= Solution
The PDF diagram specifies the prism as
$$
0\leq x\leq2,\qquad0\leq y\leq4-2x,\qquad0\leq z\leq1.
$$
Its triangular horizontal cross-section has area four. Calculate the outward <flux> on each face using its outward <normal vector>.
On $x=0$, the normal is $-\mathbf e_x$ and $F_x=0$, so the <flux> is zero. On $y=0$, the normal is $-\mathbf e_y$, giving
$$
\Phi_{y=0}=-\int_0^1\int_0^2(2x-4)^2\,dx\,dz=-\frac{32}{3}.
$$
On the sloping vertical face $y+2x=4$, the outward normal is proportional to $(2,1,0)$. Both $F_x$ and $F_y$ vanish there, while $F_z$ is tangent to the face, so the <flux> is zero.
On the bottom $z=0$, the normal is $-\mathbf e_z$ and $F_z=1$, so the <flux> is minus the base area, namely $-4$. On the top $z=1$, $F_z=0$, so the <flux> is zero. Thus all five face values and their sum are
$$
\boxed{\begin{array}{c|ccccc}
\text{face}&x=0&y=0&y+2x=4&z=0&z=1\\\hline
\text{outward flux}&0&-32/3&0&-4&0
\end{array}\qquad\Phi_{\mathrm{total}}=-\frac{44}{3}}.
$$
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