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Start with a circular region in the plane, and build a right-circular cylinder of over it. Next construct the right-triangular cone with apex at the center of one of the bases of the cylinder, and between them construct the half-ellipsoid with a vertex at that same apex. I just learned a well-known fact that the ratios of the volumes of these three solids will be 1 : 2 : 3.

But let's instead start with a base of some other (convex?) shape. You can still build the cylinder over the base by crossing it with $[0,h]$ for some height $h$, and you can still build the cone over this by taking a point in the center of one of the bases of the cylinder as the apex of the cone, and constructing the locus of all lines connecting a point in the other base to that apex. But is there a canonical way to construct the analogue of the ellipsoid in this general situation? Here's an image of what I'm thinking in the case of a square-based cylinder and cone.

enter image description here

The only requirements I can think of for this generalized ellipsoid would be that it needs cross-sections all similar to the base, and it needs to have the correct ratio with the other two volumes, 1 : 2 : 3. And I suppose it should have some relation to the fact that a usual ellipse is the solution set to a quadratic polynomial. But I don't see an obvious (unique) way to do this. And it would be nice if this construction generalized to higher dimensions.

Mike Pierce
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  • Therefore, you have a generalized cone and you desire to have a (constructive ?) definition of an ellipsoid inscribed in this cone. I am afraid that the only shape that could be a candidate for that, if you want to preserve the 1:2:3 ratios would be saussage-like... – Jean Marie Sep 12 '19 at 04:46
  • @JeanMarie Like, yeah it's going to be sausage-like, in the same sense that an elliposoid with circular cross-sections is sausage-like. But there are tons of sausage-like surfaces you could draw that preserve that 1:2:3 ration. I'm wondering if there's one that is the natural choice to be called "the generalization of an ellipsoid," in this context. – Mike Pierce Sep 13 '19 at 16:05
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    Possible hint. The $1:2:3$ ratio comes from Cavalieri's principle. applied using the areas of the horizontal slices. – Ethan Bolker Apr 25 '22 at 20:39
  • @EthanBolker What I'm observing from your hint is that those slices of this shape would still have to be elliptical. I.e. that the way the three nested solids intersect yellow slices I just drew on the image would be "the same" up to rescaling and stuff. – Mike Pierce Apr 25 '22 at 20:49
  • @MikePierce Yes, I think that's right. – Ethan Bolker Apr 25 '22 at 21:05
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    So the cross-section would still be an ellipse in every horizontal (vertical?) yellow slice. Okay :) Now the only question is if such a shape has a name – Mike Pierce Apr 25 '22 at 23:25

1 Answers1

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$\newcommand{\Reals}{\mathbf{R}}\DeclareMathOperator{\Vol}{vol}$This isn't strictly natural, but it does generalize the situations depicted and has flexibility that may allow the construction to be modified to suit.

Let $K$ be the closure of a bounded, non-empty open set in $\Reals^{n}$ (convex if desired). Let $O$ denote the origin of $\Reals^{n}$ (in $K$ if desired, but not necessarily). For each non-negative real $c$, let $cK$ be result of scaling $K$ by $c$ about the origin $O$; precisely, $$ cK = \{y \in \Reals^{n} : \text{$y = cx$ for some $x$ in $K$}\}. $$ The $n$-dimensional volume of $cK$ is $c^{n}\Vol{K}$.

Fix a positive height $h$, and for each positive exponent $r$ consider the solid of dimension $(n + 1)$ with base $K$ whose slice at height $z$ is $[(1 - z/h)^{r}K] \times \{z\}$. Geometrically, use an $r$th power function to scale $K$ to a point as the slice climbs to height $h$. By Cavalieri's principle, the volume of this solid is \begin{align*} \int_{0}^{h} \Vol[(1 - z/h)^{r}K]\, dz &= \Vol(K) \int_{0}^{h} [(1 - z/h)^{r}]^{n}\, dz \\ &= h\Vol(K) \int_{0}^{1} u^{rn}\, du \\ &= \tfrac{1}{rn+1} h\Vol(K). \end{align*} If $r = 0$ we have the cylinder $K \times [0, h]$ of volume $h\Vol(K)$. If $r = 1/2$, we have an analog of the paraboloid, of volume $\frac{1}{n/2+1} h\Vol(K)$. If $r = 1$ we have the cone, of volume $\frac{1}{n+1} h\Vol(K)$.

When $n = 2$ this reduces to $h\Vol(K)$, $\frac{1}{2} h\Vol(K)$, and $\frac{1}{3} h\Vol(K)$, giving volumes in arithmetic progression. Generally, picking values $r = k/n$ for $0 \leq k \leq n$, so that $rn + 1 = k + 1$, gives solids whose volumes are in harmonic progression: $h\Vol(K)$, $\frac{1}{2} h\Vol(K)$, $\frac{1}{3} h\Vol(K)$, ..., $\frac{1}{n+1} h\Vol(K)$.