Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z\left( {{x^2} + {y^2}} \right){\rm{d}}S = \frac{1}{5}\sqrt 2 $
Work Step by Step
We have $G\left( {u,v} \right) = \left( {u\cos v,u\sin v,u} \right)$. So,
${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {\cos v,\sin v,1} \right)$
${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( { - u\sin v,u\cos v,0} \right)$
${\bf{N}}\left( {u,{\rm{v}}} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos v}&{\sin v}&1\\
{ - u\sin v}&{u\cos v}&0
\end{array}} \right|$
${\bf{N}}\left( {u,{\rm{v}}} \right) = - u\cos v{\bf{i}} - u\sin v{\bf{j}} + u{\bf{k}}$
$||{\bf{N}}\left( {u,v} \right)|| = \sqrt {{{\left( { - u\cos v} \right)}^2} + {{\left( { - u\sin v} \right)}^2} + {u^2}} = u\sqrt 2 $
Since $f\left( {x,y,z} \right) = z\left( {{x^2} + {y^2}} \right)$ we obtain $f\left( {G\left( {u,v} \right)} \right) = {u^3}$.
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 = \sqrt 2 \mathop \smallint \limits_{u = 0}^1 \mathop \smallint \limits_{v = 0}^1 {u^4}{\rm{d}}u{\rm{d}}v = \sqrt 2 \left( {\frac{1}{5}{u^5}|_0^1} \right) = \frac{1}{5}\sqrt 2 $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z\left( {{x^2} + {y^2}} \right){\rm{d}}S = \frac{1}{5}\sqrt 2 $.
