This problem is from Harold Edwards' Advanced Calculus: A Differental Forms Approach. It is exercise $4c$ in section $1.3$.
For a unit flow in the $z$-direction find the total flow into the tetrahedron with vertices $$P=(x_0,y_0,0),\quad Q=(x_1,y_1,0),\quad R=(x_2,y_2,0),\quad S=(0,0,1)$$ that is, find the flow across each of the four sides orienting them appropriately, and add.
Just checking: Don't I only need to find the flow into the bottom side (the side $PQR$) because the other sides will either have only flow out of them or no flow in or out of them? I'm just trying to visualize this and it seems that I don't actually need to add the flow across the 4 sides. I just need to find the flow across the bottom side.
So I'd think that the flow into the tetrahedron would just be given by $$\begin{align}\text{flow into tetra } &= \left.dxdy\right|_{PQR} \\ &= [\text{area of } PQR] \\ &=\frac 12[x_0(y_1-y_2)+x_1(y_2-y_0)+x_2(y_0-y_1)]\end{align}$$
Or am I misunderstanding something?
Edit: OK. After drawing some pictures and thinking about it a bit, I can see that the other sides could contribute depending on the points $P$, $Q$, and $R$.
But now I'm stuck with something else: In addition to the area of $PQR$ given above (which I should have taken the absolute value of), I think that the areas of the projections of the other three triangles are:
$$A_{PQS}=\frac 12 \left|-y_0(x_1-x_0)+x_0(y_1-y_0)\right| \\ A_{QRS}=\frac 12 \left|-y_1(x_2-x_1)+x_1(y_2-y_1)\right| \\ A_{PRS}=\frac 12 \left|-y_0(x_2-x_0)+x_0(y_2-y_0)\right|$$
BUT, I'm pretty sure at least one side has to be on the other side of the tetrahedron from the $xy$ plane. And in fact all of them might be except the $PQR$ triangle. So how do I determine in general what the flow into the tetrahedron will be? It's going to the sum of some subset of these $4$ areas, but which ones depend on the points $P$, $Q$, and $R$ don't they? So is there any general expression I can give for this?
This exercise seems way harder than all the other ones in the text up to this point. I wonder if I'm somehow overthinking this? Is there some simple trick I'm missing?
Edit2: To Amitai Yuval: are you saying the answer should just be zero? Because that's what I'll get if I add the flow into and flow out of the tetrahedron. Even if that's what I'm supposed to get, I still wouldn't know how to orient the sides as they depend on the relative arrangements of points $P$, $Q$, and $R$.