&=8\int_0^{\pi}(1-|\cos\phi|)d\phi\\
&=8(\pi-2).
\end{align}
$$
