# Limit of $n-1$ measure of the boundary of a sphere

The measure of a sphere of radius $R$ centered in $0_{\mathbb{R}^n}$ in $\mathbb{R}^n$ is
\begin{array}{l l}\int_{B_0(R)}dx_1\ldots dx_n & =\int_0^R\rho^{n-1}d\rho \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{n-1}{\varphi_1}d\varphi_1\ldots\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos{\varphi_{n-1}}d\varphi_{n-1}\int_{0}^{2\pi} d\theta \\ & =\omega_n \int_0^R\rho^{n-1} d\rho
\end{array}
where $\omega_n$ is the $n-1$ measure of the boundary of the sphere.

We can calculate the integral of the function $e^{-||x||^2}$ over $\mathbb R^n$ as
$$\int_{\mathbb{R}^n}e^{-||x||^2}dx_1\ldots dx_n=\int_{\mathbb R^n}e^{-x_1^2-\ldots-x_1^n}dx_1\ldots dx_n=\int_{-\infty}^{+\infty}e^{-x_1^2}dx_1\cdot \ldots \cdot \int_{-\infty}^{+\infty}e^{-x_n^2}dx_n=\pi^{\frac{n}{2}}$$
This integral can also be calculated by observing that $e^{-||x||^2}$ is a radial function: in fact
$$\int_{\mathbb R^n}e^{-||x||^2}=\omega_n\int_{\mathbb R^{n}} e^{-\rho^2}\rho^{n-1}d\rho\stackrel{\rho^2=t}{=}\frac{\omega_n}{2}\int_{\mathbb R^n}e^{-t}t^{\frac{n}{2}-1}dt=\frac 12 \omega_n \Gamma\left (\frac{n}{2}\right )$$
This yields the equality
$$\omega_n=\frac{2\pi^{\frac{n}{2}}}{\Gamma\left (\frac{n}{2}\right )}=\frac{\pi^{\frac {n-1}{2}}2^{\frac{n+1}{2}}}{n!!}$$
Clearly, we have
$$\lim_{n \to \infty}\omega_n=0$$
How can we interpret such result in a geometrical way? Is there any intuitive explanation of this fact?

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An attempt at intuitive reasoning (given high dimensional results are often counter-intuitive):

Consider a unit $n$-ball centred on the origin, inscribed in an $n$-cube of sidelength $2$. The vertices of the $n$-cube lie at distance $\sqrt{n}$ from the origin, which for large $n$ makes the radius $1$ look negligible. For large $n$ we could generously truncate one of the corners of the $n$-cube without hitting the $n$-ball.

Intuitively, those corner parts become really long, and we see that most of the $n$-content of the $n$-cube will come from those regions near the vertices. As $n$ increases we therefore expect that the $n$-ball occupies a much lower proportion of the $n$-cube.

Now that we have argued that the $n$-content goes to zero, we appeal to the fact that $n$-balls are smooth, rather than fractal, so that small $n$-content must be wrapped by small $(n-1)$-content, and so the surface measure also goes to zero.

Note that the surface content $S_n = \frac{2\pi^{n/2}}{\Gamma(n/2)}$ actually increases up to $n = 7.2…$ before dropping back towards zero. (The volume content likewise peaks at some $n$.)