How to find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$?

Question

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 $dr$?

Answer 1

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,

$S=a^2(8\pi-16)$

Answer 2

$$\frac{r^2}{4a^2-r^2}+1=\frac{r^2+4a^2-r^2}{4a^2-r^2}\implies r\sqrt{\frac{r^2}{4a^2-r^2}+1}=\frac r{\sqrt{4a^2-r^2}}\implies$$

$$\left.\int_0^{2a\sin\theta}\frac r{\sqrt{4a^2-r^2}}dr=-\sqrt{4a^2-r^2}\right|_0^{2a\sin\theta}=-\sqrt{4a^2-4a^2\sin^2\theta}+2a=-2a\cos\theta+2a$$

and etc. Yet I think the final answer is different from what you said it is...