Answer
Using the spherical coordinates the surface area of a sphere of radius $R$ is $4\pi {R^2}$.
Work Step by Step
We can parametrize the surface of a sphere of radius $R$ by $G\left( {\theta ,\phi } \right) = \left( {R\cos \theta \sin \phi ,R\sin \theta \sin \phi ,R\cos \phi } \right)$. So,
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - R\sin \theta \sin \phi ,R\cos \theta \sin \phi ,0} \right)$
${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {R\cos \theta \cos \phi ,R\sin \theta \cos \phi , - R\sin \phi } \right)$
By Eq. (2), the normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = {R^2}\sin \phi {{\bf{e}}_r}$. So, $||{\bf{N}}\left( {\theta ,\phi } \right)|| = {R^2}\sin \phi $.
The domain description in spherical coordinates is
${\cal D} = \left\{ {\left( {\theta ,\phi } \right):0 \le \theta \le 2\pi ,0 \le \phi \le \pi } \right\}$
By Theorem 1, the surface area is
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {\theta ,\phi } \right)||{\rm{d}}\theta {\rm{d}}\phi $
$Area\left( S \right) = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi {R^2}\sin \phi {\rm{d}}\theta {\rm{d}}\phi $
$ = {R^2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi $
$ = {R^2}\left( {2\pi } \right)\left( { - \cos \phi } \right)|_0^\pi = 2\pi {R^2}\cdot2 = 4\pi {R^2}$
So, the surface area of a sphere of radius $R$ is $4\pi {R^2}$.
