Answer
(a) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {{\rm{e}}^{ - r}}{{\bf{e}}_r}\cdot{\rm{d}}{\bf{S}} = 18\pi {{\rm{e}}^{ - 3}}$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {{\rm{e}}^{ - r}}{{\bf{e}}_r}\cdot{\rm{d}}{\bf{S}} = \frac{\pi }{2}{{\rm{e}}^{ - 1}}$
Work Step by Step
(a) In spherical coordinates:
$x = r\cos \theta \sin \phi $, ${\ \ \ }$ $y = r\sin \theta \sin \phi $, ${\ \ \ }$ $z = r\cos \phi $
So, ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$.
The upper hemisphere of ${x^2} + {y^2} + {z^2} = 9$ can be parametrized by
$G\left( {\theta ,\phi } \right) = \left( {3\cos \theta \sin \phi ,3\sin \theta \sin \phi ,3\cos \phi } \right)$
for $0 \le \theta \le 2\pi $, $0 \le \phi \le \frac{\pi }{2}$.
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - 3\sin \theta \sin \phi ,3\cos \theta \sin \phi ,0} \right)$
${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {3\cos \theta \cos \phi ,3\sin \theta \cos \phi , - 3\sin \phi } \right)$
By Eq. (2) in Section 17.4, the outward-pointing normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = 9\sin \phi {{\bf{e}}_r}$.
Since $0 \le \phi \le \frac{\pi }{2}$, the normal vector ${\bf{N}}\left( {\theta ,\phi } \right)$ is positive value, thus it is outward-pointing.
We evaluate the integral of ${\bf{F}} = {{\rm{e}}^{ - r}}{{\bf{e}}_r}$ over the upper hemisphere by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right){\rm{d}}\theta {\rm{d}}\phi $
Since on the hemisphere $R=3$, so ${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right) = {{\rm{e}}^{ - 3}}{{\bf{e}}_r}$. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{{\rm{e}}^{ - 3}}{{\bf{e}}_r}} \right)\cdot\left( {9\sin \phi {{\bf{e}}_r}} \right){\rm{d}}\theta {\rm{d}}\phi $
$ = 9{{\rm{e}}^{ - 3}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin \phi {\rm{d}}\theta {\rm{d}}\phi = 9{{\rm{e}}^{ - 3}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin \phi {\rm{d}}\phi $
$ = - 9{{\rm{e}}^{ - 3}}\left( {2\pi } \right)\left( {\cos \phi } \right)|_0^{\pi /2} = 18\pi {{\rm{e}}^{ - 3}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {{\rm{e}}^{ - r}}{{\bf{e}}_r}\cdot{\rm{d}}{\bf{S}} = 18\pi {{\rm{e}}^{ - 3}}$.
(b) The octant $x \ge 0$, $y \ge 0$, $z \ge 0$ of the unit sphere ${x^2} + {y^2} + {z^2} = 1$ can be parametrized by
$G\left( {\theta ,\phi } \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$
for $0 \le \theta \le \frac{\pi }{2}$, $0 \le \phi \le \frac{\pi }{2}$.
By Eq. (2) in Section 17.4, the outward-pointing normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = \sin \phi {{\bf{e}}_r}$.
Since $0 \le \phi \le \frac{\pi }{2}$, the normal vector ${\bf{N}}\left( {\theta ,\phi } \right)$ is positive value, thus it is outward-pointing.
We evaluate the integral of ${\bf{F}} = {{\rm{e}}^{ - r}}{{\bf{e}}_r}$ over the upper hemisphere by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right){\rm{d}}\theta {\rm{d}}\phi $
Since on the unit sphere $R=1$, so ${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right) = {{\rm{e}}^{ - 1}}{{\bf{e}}_r}$. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{{\rm{e}}^{ - 1}}{{\bf{e}}_r}} \right)\cdot\left( {\sin \phi {{\bf{e}}_r}} \right){\rm{d}}\theta {\rm{d}}\phi $
$ = {{\rm{e}}^{ - 1}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin \phi {\rm{d}}\theta {\rm{d}}\phi = {{\rm{e}}^{ - 1}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin \phi {\rm{d}}\phi $
$ = - {{\rm{e}}^{ - 1}}\left( {\frac{\pi }{2}} \right)\left( {\cos \phi } \right)|_0^{\pi /2} = \frac{\pi }{2}{{\rm{e}}^{ - 1}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {{\rm{e}}^{ - r}}{{\bf{e}}_r}\cdot{\rm{d}}{\bf{S}} = \frac{\pi }{2}{{\rm{e}}^{ - 1}}$.