Find the surface area of the portion of the origin-centred sphere of radius $R=4$ that lies inside the cylinder $x^2 +y^2=12$ and above the $xy$ plane.
Does this question make sense? How can surface area lie inside the cylinder given that the radius of sphere is greater than radius of cylinder?
On
When the sphere is cut by the cylinder you have two spherical caps remaining. The figure below shows the 2D representation.
Now, the basics are
$$ \theta=\sin^{-1}\frac{r}{R}\\ d=R\cos\theta\\ h=R-d\\ $$
The surface area of each spherical cap is
$$S=2\pi Rh$$
For your case I find that the surface area above the $x-y$ plane is $S=16\pi$.
I have verified this calculation numerically by calculating the surface area of of surface of revolution (of the red line).
HINT: Well, you have these equations : \begin{equation} x^2+y^2+z^2 = 4 \end{equation} \begin{equation} x^2+y^2= 12 \end{equation} With the variable $z$ free for the second equation. Then you can parametrize with spheric coordinates: \begin{equation} x= 2 cos\theta sin\phi \end{equation} \begin{equation} y= 2 sin\theta sin\phi \end{equation} \begin{equation} z= 2 cos\phi \end{equation} How the surface has to be above the $xy$ plane, $\phi$ must be defined as $\phi \in [0,\frac{\pi}{2}]$ and $\theta$ staying with the typical range ,$\theta \in [0,2\pi]$. Now you wanna intersect our new surface (semi-sphere) with the cylinder. The $z$ restriction will be defined by the top of the sphere restriction. Hope this helps you dude.