Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx 0.2078$
Work Step by Step
From Exercise 8, we have $G\left( {u,v} \right) = \left( {{u^2} - {v^2},u + v,u - v} \right)$. So,
$G\left( {2,3} \right) = \left( { - 5,5, - 1} \right)$
${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2u,1,1} \right)$, ${\ \ \ }$ ${{\bf{T}}_u}\left( {2,3} \right) = \left( {4,1,1} \right)$
${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( { - 2v,1, - 1} \right)$, ${\ \ \ }$ ${{\bf{T}}_v}\left( {2,3} \right) = \left( { - 6,1, - 1} \right)$
${\bf{N}}\left( {2,3} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
4&1&1\\
{ - 6}&1&{ - 1}
\end{array}} \right| = - 2{\bf{i}} - 2{\bf{j}} + 10{\bf{k}}$
$||{\bf{N}}\left( {2,3} \right)|| = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2} + {{10}^2}} = \sqrt {108} = 6\sqrt 3 $
We estimate the area of the small patch $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$ by
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx ||{\bf{N}}\left( {2,3} \right)||\Delta u\Delta v$
Since $2 \le u \le 2.1$ and $3 \le v \le 3.2$, so $\Delta u = 0.1$ and $\Delta v = 0.2$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx 6\sqrt 3 \cdot 0.1 \cdot 0.2 = 0.2078$
