# Why does the surface area of the hypersphere go to zero as the number of dimensions goes to infinity?

Why does the surface area of the hypersphere go to zero as the number of dimensions goes to infinity? Is there an intuitive reason?

• – Hans Lundmark Aug 3 '15 at 7:56
• Why are the surface areas any more comparable than the volumes? They're in different dimensions, as for the volumes. – joriki Aug 3 '15 at 8:04
• @joriki They're all "areas"? If I have a square in two dimensions and I rotate it so that it's now embedded in three or four or five dimensions, its area is unchanged. – Neil G Aug 3 '15 at 8:05
• If I understand that remark correctly, you seem to have a wrong idea of what "surface area of the hypersphere" means. It refers to a "hyper-area", e.g. the "surface area" of a sphere in four dimensions is a three-dimensional volume. – joriki Aug 3 '15 at 8:06
• @joriki: Ah, I see now, thanks. – Neil G Aug 3 '15 at 8:08

The gist of the idea is that the converting the term $\int_{-\infty}^\infty\int_{-\infty}^\infty...\int_{-\infty}^\infty e^{-(x_1^2+x_2^2+....x_n^2)}dx_1dx_2....dx_n$ to spherical coordinates gives $S_n\int_0^\infty e^{-r^2}r^{n-1}dr=\left( \int_{-\infty}^\infty e^{-x^2}dx \right)^n$. You may massage this until you find a gamma function in the denominator of the unit hypersphere area, which blows up as the number of dimensions goes to 0 or $\infty$.
As far as the $(n-1)$ surface area of an n-sphere, looking at the surface area in $1 \rightarrow 2\rightarrow 3$ dimensions is $2\rightarrow 2\pi \rightarrow 4\pi$. This is increasing, but slowing down as n increases (based on these three data points anyway), so it may not be too surprising that at some higher number of dimension, the surface area of the unit sphere eventually hits a max and begins to decrease thereafter (although you are now comparing regions of different dimensionality too....which is it's own field of study).
Specifically, surface area of the unit sphere in n dimensions goes as $\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}$.