What is the volume of the sphere in hyperbolic space?

I'm looking for a formula to describe surface and volume of a sphere in hyperbolic 3-space. I found some results which were generalized for any dimension, but I wasn't able to understand them.

Overall, I find there's a sort of "symmetry" between spherical and hyperbolical geometry in these formulas. The circumference of circle in hyperbolic geometry is $2\pi\sinh(r)$, in spherical geometry it's $2\pi\sin(r)$. Areas are $2\pi(\cosh(r) - 1)$ and $2\pi(1 - \cos(r))$, respectively.

I then tried to generalize it into 3D. I was trying to find surface and volume of sphere in spherical space (considered as a cap of Euclidean 4D hypersphere), and eventually obtained formulas that seem to work: $2\pi(1 - \cos(2r))$ for surface and $2\pi(r - \sin(2r)/2)$ for volume. Considering the analogy between formula for area of circle and for surface of sphere, the hyperbolic formulas should be $2\pi(\cosh(2r) - 1)$ for surface and $2\pi(\sinh(2r)/2 - r)$ for volume, but this is based mostly on intuition, so I'd appreciate if you could either confirm my line of thought or give me the real result :)

(Incidentally, what is the formula for length of an equidistant curve? I used the same analogy between goniometric and hyperbolic functions to solve it for sphere: if I have a certain length a of straight line, I can construct the perpendicular lines at the endpoints which will then intersect the equidistant curve (circle on sphere, hypercycle in hyperbolic geometry) under right angles as well. If b is the distance of the equidistant from the line, the length of such defined arc of equidistant is acos(b) on sphere, so it should be acosh(b) in hyperbolic geometry -- is this correct? Finally, what would be the formula for a part of surface equidistant from plane directly "above" a part of the plane of known volume?)

• Please consider typing the math equations/symbols in LaTeX to make it more readable. Check: meta.math.stackexchange.com/questions/5020/… Commented Sep 21, 2015 at 16:30
• From en.wikipedia.org/wiki/Talk:Hyperbolic_space (still needs to be included on the main article ) The surface area of a sphere is $4\pi R^2 \sinh^2 \frac{r}{R}$ The volume of the enclosed ball is $\pi R^3 \sinh \frac{2r}{R} - 2\pi R^2r$ (references are missing) also see math.stackexchange.com/questions/1208291/length-of-a-hypercycle (an equidistant line is an hypercycle) Commented Sep 22, 2015 at 6:38
• I see. I assumed that R = 1 (in other words, that r is expressed in absolute units), but otherwise the results are the same as what I got because sinh(x)^2 = (cosh(2x) - 1)/2. So the formulation is more precise, but my idea was basically correct. Good to know. Commented Sep 22, 2015 at 6:46
• Thanks for the hypercycle link, but I notice that the question was never answered. Commented Sep 22, 2015 at 6:54

To confirm the OP's conjectures: A hyperbolic circle of hyperbolic radius $r$ has circumference $2\pi R \sinh \frac rR$ (where $R$ is the absolute length). Consequently, in hyperbolic $(n + 1)$-space an $n$-sphere of radius $r$ is isometric to a Euclidean $n$-sphere of Euclidean radius $R \sinh \frac rR$. Particularly, the area of a sphere of hyperbolic radius $r$ in hyperbolic $3$-space is $$4\pi R^2\sinh^{2} \frac rR = 2\pi R^2\bigl(\cosh (\frac {2r} R) - 1\bigr),$$ and the volume of a hyperbolic $3$-ball of radius $r$ is $$\int_{0}^{r} 2\pi R^2\bigl(\cosh(\frac {2t} R) - 1\bigr)\, dt = \pi R^2\bigl(R\sinh(\frac {2r} R) - 2r\bigr).$$

• Thanks, since the time I asked this question, I came to understand why. Any formula for surface or volume of sphere in any dimension can be rewritten to use the circumference instead of radius. And circumference of sphere is a value that doesn't depend on geometry - surface of a sphere still looks the same whether the sphere itself is embedded in spherical, Euclidean or hyperbolic space, so this set of formulas would be universal and you can just reverse-plug the circumference/radius relation to get the result for a specific geometry. Commented Jun 4, 2017 at 6:51
• What is "the absolute distance" $R$ for a space form of negative curvature $K$? Is it $\frac 1 {\sqrt[n] {-K}}$? Commented Jun 5, 2018 at 18:00
• @Tiana: I know, I was asking about its value in terms of the curvature. Commented Apr 6, 2019 at 16:51
• @AlexM. Just saw your question. You may have figured this out long ago, but $R = 1/\sqrt{-K}$, the square root regardless of the dimension. Commented Aug 29, 2021 at 11:51
• @AaronFranke The $R$s and the phrase "absolute length" were added in an edit (not by me), but it appears $R = 1/\sqrt{-K}$. Commented Mar 16 at 19:54

The volume of a sphere or a ball in hyperbolic $$n$$-space with sectional curvature $$\kappa$$ is given by $$V_\kappa(r)=\mathbf{c}_{n-1} \int_0^r \left(\frac{\sinh(\sqrt{\kappa} t)}{\sqrt{\kappa}}\right)^{n-1} \, \mathrm{d}t,$$ where $$\mathbf{c}_{n-1}:=\frac{2\pi^{n/2}}{\Gamma(n/2)}$$ is the $$n-1$$-dimensional area of a unit sphere in $$\mathbb{R}^n$$ (see Chavel (2006), Riemannian Geometry: A Modern Introduction, equation (III.4.1) with notation from (III.3.10) and (II.5.8)).

• Since the space is hyperbolic, $\kappa$ is negative, so in the formula it should be $-\kappa$ under the square root. Commented Jun 24, 2022 at 6:04

It is easy to see that the volume of a sphere of radius $$s$$ in $${\mathbb H}^n$$ (constant curvature -1) is given by $$\operatorname{vol}(S^{n-1}) \sinh^{n-1}(s)$$, where $$S^{n-1}$$ is a unit sphere in Euclidean space of dimension $$n$$. To see this observe that in Minkowski space, with $$x^2$$ given by $$x_1^2 + \cdots x_n^2 - t^2$$, the path $$x_1 = \sinh(s)$$, $$t=\cosh(s)$$ is a unit speed hyperbolic geodesic. Thus a hyperbolic sphere of radius $$s$$ centered at $$(0,\ldots,0,1)$$ projects isometrically to a sphere of radius $$\sinh(s)$$ in $$(x_1,\ldots,x_n)$$ coordinates, and in this subspace the Euclidean and Minkowski metrics agree. To get the volume of a ball of radius $$s$$, integrate with respect to $$s$$.