Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {{\rm{e}}^{ - z}}{\rm{d}}S = 4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)$
Work Step by Step
This is a cylinder with radius $2$ and of height $4$. So, we can parametrize it by
$G\left( {\theta ,z} \right) = \left( {2\cos \theta ,2\sin \theta ,z} \right)$
So,
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - 2\sin \theta ,2\cos \theta ,0} \right)$
${{\bf{T}}_z} = \frac{{\partial G}}{{\partial z}} = \left( {0,0,1} \right)$
${\bf{N}}\left( {\theta ,z} \right) = {{\bf{T}}_\theta } \times {{\bf{T}}_z} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - 2\sin \theta }&{2\cos \theta }&0\\
0&0&1
\end{array}} \right|$
${\bf{N}}\left( {\theta ,z} \right) = 2\cos \theta {\bf{i}} + 2\sin \theta {\bf{j}}$
$||{\bf{N}}\left( {\theta ,z} \right)|| = \sqrt {{{\left( {2\cos \theta } \right)}^2} + {{\left( {2\sin \theta } \right)}^2}} = 2$
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {\theta ,z} \right)} \right)||{\bf{N}}\left( {\theta ,z} \right)||{\rm{d}}\theta {\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 2\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{\rm{d}}\theta {\rm{d}}z = - 4\pi \left( {{{\rm{e}}^{ - z}}|_0^4} \right) = 4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {{\rm{e}}^{ - z}}{\rm{d}}S = 4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)$.
