I am trying to find the area of a surface, but I can't describe the domain of integration correctly. The surface is part of the cylinder $x^2+z^2=a^2$ inside $x^2+y^2=a^2$. Here is what I have done so far:
I tried to parametrize the cylinder $S:\{(x,y,z)=(a\cos{u},v,a\sin{u}):0 \le a^2\cos^2{u}+v^2 \le a^2\}$. But I can't integrate using this condition, it gets too hairy. Is there another way to visualize this problem?
The cylinder you want to find the surface area is $z=f(x,y)=\sqrt{a^2-x^2}$ so the differential of surface area is $dS=\sqrt{1+\left(\frac{-2x}{\sqrt{a^2-x^2}}\right)^2}dxdy$. Note that we are only taking the upper half of the surface, so will double our answer in the end.
$$ \begin{aligned} \iint_SdS&=\iint_D \sqrt{1+\left(\frac{-x}{\sqrt{a^2-x^2}}\right)^2} dx dy \\ &=\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \frac{a}{\sqrt{a^2-x^2}} dy dx \\ &=\int_{-a}^a2adx = 4a^2. \end{aligned}$$ Then we double the answer to get $8a^2$ for the surface area desired.