I needed to find the volume of what Wikipedia calls a truncated prism, which is a prism (with triangle base) that is intersected with a halfspace such that the boundary of the halfspace intersects the three vertical edges of the prism at heights $h_1, h_2, h_3$.
I was able to find the formula $$ V=A\frac{h_1+h_2+h_3}{3}, $$ (where $A$ is the area of the triangle base) online, but without proof.
I was also able to prove this formula myself, but with a really nasty proof. (I integrated the area of the horizontal cross-sections; after passing the first intersection with the hyperplane at height $h_1$ these cross-sections have the form of the base triangle minus a quadratically increasing triangle, then after crossing the first intersection at height $h_2$ they have the form of a quadratically shrinking triangle)
Do you know of an elegant proof of the volume formula?
PS: Wikipedia cites 'William F. Kern, James R Bland,Solid Mensuration with proofs, 1938, p.81' for the name truncated prism, but I cannot find this book.
You see this as follows:
First we can reduce to the case where $h_1=0$ by observing that volume of the piece up to height $h_m=\min(h_1,h_2,h_3)$ is exactly $h_m \cdot A$.
Assume now $h_0=0$ and $h_1, h_2 \geq 0$. This denote the corners of the triangle $D,E,F$ where $D$ is the one with no vertical edge attached on it.
We denote the face bordered by the edges with lengths $h_2$ and $h_3$ by $B$. We observe, that the remaining truncated prism is actually a pyramid over $B$ with apex $D$.
That's the main observation, the rest is relatively simple calculation.
Denote now the height of triangle $A$ on vertex $D$ by $h_D$ and the triangle side opposite of $D$ by $d$. We get the area of $A = \tfrac{1}{2} d h_D$. Note that $h_D$ is perpendicular to $B$
For the volume $V$ we have $$ V = \tfrac{1}{3} B h_D = \tfrac{1}{3} \frac{d (h_2 + h_3)}{2} h_D = \tfrac{1}{3} \frac{d h_D}{2} (h_2 + h_3) = \tfrac{1}{3} A (h_2 + h_3) $$