Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} y{\rm{d}}S = \frac{1}{{108}}\left( {19\sqrt {19} - 1} \right)$
Work Step by Step
We have $G\left( {u,v} \right) = \left( {u,{v^3},u + v} \right)$. So,
${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {1,0,1} \right)$
${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( {0,3{v^2},1} \right)$
${\bf{N}}\left( {u,{\rm{v}}} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&1\\
0&{3{v^2}}&1
\end{array}} \right|$
${\bf{N}}\left( {u,{\rm{v}}} \right) = - 3{v^2}{\bf{i}} - {\bf{j}} + 3{v^2}{\bf{k}}$
$||{\bf{N}}\left( {u,v} \right)|| = \sqrt {{{\left( { - 3{v^2}} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {3{v^2}} \right)}^2}} = \sqrt {1 + 18{v^4}} $
Since $f\left( {x,y,z} \right) = y$ we obtain $f\left( {G\left( {u,v} \right)} \right) = {v^3}$.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {u,v} \right)} \right)||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{u = 0}^1 \mathop \smallint \limits_{v = 0}^1 {v^3}\sqrt {1 + 18{v^4}} {\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{u = 0}^1 {\rm{d}}u\mathop \smallint \limits_{v = 0}^1 {v^3}\sqrt {1 + 18{v^4}} {\rm{d}}v$
$ = \mathop \smallint \limits_{v = 0}^1 {v^3}\sqrt {1 + 18{v^4}} {\rm{d}}v$
Let $t = 1 + 18{v^4}$. So, ${\rm{d}}t = 72{v^3}{\rm{d}}v$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{1}{{72}}\mathop \smallint \limits_{t = 1}^{19} \sqrt t {\rm{d}}t = \frac{1}{{72}}\left( {\frac{2}{3}{t^{3/2}}|_1^{19}} \right) = \frac{1}{{108}}\left( {19\sqrt {19} - 1} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} y{\rm{d}}S = \frac{1}{{108}}\left( {19\sqrt {19} - 1} \right)$.
