Volume of a "hyperbolic cube"-shaped solid in Euclidean space?

Question

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

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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.)

Answer 1

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

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

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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):

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

Answer 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.

Investigation

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".

Images

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

Calculations

Parametrization

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 $\rho^2=x^2+y^2+z^2$.

Perimeter

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.

Volume

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: