Question

Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=x2+y2−−−−−−√

Answer 1

Parameterize the part of the sphere by

$\mathbf s(u,v)=(9\cos u\sin v,9\sin u\sin v,9\cos v)$

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi4$. Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=81\sin v\,\mathrm du\,\mathrm dv$

So the area of $S$ (the part of the sphere above the cone) is given by

$\displaystyle\iint_S\mathrm dS=81\int_{v=0}^{v=\pi/4}\int_{u=0}^{u=2\pi}\sin v\,\mathrm du\,\mathrm dv=81(2\pi)\left(1-\dfrac1{\sqrt2}\right)=(162-81\sqrt2)\pi$