Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S \approx 18.672$
Work Step by Step
We can parametrize the surface as $G\left( {x,z} \right) = \left( {x,9 - {z^2},z} \right)$. So,
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0,0} \right)$
${{\bf{T}}_z} = \frac{{\partial G}}{{\partial z}} = \left( {0, - 2z,1} \right)$
${\bf{N}}\left( {x,z} \right) = {{\bf{T}}_x} \times {{\bf{T}}_z} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&0\\
0&{ - 2z}&1
\end{array}} \right|$
${\bf{N}}\left( {x,z} \right) = - {\bf{j}} - 2z{\bf{k}}$
$||{\bf{N}}\left( {x,z} \right)|| = \sqrt {0 + {{\left( { - 1} \right)}^2} + {{\left( { - 2z} \right)}^2}} = \sqrt {1 + 4{z^2}} $
Since $f\left( {x,y,z} \right) = 1$, we obtain $f\left( {G\left( {x,z} \right)} \right) = 1$.
From $0 \le x \le z \le 3$, we obtain the description:
${\cal D} = \left\{ {\left( {x,z} \right):0 \le z \le 3,0 \le x \le z} \right\}$
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,z} \right)} \right)||{\bf{N}}\left( {x,z} \right)||{\rm{d}}x{\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{x = 0}^z \sqrt {1 + 4{z^2}} {\rm{d}}x{\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{z = 0}^3 \left( {\sqrt {1 + 4{z^2}} } \right)\left( {x|_0^z} \right){\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{z = 0}^3 \left( {z\sqrt {1 + 4{z^2}} } \right){\rm{d}}z$
Write $t = 1 + 4{z^2}$. so ${\rm{d}}t = 8z{\rm{d}}z$. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{1}{8}\mathop \smallint \limits_{z = 1}^{37} \sqrt t {\rm{d}}t$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{1}{8}\cdot\frac{2}{3}\left( {{t^{3/2}}|_1^{37}} \right) = \frac{1}{{12}}\left( {\sqrt {{{37}^3}} - 1} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S \approx 18.672$.
