# Surface area of cap of hypersphere

Let $$S = \mathbb{S}^{n-1}$$, where this denotes $$\mathbb{S}^{n-1} = \left\{ x \in \mathbb{R}^n: \sum_{i=1}^n x_i^2 = 1 \right\}.$$

Let $$x_0 \in S$$ be fixed. I want to find the area of the region $$R_n = \left \{ x \in S: ||x - x_0||_2 \leq r_n \right \}.$$ More specifically, I would like to know the ratio of the area of this region over the total area of $$S$$, i.e., $$|R_n|/|S|$$. If this is too difficult, a lower bound (for example, an estimate of the order of magnitude in terms of $$r_n$$) would be great.

I suspect that the this area should be $$O(r_n^{n-1}),$$ at least when $$r_n$$ is relatively small. If you think about when $$n=2$$ and you have a circle, then this region is locally similar to a line segment of length $$r_n$$. If you have $$n=3$$ and you're dealing with a spherical shell, then this region is locally a circle of radius $$r_n$$, and the area should be approximately equal to $$\pi r_n^2$$. Any references or more precise computation would be great!

• The integral looks doable in general, though I haven't written out the details. The volume does asymptotically approach the volume of an $(n-1)$-ball of radius $r_{n}$ as $r_{n}\to0$. <> Interestingly, when $n = 3$ the area is exactly $\pi r^{2}$ for $0 \leq r \leq 2$ (!!), see for example here. Sep 19 at 12:14