Volume of the unit ball through areas of spheres How can the following formula be proved?
$$
|B_n| = \int_0^1 r^{n-1} \sigma_n \, dr = \frac{\sigma_n}{n}.
$$
Here, $|B_n|$ is volume of the unit ball in $\mathbb R^n$ and $\sigma_n$ is area of the unit sphere in $\mathbb R^n$.
The expression under the integral represents area of the sphere with radius $r$.
 A: The surface area of any $n$ dimensional sphere is proportional to $r^{n-1}$. You can get this using the surface element $ds=r^{n-1}d\Omega_n$, where $\Omega_n$ is the $n$-dimensional solid angle.  Let's say the constant of proportionality is $C_n$. You can consider the sphere made out of infinitesimal spherical shells of thickness $dr$, so $$|B_n|=\int_0^1C_nr^{n-1}dr=C_n\frac{r^{n}}{n}\Bigg|_0^1=\frac{C_n}{n}=\frac{\sigma_n}{n}$$
A: Apply the divergence theorem on the unit ball with the identity vector field. The integral on the sphere will be the integral of the constant function $1$, and the integral in the ball will be the integral of the divergence of such field, namely the constant function $n$.
Explicitly, let $V:\mathbb{R}^n \to \mathbb{R}^n$ be given by $V(x)=x$. Then $\mathrm{div}V=n$. Since the outer normal vector field $\eta$ on the unit sphere coincides with the restriction of $V$, we have that $\langle V , \eta \rangle=\langle \eta, \eta \rangle=1$ . It follows that
$$\int_B n= \int_S 1 $$
$$ \implies n\mathrm{vol}(B)=\mathrm{vol}(S).$$
