Answer
The flow rate through ${\cal T}$:
(a) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 1$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 1$
Work Step by Step
(a) The triangular region ${\cal T}$ with vertices $\left( {1,0,0} \right)$, $\left( {0,1,0} \right)$, and $\left( {0,0,1} \right)$ as in Figure 16 has equation: $x+y+z=1$. By writing $y=1-x-z$, we can parametrize ${\cal T}$ by
$G\left( {x,z} \right) = \left( {x,1 - x - z,z} \right)$, for $x,y,z \ge 0$
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 downward $z$-direction. To use the upward-pointing direction, we set ${\bf{N}}\left( {x,z} \right) = {\bf{i}} + {\bf{j}} + {\bf{k}}$.
As can be seen in Figure 16, $x,y,z \ge 0$, so
$1 - x - z \ge 0$
$x + z \le 1$
From here we obtain the domain of the triangular region:
${\cal D} = \left\{ {\left( {x,z} \right):0 \le x \le 1,0 \le z \le 1 - x} \right\}$
Since ${\bf{v}} = 2{\bf{k}}$, so
${\bf{v}}\left( {G\left( {x,z} \right)} \right) = \left( {0,0,2} \right)$
${\bf{v}}\left( {G\left( {x,z} \right)} \right)\cdot{\bf{N}}\left( {x,z} \right) = 2$
We calculate the flow rate through ${\cal T}$ by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\left( {G\left( {x,z} \right)} \right)\cdot{\bf{N}}\left( {x,z} \right){\rm{d}}x{\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 2\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - x} {\rm{d}}z{\rm{d}}x = 2\mathop \smallint \limits_{x = 0}^1 \left( {z|_0^{1 - x}} \right){\rm{d}}x$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 2\mathop \smallint \limits_{x = 0}^1 \left( {1 - x} \right){\rm{d}}x = 2\left( {x - \frac{1}{2}{x^2}} \right)|_0^1 = 1$
So, the flow rate through ${\cal T}$ is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{\cal T}} = 1$.
(b) The projection of ${\cal T}$ onto the $xy$-plane [the triangle with vertices $\left( {0,0,0} \right)$, $\left( {1,0,0} \right)$, and $\left( {0,1,0} \right)$] can be parametrized by
$G\left( {x,y} \right) = \left( {x,y,0} \right)$, for $x,y \ge 0$
As can be seen in Figure 16, in this case the triangular domain is
${\cal D} = \left\{ {\left( {x,y} \right):0 \le x \le 1,0 \le y \le 1 - x} \right\}$
So,
${{\bf{T}}_x} = \left( {1,0,0} \right)$
${{\bf{T}}_y} = \left( {0,1,0} \right)$
${\bf{N}}\left( {x,y} \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&0\\
0&1&0
\end{array}} \right) = {\bf{k}}$
Notice that this normal vector is already upward-pointing.
Since ${\bf{v}} = 2{\bf{k}}$, so
${\bf{v}}\left( {G\left( {x,y} \right)} \right) = \left( {0,0,2} \right)$
${\bf{v}}\left( {G\left( {x,y} \right)} \right)\cdot{\bf{N}}\left( {x,y} \right) = 2$
We calculate the flow rate through ${\cal T}$ by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\left( {G\left( {x,y} \right)} \right)\cdot{\bf{N}}\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 2\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} {\rm{d}}y{\rm{d}}x = 2\mathop \smallint \limits_{x = 0}^1 \left( {1 - x} \right){\rm{d}}x$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 2\left( {x - \frac{1}{2}{x^2}} \right)|_0^1 = 1$
So, the flow rate through the projection of ${\cal T}$ onto the $xy$-plane is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 1$.
