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The volume of a straight cone in $\mathbb R^3$ is usually find adding the circular sections orthogonal to the height. If the base has radius $R$ and the height is $h$ we have: $$ V_{C3}=\int_0^h \pi r^2 dz=\pi \int_0^h \frac{R^2}{h^2}z^2 dz=\pi \frac{R^2}{h^2} \frac{1}{3} h^3= \frac{1}{3}\pi R^2 h $$ tha same result can be foud using the Pappus centroid theorem, rotating a right triangle with legs $h$ and $R$ around the cathetus $h$. Since the distance of the centroid from the axis of rotation is $R/3$, the theorem gives $$V_{C3}=Ad=\frac{1}{2}Rh \cdot 2\pi\frac{R}{3}=\frac{1}{3}\pi R^2h$$

If we generalize such results to a hypercone in $\mathbb R^4$, the first procedure gives the result $$ V_{C4}=\int_0^h\frac{4}{3}\pi r^3dz=\frac{4}{3}\int_0^h\left(\frac{R}{h}z \right)^3dz=\frac{1}{4}\cdot\frac{4}{3}\pi R^3h $$ in accord to what we can find in Wikipedia (and in other internet resources). And this result can be generalized to $n-$dimension : $$ V_{Cn}=\frac{1}{n}V_{S(n-1)}h $$
($V_{S(n-1)}$ is the volume of the hypersphere of dimension $n-1$).

To generalize te second procedure I use the generalized Pappus Theorem as formulated in this page of Wikipedia:

Volume of $n$-solid of revolution of species $p$ = (Volume of generating $(n-p)$ solid) $\times$ (Surface area of $p-$sphere traced by the $p-$th centroid of the generating solid)

so, using the same triangle but rotating it on the surface of a $3-$sphere, I found: $$ V_{C4}=Rh \cdot 4\pi\left( \frac{R}{3} \right)^2=\frac{4}{9}\pi R^3h$$ that is a different volume (and in $n-$dimension I have an analogous result).

Where is my reasoning wrong? I suspect that I have a misinterpretation of the generalized Pappus Theorem or that the rotation of a triangle around a $(n-1)-$sphere gives a different kind of cone with respect to the usual cone in n-dimension. If so, where I can find a classification of such kind of cones?

Emilio Novati
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  • (+1) Just a couple of immediate impressions (time doesn't permit more at the moment): 1. The volume of both the unit sphere and unit ball in four-space are rational multiples of $\pi^2$, so something is awry with the cone volume (maybe just a missing factor of $\pi$?). 2. Rotations (orthogonal transformations of determinant $1$) in dimension greater than three work differently than three-dimensional intuition suggests. That's not to say Wikipedia is wrong, only that the language of "rotation" may be misleading. – Andrew D. Hwang Jan 20 '22 at 16:09
  • thank you @Andrew. The base of the 4-cone is a 3-spere so the factor $\pi$ (and not $\pi^2$) is correct. – Emilio Novati Jan 20 '22 at 16:15
  • My oops; you're right, of course. – Andrew D. Hwang Jan 20 '22 at 20:38
  • @EmilioNovati Yours doesn't seem to be the same hypercone defined in wikipedia as "the hypervolume swept out by a tetrahedron standing upright in the 4th direction as it rotates freely about its base in the 3D hyperplane on which it rests". That said, all this is not exactly my cup of tea, and I am not sure what a p-th centroid even is. – dxiv Jan 22 '22 at 09:52
  • Thank you @dxiv. I also think that the problem is that. But I don't quite understand what it means "...it rotates freely about its base in the 3D hyperplane ..." and I can't find a source where the classification of cones in n-dimension is explained and discussed. – Emilio Novati Jan 22 '22 at 15:11
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    In the hope it's helpful, the wikipedia page on Pappus's theorem references a 1958 book by D.M.Y. Sommerville on $n$-dimensional geometry; the link here points to a copy from Osmania University at the Internet Archive. (Still trying to Really Understand understand the definitions myself....) – Andrew D. Hwang Jan 22 '22 at 16:09
  • that seems useful. Tank you – Emilio Novati Jan 22 '22 at 16:55

1 Answers1

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tl; dr: The formulas work out for a cone of height $h$ and base radius $R$ in four-space. The volume is indeed \begin{align*} \tfrac{1}{3}\pi R^{3}h &= (\tfrac{1}{2}Rh)(\tfrac{2}{3}\pi R^{2}) \\ &= (\text{area of generating triangle})(\text{area of sphere through the triangle's $p$-centroid}) \end{align*} for a suitable "$p$-centroid." This point is not the usual geometric centroid, however: Its location depends on $p$.


