Answer
The flow rate is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{1}{8}$
Work Step by Step
The net given by $y = 1 - x - z$, for $x,y,z \ge 0$ can be parametrized by
$G\left( {x,z} \right) = \left( {x,1 - x - z,z} \right)$,
Since $x,y,z \ge 0$, so
$1 - x - z \ge 0$
$x + z \le 1$
From here we obtain the domain of the net:
${\cal D} = \left\{ {\left( {x,z} \right):0 \le x \le 1,0 \le z \le 1 - x} \right\}$
So,
${{\bf{T}}_x} = \left( {1, - 1,0} \right)$
${{\bf{T}}_z} = \left( {0, - 1,1} \right)$
${\bf{N}}\left( {x,z} \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&{ - 1}&0\\
0&{ - 1}&1
\end{array}} \right) = - {\bf{i}} - {\bf{j}} - {\bf{k}}$
Notice that the normal vector obtained above is in the negative $y$-direction. To use the positive orientation, we set ${\bf{N}}\left( {x,z} \right) = {\bf{i}} + {\bf{j}} + {\bf{k}}$.
Evaluate:
${\bf{v}}\left( {G\left( {x,z} \right)} \right) = \left( {z\left( {1 - x - z} \right),xz,x\left( {1 - x - z} \right)} \right)$
${\bf{v}}\left( {G\left( {x,z} \right)} \right)\cdot{\bf{N}}\left( {x,z} \right) = z\left( {1 - x - z} \right) + xz + x\left( {1 - x - z} \right)$
$ = - {z^2} - {x^2} + \left( {1 - x} \right)z + x$
We calculate the flow rate through the net by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\left( {G\left( {x,z} \right)} \right)\cdot{\bf{N}}\left( {x,z} \right){\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - x} \left( { - {z^2} - {x^2} + \left( {1 - x} \right)z + x} \right){\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{x = 0}^1 \left( { - \dfrac{1}{3}{z^3} - {x^2}z + \dfrac{1}{2}\left( {1 - x} \right){z^2} + xz} \right)|_0^{1 - x}{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( { - \dfrac{1}{3}{{\left( {1 - x} \right)}^3} - {x^2}\left( {1 - x} \right) + \dfrac{1}{2}\left( {1 - x} \right){{\left( {1 - x} \right)}^2} + x\left( {1 - x} \right)} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( { - \dfrac{1}{3}{{\left( {1 - x} \right)}^3} - {x^2}\left( {1 - x} \right) + \dfrac{1}{2}\left( {1 - x} \right){{\left( {1 - x} \right)}^2} + x\left( {1 - x} \right)} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {\dfrac{{5{x^3}}}{6} - \dfrac{{3{x^2}}}{2} + \dfrac{x}{2} + \dfrac{1}{6}} \right){\rm{d}}x$
$ = \left( {\dfrac{5}{{24}}{x^4} - \dfrac{1}{2}{x^3} + \dfrac{1}{4}{x^2} + \dfrac{1}{6}x} \right)|_0^1$
$ = \dfrac{5}{{24}} - \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{1}{8}$
So, the flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{1}{8}$.
