Find surface area of part of cylinder.

Question

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.

Answer 1

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.