Answer
Using the cylindrical coordinates, the surface area of a sphere of radius $R$ is $4\pi {R^2}$.
Work Step by Step
The equation of a sphere of radius $R$ is ${x^2} + {y^2} + {z^2} = {R^2}$.
In cylindrical coordinates ${x^2} + {y^2} = {r^2}$, so we can parametrize the sphere as
$\left( {r\cos \theta ,r\sin \theta , \pm \sqrt {{R^2} - {r^2}} } \right)$
We shall consider the $z$-positive axis, that is the upper hemisphere of the sphere. So,
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,\sqrt {{R^2} - {r^2}} } \right)$
${{\bf{T}}_r} = \dfrac{{\partial G}}{{\partial r}} = \left( {\cos \theta ,\sin \theta ,\dfrac{{ - r}}{{\sqrt {{R^2} - {r^2}} }}} \right)$
${{\bf{T}}_\theta } = \dfrac{{\partial G}}{{\partial \theta }} = \left( { - r\sin \theta ,r\cos \theta ,0} \right)$
${\bf{N}}\left( {r,\theta } \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{\sin \theta }&{\dfrac{{ - r}}{{\sqrt {{R^2} - {r^2}} }}}\\
{ - r\sin \theta }&{r\cos \theta }&0
\end{array}} \right|$
${\bf{N}}\left( {r,\theta } \right) = \dfrac{{{r^2}\cos \theta }}{{\sqrt {{R^2} - {r^2}} }}{\bf{i}} + \dfrac{{{r^2}\sin \theta }}{{\sqrt {{R^2} - {r^2}} }}{\bf{j}} + r{\bf{k}}$
$||{\bf{N}}\left( {r,\theta } \right)|| = \sqrt {\dfrac{{{r^4}{{\cos }^2}\theta }}{{{R^2} - {r^2}}} + \dfrac{{{r^4}{{\sin }^2}\theta }}{{{R^2} - {r^2}}} + {r^2}} = \sqrt {\dfrac{{{r^4}}}{{{R^2} - {r^2}}} + {r^2}} $
$||{\bf{N}}\left( {r,\theta } \right)|| = \dfrac{{rR}}{{\sqrt {{R^2} - {r^2}} }}$
The domain description in cylindrical coordinates is
${\cal D} = \left\{ {\left( {r,\theta } \right):0 \le r \le R,0 \le \theta \le 2\pi } \right\}$
By Theorem 1, the surface area of the upper hemisphere is
$area = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {r,\theta } \right)||{\rm{d}}r{\rm{d}}\theta $
$area = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \dfrac{{rR}}{{\sqrt {{R^2} - {r^2}} }}{\rm{d}}r{\rm{d}}\theta = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^R \dfrac{{rR}}{{\sqrt {{R^2} - {r^2}} }}{\rm{d}}r$
$area = 2\pi R\mathop \smallint \limits_{r = 0}^R \dfrac{r}{{\sqrt {{R^2} - {r^2}} }}{\rm{d}}r$
Write $t = {R^2} - {r^2}$. So, $dt = - 2rdr$. The integral becomes
$area = - \pi R\mathop \smallint \limits_{{R^2}}^0 {t^{ - 1/2}}{\rm{d}}t = - 2\pi R\left( {{t^{1/2}}} \right)|_{{R^2}}^0 = 2\pi {R^2}$
The surface area of the sphere is twice the surface area of the upper hemisphere, so
$Area\left( S \right) = 2\cdot2\pi {R^2} = 4\pi {R^2}$
