Question

Find the area of the part of the sphere x2 + y2 + z2 = 100z that lies inside the paraboloid z = x2 + y2.

Answer 1

$x^2+y^2+z^2=100z\iff x^2+y^2+(z-50)^2=50^2$

The surface of interest can be parameterized by

$\mathbf r(u,v)=\begin{cases}x(u,v)=50\cos u\sin v\\y(u,v)=50\sin u\sin v\\z(u,v)=50(\cos v+1)\end{cases}$

with $0\le u\le2\pi$ and $0\le v\le\arccos\dfrac{49}{50}$.

The area of the surface $S$ is given by the surface integral

$\displaystyle\iint_S\mathrm dA=\iint_T\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv$

We have

$\|\mathbf r_u\times\mathbf r_v\|=50^2\sin v$

so the area is given by the double integral

$\displaystyle50^2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\arccos(49/50)}\sin v\,\mathrm dv\,\mathrm du=100\pi$