Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \sqrt {{x^2} + {y^2}} {\rm{d}}S = \frac{2}{3}\pi \left( {2\sqrt 2 - 1} \right)$
Work Step by Step
We have $G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,\theta } \right)$. So,
${{\bf{T}}_r} = \frac{{\partial G}}{{\partial r}} = \left( {\cos \theta ,\sin \theta ,0} \right)$
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - r\sin \theta ,r\cos \theta ,1} \right)$
${\bf{N}}\left( {r,\theta } \right) = {{\bf{T}}_r} \times {{\bf{T}}_\theta } = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{\sin \theta }&0\\
{ - r\sin \theta }&{r\cos \theta }&1
\end{array}} \right|$
${\bf{N}}\left( {r,\theta } \right) = \sin \theta {\bf{i}} - \cos \theta {\bf{j}} + r{\bf{k}}$
$||{\bf{N}}\left( {r,\theta } \right)|| = \sqrt {{{\left( {\sin \theta } \right)}^2} + {{\left( {\cos \theta } \right)}^2} + {r^2}} = \sqrt {1 + {r^2}} $
Since $f\left( {x,y,z} \right) = \sqrt {{x^2} + {y^2}} $ we obtain $f\left( {G\left( {r,\theta } \right)} \right) = r$.
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( {r,\theta } \right)} \right)||{\bf{N}}\left( {r,\theta } \right)||{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 r\sqrt {1 + {r^2}} {\rm{d}}r{\rm{d}}\theta $
Let $t = 1 + {r^2}$. So, ${\rm{d}}t = 2r{\rm{d}}r$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{t = 1}^2 \sqrt t {\rm{d}}t{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{1}{2}\left( {2\pi } \right)\left( {\frac{2}{3}{t^{3/2}}|_1^2} \right) = \frac{2}{3}\pi \left( {2\sqrt 2 - 1} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \sqrt {{x^2} + {y^2}} {\rm{d}}S = \frac{2}{3}\pi \left( {2\sqrt 2 - 1} \right)$.
