I'm searching for a formula for the volume (and, if available, also other information like area, etc.) of a so-called "hyperbolic cube":

Hyperbolic Cube

I couldn't find anything in Wikipedia, and also MathWorld's "Hyperbolic Cube" entry has no information about these solids apart from the picture.

Clarification by @Blue.

The goal is to find analogies for (what MathWorld calls) the "hyperbolic octahedron" (aka, a symmetric "astroidal ellipsoid"), with Cartesian equation $x^{2/3}+y^{2/3}+z^{2/3}=1$. That solid has volume 0.359038 (with "apparently" no known exact expression) and surface area $17\pi/12$.

OP wants the Euclidean volume (and surface area, etc) of a pointy-cornered, curvy-edged solid in Euclidean space, not the hyperbolic volume of the corresponding solid in hyperbolic space.

(The use of the hyperbolic-geometry tag in the original version of this question was in error. But then, MathWorld's use of "hyperbolic" to describe these solids is somewhat misleading. I've edited the question and title to (hopefully) avoid further confusion with this terminology.)

  • 2
    $\begingroup$ A quick web search leads to a Conformal Geometry and Dynamics journal article "Volume Formulae for Regular Hyperbolic Cubes" (PDF link). See Theorem 1, using $n=3$. (Volume in hyperbolic geometry is complicated.) $\endgroup$
    – Blue
    Dec 17, 2019 at 21:49
  • 2
    $\begingroup$ @Blue You should make that an answer. $\endgroup$
    – Lee Mosher
    Dec 18, 2019 at 2:42

2 Answers 2


What shape is it?

Before we can do calculations, we need to know the exact shape we're dealing with. Ideally, this would include an explanation as to why it's called "Hyperbolic Polyhedron" in the first place.


Firstly, I visited the MathWorld page for this "Hyperbolic Cube", and downloaded the Mathematica notebook file. It doesn't have much except for deprecated code for an approximate 3D model attributed to Michael Trott, and the image in the article with the name "Rivin" next to it. This almost certainly refers to our very own Igor Rivin.

I did a web search for Michael Trott hyperbolic polyhedron, hoping to find the deprecated code. Instead I stumbled upon an answer by Igor Rivin on Mathematica StackExchange, explaining how he created an analogous dodecahedron.

Hyperbolic geometry relation

Basically, inscribe a regular Euclidean cube (or whichever polyhedron) in a sphere, and interpret that as a sort of "ideal cube" (with vertices at infinity) in the Beltrami-Klein model of 3D hyperbolic space. Then represent the same shape in the Poincaré ball model, and you get the desired object.

Euclidean explanation

Alternatively, so that you don't need familiarity with models of hyperbolic space: Take a cube inscribed in a sphere (bounding a ball). And for each face, consider the (large) sphere passing through its vertices that also intersects the original sphere at right angles. The left-over region of the original ball that is not in any of the 6 large spheres is the "hyperbolic cube".


The hyperbolic cube can be seen in these images I made with Mathematica:

hyperbolic cube inscribed in cube inscribed in sphere hyperbolic cube alone



As Igor Rivin mentioned, there is a tidy formula on Wikipedia we can use to transform the points of the original cube to the hyperbolic cube. Suppose the bounding sphere has radius 1 and is centered at the origin. Then one inscribed cube would be $\left\{(x,y,z)\left||x|,|y|,|z|\le\dfrac{1}{\sqrt{3}}\right.\right\}$. The corresponding hyperbolic cube is then $\left\{\dfrac{(x,y,z)}{1+\sqrt{1-\rho^2}}\left||x|,|y|,|z|\le\dfrac{1}{\sqrt{3}}\right.\right\}$ where $\rho^2=x^2+y^2+z^2$.


The perimeter of the hyperbolic cube is 12 times the arc length of one circular arc "edge", or 24 times the arc length of half of an "edge".

A half-edge is parametrized by $\left\{\left.\dfrac{\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},t\right)}{1+\sqrt{\frac13-t^2}}\right|0\le t\le\dfrac{1}{\sqrt{3}}\right\}$. Using the formula for the length of a parametized curve from multivariable calculus, we find that the length of a half-edge is $$\int_0^{1/\sqrt{3}}\sqrt{\dfrac{9t^2-12+6\sqrt{3-9t^2}}{27(t^4+t^6)-4}}\,\mathrm dt\text{.}$$ This isn't tidy enough for Mathematica to produce an exact value, but it suggests a certain trigonometric substitution: $t=\dfrac{1}{\sqrt{3}}\sin u$, so $\mathrm dt=\dfrac{1}{\sqrt{3}}\cos u\,\mathrm du$. For the indefinite integral, we now have: $$\int \cos u\sqrt{\dfrac{2\sqrt{3}\cos u+\sin^2 u-4}{3\sin^4 u+\sin^6 u-4}}\,\mathrm du\text{.}$$ This is something Mathematica can handle, and it produces something equal to: $$\dfrac{(\sqrt{3}-1)\arctan\left(\dfrac{\tan(\frac{u}{2})}{\sqrt{2+\sqrt{3}}}\right)\sqrt{(7-4\sqrt{3}\cos u+\cos(2u))(2-\sqrt{3})}}{3-\sqrt{3}-(\sqrt{3}-1)\cos u}$$ This is $0$ at $u=t=0$, so we just need to evaluate it at $u=\arcsin(1)$ (corresponding to $t=\dfrac{1}{\sqrt{3}}$). This can be written as $\dfrac{2(\mathrm{arccot}\sqrt{2+\sqrt{3}})\sqrt{6-3\sqrt{3}}}{3-\sqrt{3}}$, which simplifies to $\sqrt{2}\,\mathrm{arccot}\sqrt{2+\sqrt{3}}$ or $\sqrt{2}\arctan\sqrt{2-\sqrt{3}}\approx0.675511$. So the total perimeter is $24\sqrt{2}\arctan\sqrt{2-\sqrt{3}}\approx16.2123$

Surface Area

The surface area is 6 times the area of one "face", or 48 times the area of one eighth of a face, because of all of the symmetry. Using a parametrization, we can use a similar forumla for surface area. However, after similar steps to the perimeter, I was not able to wrangle the outer integral into a form Mathematica could evaluate exactly.

