The ancient result is that a right-circular cone of height $h$ and base-radius $r$ will have volume $\frac{1}{3} \pi r^2h$, which is $1/3$ the volume of the cylinder with same base and height. And the same is true if you start with a square-based pyramid: it's volume will be $1/3$ the volume of the rectangular prism that it lives in. Here's a relevant MathSE question for that.

What is the most general statement of this fact that we currently know to be true?

Thinking of the locally two-dimensional case, I'm pretty sure that this fact still holds if you start with any "nice" bounded planar region, build a generalized cylinder with bases and cross-sections congruent to that region, then take any point on a base of that cylinder as the apex of your cone. And then this generalizes even further due to Cavalieri's principle. But is this true? And in what generality is is true? Like, what conditions must we have on our base region? Furthermore, is there a version of this in higher dimensional spaces? Is there a version of this in spaces that are not $\mathbb{R}^n$?

  • $\begingroup$ given a planar shape with area S and a point in space with distance H from the plane of the planar shape, the volume of the object created from connecting the point to the contour of the planar shape is given by SH/3. Is this general enough to your taste?:) $\endgroup$ – Moti Aug 19 '19 at 17:45
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    $\begingroup$ I believe that this can be extended to higher spaces by having a proper n-1 dimension object and a point in the n space and the n dimension created volume will be by creating the n dimension from connecting the point to the "surface" of the n-1 dimension object divided by n. It is true for 2D and 3D. $\endgroup$ – Moti Aug 19 '19 at 17:49
  • $\begingroup$ @Moti nah, way more general. Like, are we sure that what happens in your second comment actually happens in higher dimensions? Also, what are we qualifying as a "planar shape?" What if the boundary of the shape is not rectifiable? I'm pretty sure that's fine. But is there any way that you can mess with the "shape," to where this will no longer hold true? $\endgroup$ – Mike Pierce Aug 19 '19 at 21:05
  • $\begingroup$ The only requirement from the n-1 dimension shape that it is defined by n-2 elements and that its n-1 volume is well defined. for example - 1, 2, and 3 dimension - the n-2 line has to be a straight line, well defined in as 1 dimension. $\endgroup$ – Moti Aug 20 '19 at 1:45

The $n$-dimensional prism still has volume formula $$V_n=V_{n-1}\cdot h$$ while the $n$-dimensional pyramid has volume formula $$V_n=\frac1n\cdot V_{n-1}\cdot h$$ where $V_{n-1}$ is the $(n-1)$-dimensional volume of the base facet and $h$ is the orthogonal height.

For $n=1$ you'd simply use $V_0=1$ and you see that $V_1=h$ in either formula. The case $n=2$ also is easy by planar geometry, and $n=3$ was the starting point of your question $-$ and as such admitted to be correct either.

A visualization of the idea on how this generalizes for instance is outlined here, by dissecting the according $n$-dimensional hypercube by its body diagonal into $n$ copies of a pyramid on a $(n-1)$-dimensional hypercube (q.e. a facet of the former).

--- rk

  • $\begingroup$ Yeah, I kinda figured that the $1/3$ wouldn't stick around in higher dimensions. You wouldn't happen to know if you can use the case of the $n$-dimensional pyramid as a basis for answering this question for different bases, would you? Like suppose you start with a generalized cylinder with $n$-dimensional base region (facet?) $R$. Can you show that the generalized cone of facet $R$ will have hyper-volume $1/(n+1)$ that of the generalized cylinder by chopping up $R$ into $n$-cubes, looking at the volumes of these pyramids, then taking a limit as the number of $n$-cubes approaches infinity? $\endgroup$ – Mike Pierce Aug 19 '19 at 21:08
  • $\begingroup$ @MikePierce Chopping into little cubes is hard. What you want is the $n$-dimensional analog of Cavalieri's principle, as your questions suggests. That's where the $1/n$ comes from. It starts with the formula for the area of a triangle when $n=2$. $\endgroup$ – Ethan Bolker Aug 19 '19 at 21:11

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