<s>[INST] How to find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$? 
Find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$. The correct answer should be $(8\pi-16)a^2$.

This is an illustration:

First, there's the following formula to calculate surface area:
$$
\int\int_R\sqrt{Z_x^2+Z_y^2+1}\cdot dA
$$
In our case it's:
$$
\int\int_R \sqrt{\frac{x^2+y^2}{4a^2-x^2-y^2}+1}dA
$$
We can move into polar coordinates (and multiply the original operand by $r$ as the Jacobian in polar), then:
$$
\int\int_R\ r\sqrt{\frac{r^2}{4a^2-r^2}+1}drd\theta
$$
Now the only thing left is to find the bounds of $r$ and $\theta$. 
The projection of the surface to $xy$ plane will be a circle with the center in $(0,a)$ with the radius $a$ so it's symmetric around $y$ axis. Then $0\le\theta\le\pi$. 
Because $x^2+y^2=2ay$ then $r=2a\sin\theta$, therefore we now have the integral:
$$
\int_0^{\pi}\int_0^{2a\sin\theta}r\sqrt{\frac{r^2}{4a^2-r^2}+1}drd\theta
$$
And this is where I'm stuck. How can I integrate $dr$?
 [/INST] After some chat and subsequent corrections, all is clear now. Continuing from the end of the question,
$$S/2=\int_0^{\pi}\int_0^{2a\sin\theta}r\sqrt{\dfrac{r^2}{4a^2-r^2}+1}drd\theta=$$
$$=\int_0^{\pi}\int_0^{2a\sin\theta}\frac {2ar}{\sqrt{4a^2-r^2}}drd\theta=2a\int_0^{\pi}\left(-\sqrt{4a^2-4a^2\sin^2\theta}+2a\right)d\theta=$$
$$=4a^2\int_0^\pi(-\vert\cos\theta\vert+1)d\theta=4a^2(\pi-2)$$
This is the area of the upper part, but upper and lower are symmetric, so,
$S=a^2(8\pi-16)$
</s>