Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{{54}}\left( {10\sqrt {10} - 1} \right)$
Work Step by Step
We can parametrize the surface by $G\left( {x,y} \right) = \left( {x,y,{x^3}} \right)$. So,
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0,3{x^2}} \right)$
${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {0,1,0} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{3{x^2}}\\
0&1&0
\end{array}} \right| = - 3{x^2}{\bf{i}} + {\bf{k}}$
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {1 + 9{x^4}} $
Since $f\left( {x,y,z} \right) = z$, we obtain $f\left( {G\left( {x,y} \right)} \right) = {x^3}$.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^1 {x^3}\sqrt {1 + 9{x^4}} {\rm{d}}x{\rm{d}}y$
Let $t = 1 + 9{x^4}$. So, $dt = 36{x^3}dx$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{1}{{36}}\mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{t = 1}^{10} \sqrt t {\rm{d}}t{\rm{d}}y$
$ = \frac{1}{{36}}\cdot\frac{2}{3}\left( {{t^{3/2}}|_1^{10}} \right) = \frac{1}{{54}}\left( {10\sqrt {10} - 1} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{{54}}\left( {10\sqrt {10} - 1} \right)$.
