# what is the surface area of a cap on a hypersphere?

According to mathworld, let the sphere have radius $R$, then the surface area a spherical cap of height $h$ and base radius $a$ is given by $$S=2\pi Rh=2\pi(a^2+h^2).$$

What is this value for an n-dimensional hypersphere?
If it helps simplify the problem we can assume $R=1$ and $a=0.5$.

Many thanks.

• The terms "$n$-dimensional" and "hyper<object>" have different meanings to mathematicians than to the public at large. To clarify, by "$n$-dimensional hypersphere" do you mean (1) An $n$-dimensional sphere (i.e., a sphere in $\mathbf{R}^{n+1}$), or (2) A sphere in $\mathbf{R}^{n}$? Commented Apr 17, 2017 at 11:12
• @AndrewD.Hwang thanks for the clarification, I wanted to mean a sphere in $R^n$ Commented Apr 17, 2017 at 12:02
• There are easier forms for asymptotic approximations of these values, when n is large. For a sphere of radius 1, the spherical cap of height a has volume/surface area (1-a^2)^[n/2 + o(n)].
– TMM
Commented Apr 17, 2017 at 12:12
• @TMM thank you that's very useful, do you have any idea about the asymptotic approximation of the ratio between the cap and the total surface area? Commented Apr 17, 2017 at 12:30
• According to that link, the surface area is half that: $S = \pi(a^2+h^2)$. Commented Dec 13, 2022 at 1:31

The solution requires special classes of functions, namely regularized incomplete beta functions $I_{\sin^2 \Phi} (\frac{n-1}{2}, \frac{1}{2})$ and Gamma functions. The result and its derivation can e.g. be found in the very nice article by S. Li here. Let $\Phi$ be the angle of the cap, so $a = R\sin \Phi$.
Then the surface area of the cap is $$A(\Phi) = R^{n-1} \frac{\pi^{n/2}}{\Gamma(n/2)} I_{\sin^2\Phi} (\frac{n-1}{2}, \frac{1}{2})$$
• Thanks for the answer, I just saw that paper as well, I think one crucial factor is the regularized incomplete beta functions, however after some googling I still don't have a clue what is the function like, especially how it grows wrt n with different $\phi$ values, could you explain a bit? Commented Apr 17, 2017 at 14:15
• For constant angles $\Phi \in (0, \pi/2)$, the ratio of cap surface area to total area is exponentially small. Commented Mar 29, 2019 at 21:49