Let $S$ be the portion of the sphere $x^2+y^2+z^2=9$, where $1\leq x^2+y^2\leq4$ and $z\geq0$. Calculate the surface area of $S$
Ok i'm really confused with this one. I know i have to apply the surface area formula but and possibly spherical coordinates but i can't seem how to get the integral out.
The shape. I thought of using spherical system but after doing i ended up with a 3 coordinate system. I'm not even sure how to begin with this one.
Best Answer
The surface $S$ in question is a spherical zone. Its area can be found by elementary means: If $R$ is the radius of the sphere and $h$ is the $z$-height of the zone then the area $\omega(S)$ is given by $$\omega(S)=2\pi R h\ .$$ As $R=3$ and $h$ is easily computed as $h=\sqrt{9-1}-\sqrt{9-4}$ we obtain $$\omega(S)=6\pi(\sqrt{8}-\sqrt{5})\ .$$ Now for the integral: Use polar coordinates in the $(x,y)$-plane as parameter variables. Then $S$ is produced by $$S:\quad(r,\phi)\mapsto{\bf x}(r,\phi):=(r\cos\phi, \>r\sin\phi, \>\sqrt{9-r^2})\qquad(0\leq r\leq 2, \ 0\leq\phi\leq 2\pi)\ .$$ Then $${\bf x}_r=\bigl(\cos\phi,\sin\phi,-{r\over\sqrt{9-r^2}}\bigr),\quad{\bf x}_\phi=(-r\sin\phi,r\cos\phi,0)$$ and $${\bf x}_r\times{\bf x}_\phi=\left({r^2\cos\phi\over\sqrt{9-r^2}}, \ {r^2\sin\phi\over\sqrt{9-r^2}}, \ r\right)\ .$$ Therefore the area element becomes $${\rm d}\omega=|{\bf x}_r\times{\bf x}_\phi|\>{\rm d}(r,\phi)={3r\over\sqrt{9-r^2}}\>{\rm d}(r,\phi)\ ,$$ and we obtain $$\omega(S)=\int_0^{2\pi} \int_1^2{3r\over\sqrt{9-r^2}}\>dr\>d\phi=2\pi\ \left(-3\sqrt{9-r^2}\right)\biggr|_1^2=6\pi(\sqrt{8}-\sqrt{5})\ ,$$ as before.