Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S = \frac{1}{2}\sqrt 3 \pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = {z^2}$.
We can parametrize the plane by $G\left( {x,y} \right) = \left( {x,y, - x - y} \right)$. So,
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0, - 1} \right)$
${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {0,1, - 1} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{ - 1}\\
0&1&{ - 1}
\end{array}} \right|$
${\bf{N}}\left( {x,y} \right) = {\bf{i}} + {\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,z} \right)|| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 $
Since $x + y + z = 0$, so $f\left( {G\left( {x,y} \right)} \right) = {\left( { - x - y} \right)^2} = {x^2} + 2xy + {y^2}$.
From the cylinder ${x^2} + {y^2} = 1$, we obtain the polar description of the domain in the $xy$-plane ($z = 0$):
${\cal D} = \left\{ {\left( {r,\theta } \right):0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \sqrt 3 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + 2xy + {y^2}} \right){\rm{d}}x{\rm{d}}y$
We evaluate this integral in polar coordinates, so
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S$
$ = \sqrt 3 \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {{r^2}{{\cos }^2}\theta + 2{r^2}\cos \theta \sin \theta + {r^2}{{\sin }^2}\theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \sqrt 3 \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{{\cos }^2}\theta + 2\cos \theta \sin \theta + {{\sin }^2}\theta } \right){\rm{d}}\theta $
$ = \sqrt 3 \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {1 + \sin 2\theta } \right){\rm{d}}\theta $
$ = \sqrt 3 \left( {\frac{1}{4}{r^4}|_0^1} \right)\left( {\theta - \frac{1}{2}cos2\theta } \right)|_0^{2\pi } = \frac{1}{4}\sqrt 3 \left( {2\pi } \right) = \frac{1}{2}\sqrt 3 \pi $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S = \frac{1}{2}\sqrt 3 \pi $.
