Answer
(a) $Area\left( S \right) = 6\pi \left( {2\sqrt 2 - \sqrt 5 } \right)$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} {z^{ - 1}}{\rm{d}}S = 3\pi \left( {\ln 8 - \ln 5} \right)$
Work Step by Step
(a) We can parametrize the sphere ${x^2} + {y^2} + {z^2} = 9$, where the domain is ${\cal D}:1 \le {x^2} + {y^2} \le 4$ in polar coordinates:
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta , \pm \sqrt {9 - {r^2}} } \right)$
Since $z \ge 0$, we choose
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,\sqrt {9 - {r^2}} } \right)$,
where the domain is ${\cal S} = \left\{ {\left( {r,\theta } \right):0 \le \theta \le 2\pi ,1 \le r \le 2} \right\}$.
${{\bf{T}}_r} = \dfrac{{\partial G}}{{\partial r}} = \left( {\cos \theta ,\sin \theta , - \dfrac{r}{{\sqrt {9 - {r^2}} }}} \right)$
${{\bf{T}}_\theta } = \dfrac{{\partial G}}{{\partial \theta }} = \left( { - r\sin \theta ,r\cos \theta ,0} \right)$
${\bf{N}}\left( {r,\theta } \right) = {{\bf{T}}_r} \times {{\bf{T}}_\theta } = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{\sin \theta }&{ - \dfrac{r}{{\sqrt {9 - {r^2}} }}}\\
{ - r\sin \theta }&{r\cos \theta }&0
\end{array}} \right|$
${\bf{N}}\left( {r,\theta } \right) = \dfrac{{{r^2}\cos \theta }}{{\sqrt {9 - {r^2}} }}{\bf{i}} + \dfrac{{{r^2}\sin \theta }}{{\sqrt {9 - {r^2}} }}{\bf{j}} + r{\bf{k}}$
$||{\bf{N}}\left( {r,\theta } \right)|| = \sqrt {\dfrac{{{r^4}{{\cos }^2}\theta + {r^4}{{\sin }^2}\theta + 9{r^2} - {r^4}}}{{9 - {r^2}}}} = \dfrac{{3r}}{{\sqrt {9 - {r^2}} }}$
Evaluate the area of $S$:
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} ||{\bf{N}}\left( {r,\theta } \right)||{\rm{d}}r{\rm{d}}\theta $
$Area\left( S \right) = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 1}^2 \dfrac{{3r}}{{\sqrt {9 - {r^2}} }}{\rm{d}}r{\rm{d}}\theta $
Write $t = 9 - {r^2}$. So, $dt = - 2rdr$. The integral becomes
$Area\left( S \right) = - \dfrac{3}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{t = 8}^5 {t^{ - 1/2}}{\rm{d}}t{\rm{d}}\theta $
$ = - 3\left( {2\pi } \right)\left( {{t^{1/2}}|_8^5} \right) = - 6\pi \left( {\sqrt 5 - 2\sqrt 2 } \right)$
So, $Area\left( S \right) = 6\pi \left( {2\sqrt 2 - \sqrt 5 } \right)$.
(b) Write $f\left( {x,y,z} \right) = {z^{ - 1}}$. Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} {z^{ - 1}}{\rm{d}}S$ in polar coordinates.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} f\left( {G\left( {r,\theta } \right)} \right)||{\bf{N}}\left( {r,\theta } \right)||{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} {z^{ - 1}}{\rm{d}}S = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 1}^2 \left( {\dfrac{1}{{\sqrt {9 - {r^2}} }}} \right)\left( {\dfrac{{3r}}{{\sqrt {9 - {r^2}} }}} \right){\rm{d}}r{\rm{d}}\theta $
$ = 3\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 1}^2 \dfrac{r}{{9 - {r^2}}}{\rm{d}}r{\rm{d}}\theta $
Write $t = 9 - {r^2}$, so $dt = - 2rdr$. The integral becomes
$ = - \dfrac{3}{2}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{t = 8}^5 {t^{ - 1}}{\rm{d}}t} \right)$
$ = - \dfrac{3}{2}\left( {2\pi } \right)\left( {\ln t|_8^5} \right) = - 3\pi \left( {\ln 5 - \ln 8} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal S}^{} {z^{ - 1}}{\rm{d}}S = 3\pi \left( {\ln 8 - \ln 5} \right)$.
