Not fully understanding the main answer to a previous question: Why is the volume of a cone one third of the volume of a cylinder?

Question

In Larry Wang's answer to this question, "Why is the volume of a cone one third of the volume of a cylinder?", the nice animation shows that the volume of a square-based pyramid that fits just inside a cube has $1/3$ the volume of the cube. Now, looking at the light blue pyramid in the animation for example, you can look at the front and plan views of a horizontal cross-section of the pyramid and using similar triangles/proportional lengths of parallel sides, be convinced with Cavalieri's principle, that the apex of the light blue pyramid can be moved around on the upper face of the cube without changing the volume of the pyramid.

However, all this shows is that the volume of any square-based pyramid that fits just inside a cube has $1/3$ the volume of the cube. This does not show that the volume of any square-based pyramid that fits just inside a cuboid with a square face has $1/3$ the volume of the cuboid. So the proof with the cone only works for height $h = r\sqrt\pi$ right?

Answer 1

As long as you keep one edge length constant, you can multiply a second edge length by $k$ and divide the third edge length by $k$ to create a new cuboid with the same volume, and by Cavalieri's principle, using planes orthogonal to the edge whose length did not change, show that the volume of a pyramid with base on one face of the cuboid and apex at a vertex of the cuboid does not change.

Now start with a cube of side length $\sqrt[3]{Hs^2}$ for arbitrary positive $H$ and $s$ and construct a pyramid with apex at one vertex of the cube, using one of the opposite faces as the base of the pyramid. The volume of the cube is $Hs^2$ and the volume of the pyramid is $\frac13Hs^2$.

Multiply the edge of the cube perpendicular to the pyramid's base by $\sqrt[3]{H/s}$ and divide one of the edges of the base by $\sqrt[3]{H/s}$. You now have a cuboid of volume $Hs^2$ with edges $\sqrt[3]{H^2s}$, $s$, and $\sqrt[3]{Hs^2}$ containing a pyramid of volume $\frac13Hs^2$ with base $\sqrt[3]{Hs^2}$ by $s$ and height $\sqrt[3]{H^2s}$.

Multiply the edge of the cuboid of length $\sqrt[3]{H^2s}$ by $\sqrt[3]{H/s}$ and divide the edge of length $\sqrt[3]{Hs^2}$ by $\sqrt[3]{H/s}$. Now you have a cuboid of volume $Hs^2$ with edges $H$, $s$, and $s$ containing a pyramid of volume $\frac13Hs^2$ with base $s$ by $s$ and height $H$.

Therefore the square pyramid has $\frac13$ the volume of the cuboid with the same base and the same height, regardless of the height.

If you combine a generalized version of this construction with the fact that the volumes of similar objects are in proportion as the cubes of their linear measurements, you can conclude that "stretching" two objects linearly in one direction preserves the ratio of the volumes of the objects.