# derivation of surface area measure on sphere

The sphere of radius $r$ can be described parametrically by spherical coordinates (what else ;) ) as follows:

 $x=r\sin u\sin v$
 $y=r\sin u\cos v$
 $z=r\cos u$

Then, using trigonometric identities to simplify expressions we find that

 $\frac{\partial(x,y)}{\partial(u,v)}=\left|\begin{matrix}r\cos u\sin v&r\sin u% \cos v\\ r\cos u\cos v&-r\sin u\sin v\end{matrix}\right|=-r^{2}\cos u\sin u$
 $\frac{\partial(y,z)}{\partial(u,v)}=\left|\begin{matrix}r\cos u\cos v&-r\sin u% \sin v\\ -r\sin u&0\end{matrix}\right|=-r^{2}\sin^{2}u\sin v$
 $\frac{\partial(z,x)}{\partial(u,v)}=\left|\begin{matrix}-r\sin u&0\\ r\cos u\sin v&r\sin u\cos v\end{matrix}\right|=r^{2}\sin^{2}u\cos v$

and hence, using more trigonometric identities, we find that

 $\sqrt{\left(\frac{\partial(x,y)}{\partial(u,v)}\right)^{2}+\left(\frac{% \partial(y,z)}{\partial(u,v)}\right)^{2}+\left(\frac{\partial(z,x)}{\partial(u% ,v)}\right)^{2}}=$
 $\sqrt{r^{4}\cos^{2}u\sin^{2}u+r^{4}\sin^{4}u\sin^{2}v+r^{4}\sin^{4}u\cos^{2}v}% =r^{2}\sin u.$

This means that, on a sphere

 $d^{2}A=r^{2}\sin u\>du\,dv.$

Note that in the case of a unit sphere, ($r=1$) this agrees with the formula presented in the second paragraph of subsection 2 of the main entry.