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 $\mathbf{R}^{n}$? – Andrew D. Hwang Apr 17 '17 at 11:12
• @AndrewD.Hwang thanks for the clarification, I wanted to mean a sphere in $R^n$ – dontloo Apr 17 '17 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 Apr 17 '17 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? – dontloo Apr 17 '17 at 12:30

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? – dontloo Apr 17 '17 at 14:15
• For constant angles $\Phi \in (0, \pi/2)$, the ratio of cap surface area to total area is exponentially small. – Chris Jones Mar 29 '19 at 21:49