$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}\DeclareMathOperator{\Vol}{Vol}$Generalities: To start, here's a brief account of "volumes of rotation of species $p$." (This is doubtless the content of Sommerville's book, but Sommerville's notation is not how I visualize the geometry.) Let $p$ and $m$ be positive integers, and let $n = m + p + 1$. To describe what is meant by "revolving a region $D \subset \Reals^{m+1}$ about $p$-dimensional spheres," let $SO(p+1)$ denote the group of Euclidean rotations of $\Reals^{p+1}$, and decompose Cartesian $n$-space as \begin{align*} \Reals^{n} &= \Reals^{m} \times \Reals \times \Reals^{p} \\ &= \Reals^{m} \times \underbrace{\Reals^{p+1}}_{\text{$SO(p+1)$ acts}} \\ &= \underbrace{\Reals^{m+1}}_{\text{$D$ lives here}} \times \Reals^{p}. \end{align*} If $\Vec{x} \in \Reals^{m}$, $\Vec{y}_{0} \in \Reals^{p}$, $r$ is real, and $\Vec{y} = (r, \Vec{y}_{0}) \in \Reals^{p+1}$, the general element of $\Reals^{n}$ may be written $$ (\Vec{x}, \Vec{y}) = (\Vec{x}, r, \Vec{y}_{0}). $$ We're arranging that $D$ is contained in the half-space of $\Reals^{m+1}$ where $r \geq 0$ and $\Vec{y}_{0} = \Vec{0}$. A rotation $A \in SO(p+1)$ acts by $$ (\Vec{x}, \Vec{y}) \mapsto (\Vec{x}, A\Vec{y}); $$ the orbit of the point $(\Vec{x}, r, \Vec{0}) \in D$ is a $p$-sphere of radius $r$, specifically $\{\Vec{x}\} \times S^{p}(r) \subset \{\Vec{x}\} \times \Reals^{p+1}$.

Under this action, a volume element $d\Vec{x}\, dr$ of $D$ at $(\Vec{x}, r, \Vec{0})$ sweeps an infinitesimal volume $$ \Vol_{p} S^{p}(r)\, d\Vec{x} = (\Vol_{p} S^{p}) r^{p}\, d\Vec{x}. $$ The volume swept by revolving $D$ is the integral over $D$, $$ \Vol_{n}[SO(p+1)(D)] = \Vol_{p} S^{p} \int_{D} r^{p}\, d\Vec{x}. $$

This framework reduces to that of calculus if $p = m = 1$: In Cartesian coordinates $(x, r, y)$, the region $D$ lies in the half-plane $r \geq 0$ and $y = 0$; the group $SO(p + 1) = SO(2)$ revolves the $(r, y)$-plane about $\Reals^{m}$, a.k.a. the $x$-axis.

If we define $$ \bar{r}^{p} = \int_{D} r^{p}\, d\Vec{x}\bigg/\int_{D} d\Vec{x} = \frac{\Vol_{n}[SO(p+1)(D)]}{\Vol_{p} S^{p} \Vol_{m+1}(D)}, $$ then $\bar{r}$ is (by fiat, see Note at the bottom) the distance from the $p$-centroid to $\Reals^{m}$. Pappus' theorem is a formal triviality: \begin{align*} \Vol_{n}[SO(p+1)(D)] &= (\Vol_{p} S^{p}) \bar{r}^{p} \Vol_{m+1}(D) \\ &= \Vol_{p} S^{p}(\bar{r}) \Vol_{m+1}(D), \end{align*} the $p$-dimensional volume of the sphere of radius $\bar{r}$ times the $(m + 1)$-dimensional volume of $D$.


The four-dimensional cone: Here $m = 1$ and $p = 2$. The region $D$ may be taken to be the triangle $0 \leq x \leq h$ and $0 \leq r \leq Rx/h$. The $2$-centroid (really, its distance $\bar{r}$ from the axis) satisfies $$ \bar{r}^{2} = \frac{1}{\frac{1}{2}Rh} \int_{0}^{h} \int_{0}^{Rx/h} r^{2}\, dr\, dx = \frac{2R^{2}}{3h^{4}} \int_{0}^{h} x^{3}\, dx = \frac{R^{2}}{6}. $$ The sphere of radius $\bar{r}$ has area $4\pi \bar{r}^{2} = \frac{2}{3}\pi R^{2}$, so as claimed above, \begin{align*} (\text{area of generating triangle})(\text{area of sphere through $p$-centroid}) &= (\tfrac{1}{2}Rh)(\tfrac{2}{3}\pi R^{2}) \\ &= \tfrac{1}{3}\pi R^{3}h \\ &= \text{volume of the (hyper-)cone}. \end{align*}


Note: I don't see a natural mechanical interpretation of $\bar{r}$ Probabilistically, $\bar{r}^{p}$ is the mean of $r^{p}$, the $p$th power of the distance function from the "axis" (i.e., from $\Reals^{m}$). That is, $\bar{r}$ is the $p$-norm of the distance function $r$ over $D$, computed with respect to $(m + 1)$-dimensional Lebesgue measure.

  • Thank you. I'm studying your answer. Especially what is $\bar r$ and what has to do with the usual centroid of a figure (a simplex in our case). – Emilio Novati Jan 23 '22 at 21:17
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    The "$p$-centroid" of a region (such as the triangle) is a point $(\bar{\mathbf{x}}, \bar{r})$ in $\mathbf{R}^{m+1}$; that is, $\bar{r}$ is the distance from the $p$-centroid to the "axis" $\mathbf{R}^{m}$. When $p = 1$ we recover the ordinary centroid: The region being revolved lies in the $(x, r)$-plane, $\bar{r}$ is the $r$-component of the usual centroid. <> In a sense this answer effectively uses Pappus's theorem to define the $p$-centroid, but that also seems to be what Sommerville does. – Andrew D. Hwang Jan 23 '22 at 22:37