Answer
The surface area of S:
$Area\left( S \right) = 16$
Work Step by Step
We have
${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2,0,1} \right)$, ${\ \ \ \ \ }$ ${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( {4,0,3} \right)$
So,
${\bf{N}}\left( {u,v} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
2&0&1\\
4&0&3
\end{array}} \right| = - 2{\bf{j}}$
$||{\bf{N}}\left( {u,v} \right)|| = 2$
The surface area of S:
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$
$Area\left( S \right) = 2\mathop \smallint \limits_{u = 0}^2 \mathop \smallint \limits_{v = 0}^4 {\rm{d}}u{\rm{d}}v = 16$
