Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{6}\sqrt 3 $
Work Step by Step
We have $f\left( {x,y,z} \right) = z$.
We can parametrize the plane by $G\left( {x,y} \right) = \left( {x,y,1 - x - y} \right)$. 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 $x + y + z = 1$, so $f\left( {G\left( {x,y} \right)} \right) = 1 - x - y$.
From the plane $x + y + z = 1$, we obtain the description of the domain in the $xy$-plane ($z = 0$):
${\cal D} = \left\{ {\left( {x,y} \right):0 \le x \le 1,0 \le y \le 1 - x} \right\}$
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \sqrt 3 \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} \left( {1 - x - y} \right){\rm{d}}x{\rm{d}}y$
$ = \sqrt 3 \mathop \smallint \limits_{x = 0}^1 {\rm{d}}x\mathop \smallint \limits_{y = 0}^{1 - x} \left( {1 - x - y} \right){\rm{d}}y$
$ = \sqrt 3 \mathop \smallint \limits_{x = 0}^1 {\rm{d}}x\left( {y - xy - \frac{1}{2}{y^2}} \right)|_0^{1 - x}$
$ = \sqrt 3 \mathop \smallint \limits_{x = 0}^1 {\rm{d}}x\left[ {1 - x - x\left( {1 - x} \right) - \frac{1}{2}{{\left( {1 - x} \right)}^2}} \right]$
$ = \sqrt 3 \mathop \smallint \limits_{x = 0}^1 \left( {1 - 2x + {x^2} - \frac{1}{2} + x - \frac{1}{2}{x^2}} \right){\rm{d}}x$
$ = \sqrt 3 \mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{2}{x^2} - x + \frac{1}{2}} \right){\rm{d}}x$
$ = \sqrt 3 \left( {\frac{1}{6}{x^3} - \frac{1}{2}{x^2} + \frac{1}{2}x} \right)|_0^1 = \frac{1}{6}\sqrt 3 $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{6}\sqrt 3 $.
