Find surface area of part of cylinder. I ran into trouble when I'm trying to find a surface area of parts of the cylinder $x^2+z^2=4$ bounded by another cylinder $x^2+y^2=4$, I simply used a traditional way of double integral, change into polar coordinate calculate
$$
\iint\limits_{x^2+y^2=4}
  \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+
        \left(\frac{\partial z}{\partial y}\right)^2+1} \,dx\,dy
 = 
\int_0^{2\pi}\int_{0}^{2} \frac{2r}{\sqrt{4-(r\cos\theta)^2}} \,dr\,d\theta
$$ and eventually this integral diverges. Could anyone tell me where I was wrong ?  thanks a lot.
 A: Per my comment, the integral simplifies down to
$$\int_0^{2\pi} \frac{4}{1+|\sin\theta|}\:d\theta$$
after doing the $r$ integral. The easiest way to compute this integral then would be to exploit symmetries
$$ = \int_0^\pi \frac{8}{1+\sin\theta}\:d\theta = \int_0^\pi \frac{8}{\cos^2\left(\frac{\theta}{2}\right)+\sin^2\left(\frac{\theta}{2}\right)+2\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}\:d\theta$$
$$= \int_0^\pi \frac{8}{\left[\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)\right]^2}\:d\theta = 16\int_0^\pi \frac{\frac{1}{2}\sec^2\left(\frac{\theta}{2}\right)}{\left[1+\tan\left(\frac{\theta}{2}\right)\right]^2}\:d\theta$$
$$ = \frac{-16}{1+\tan\left(\frac{\theta}{2}\right)}\Biggr|_0^{\pi^-} = 16$$
And then the final answer for the problem would be $32$, doubled to account for both sides of the cylinder.
A: Easier to keep this in rectangular coordinates.
Limits
$x^2 + y^2 = 4\\
x^2 + z^2 = 4\\
y^2 = z^2\\
y = \pm z$
This suggests that integrating with respect to y and z makes more sense than integrating with respect to x and y.
$x = \pm \sqrt {4 - z^2}\\
dS = (1, - \frac {\partial x}{\partial y}, - \frac {\partial x}{\partial z})\\
dS =(1,0, -\frac {z}{\sqrt {4-z^2}})\\
\|dS\| = \frac {2}{\sqrt {4-z^2}}$
$4\int_{0}^2 \int_{-z}^{z} \frac {2}{\sqrt {4-z^2}} \ dy\ dz$
we are multiplying by 4 because over the triangle in the yz plane, there is a surface above the plane and a complimentary surface below the plane.  There is then an identical surface that appears when $z < 0$