The surface area of the whole hyperbolic cube seems to be about $4.025$ (maybe as precise as $4.02530846936$?). If that more precise estimate is accurate, then the surface area is not a rational multiple of $\pi$ with a relatively small denominator (as claimed for the hyperbolic octahedron), since $\dfrac{132}{103}\pi$ and $\dfrac{41}{32}\pi$ are too far off.


I have not even tried to calculate the volume exactly, but Mathematica suggests it is about $0.350578$. At first, I thought that seemed low since the original cube has volume $\dfrac{8}{3\sqrt{3}}\approx1.5396$. But the volume of a (euclidean) cube with side length half of the original is $\dfrac{1}{3\sqrt{3}}\approx0.19245$. You can see it inside of the hyperbolic cube in this image:

small cube inside of hyperbolic cube inside of the original cube

  • 1
    $\begingroup$ +1. I abandoned this question due to uncertainty about the figure, and lack of guidance from OP. A source of my confusion is that MathWorld's "hyperbolic octahedron" is not an ideal Poincaré octahedron; as noted in my edit to the question, it's an "astroidal ellipsoid" with equation $x^{2/3}+y^{2/3}+z^{2/3}=1$. This led me to wonder if the "hyperbolic cube" might also be something else. I guess your Rivin-sleuthing solves that mystery, but it shows that MathWorld isn't consistent about what it defines as a "hyperbolic" polyhedron. $\endgroup$
    – Blue
    Feb 1, 2020 at 23:28
  • $\begingroup$ @Blue I suspect, but it will take time to check, that the image at the top of the hyperbolic octahedron page is from Rivin making an ideal Poincaré octahedron for Mathematica 2, but that someone since then (Eric W. Weisstein in 2014?) conflated that with the differently-curved "astroidal ellipsoid", so that the page is extra confusing. $\endgroup$
    – Mark S.
    Feb 1, 2020 at 23:40
  • $\begingroup$ "... so that the page is extra confusing." ... Indeed. You'd think Wolfram MathWorld get this entire topic straight, considering the connection with the Wolfram Mathematica logo. :) $\endgroup$
    – Blue
    Feb 1, 2020 at 23:45

Volume in hyperbolic geometry is complicated.

A quick web search leads to a 1998 Conformal Geometry and Dynamics journal article "Volume Formulae for Regular Hyperbolic Cubes" (PDF link via ams.org) by T. H. Marshall. From page 26:

Let $C_n(\lambda)$ be the regular hyperbolic $n$-cube, represented in the Klein model by a cube centred at the origin with Euclidean edge length $2\lambda/\sqrt{n}$. The parameter $\lambda$ thus lies in $(0,1]$ and $\lambda = 1$ gives the ideal regular cube. In hyperbolic terms, $\lambda = (\tanh d)\sqrt{n}$, where $d$ is the hyperbolic distance from the centre of $C_n(\lambda)$ to the centre of any of its faces. [...]

Theorem 1. $$\text{Volume}(C_n(\lambda)) = \frac{2^{n+1}\sqrt{n}}{\lambda \Gamma(\frac{n+1}{2})}\;\int_0^\infty \left(\;\exp(-u^2/\lambda^2)\;h(u)\;\right)^n\;du,$$ where $$h(u) =\int_0^u \exp(x^2)\;dx$$

For $n=3$, the formula (in terms of distance $d$ as described above) reduces to

$$V = 16\coth d\;\int_0^\infty \;\exp(-u^2\coth^2d)\;\left(h(u)\right)^3\;du$$

Of course, this assumes you're looking for the hyperbolic volume of a hyperbolic cube in hyperbolic space. If you just want the "regular" volume of a curvy, pointy-cornered cube-like object in Euclidean space, that's a whole other thing. To get at that, though, we'd need to know exactly how the curves and pointy ends are determined; perhaps with the Poincaré ball model, like this icosahedron (without the golden honeycomb):

enter image description here

(image credit: Claudio Rocchini CC BY-SA 3.0; via Wikimedia Commons)

  • $\begingroup$ I don‘t understand what you mean with hyperbolic volume in hyperbolic space. I was expecting formulas in analogy to the ones given at mathworld.wolfram.com/HyperbolicOctahedron.html. $\endgroup$
    – asmaier
    Dec 24, 2019 at 13:24
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    $\begingroup$ You included the hyperbolic-geometry tag with your question, so I thought you wanted an answer within hyperbolic geometry ... although the last section of my answer expresses a certain suspicion in this regard. Now that I know that you really do want the "regular" volume of a pointy-cornered cube-like object in Euclidean space, I've taken the liberty of updating your question to make that clear. $\endgroup$
    – Blue
    Dec 24, 2019 at 18:19
  • $\begingroup$ I believe I found evidence that your suspicion was correct: the Poincaré ball model was the source of the curves and pointy ends. Briefly, the Mathematica notebook on the page for "hyperbolic cube" suggests that Igor Rivin was a part of making these shapes, and on Mathematica SE he explains how the Poincaré ball model is used. My answer here contains a bit more detail. $\endgroup$
    – Mark S.
    Feb 1, 2020 at 22:34

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