How do I evaluate the Lebesgue measure of a ball? I remember I saw a post related to this question somewhere here, but I cannot find this now.
How do I evaluate the Lebesgue measure of a ball?
What I only remember is that the proof used the gamma function.
 A: Here's an alternative approach which I found some time ago in those lecture notes by Govind Menon. It is entirely based on the formula of integration in spherical coordinates:
\begin{equation}\tag{Sph}
\int_{\mathbb{R}^n} f(x)\, dx= \int_0^\infty \left( \int_{S(r)} f\, dS\right)\, dr, 
\end{equation}
where $S(r)$ denotes the surface of the ball centered at the origin and having radius $r$. We denote the latter ball by $B(r)$.
We begin by applying formula (Sph) to the integral expression for the volume of $B(R)$: 
\begin{equation}
\begin{split}
\mathrm{vol}(B(R))&=\int_{\mathbb{R}^n} \mathbf{1}_{(\lvert x\rvert \le R)}\, dx \\
&=\int_0^R \lvert S(r) \rvert \, dr, 
\end{split}
\end{equation}
where $\lvert S(r)\rvert$ denotes the surface area of $S(r)$. This computation shows that it is enough for us to compute $\lvert S(r)\rvert$, since the sought volume $\mathrm{vol}(B(R))$ is just its antiderivative. To compute $\lvert S(r)\rvert$ we will evaluate the $n$-dimensional Gaussian integral 
\begin{equation}
I_n=\int_{\mathbb{R}^n} e^{-\frac{x_1^2+\ldots + x_n^2}{2}}\,dx_1\ldots dx_n
\end{equation}
in two different ways. First, using the value of $I_1=\sqrt{2\pi}$ (this we take as granted), and factorizing the integral defining $I_n$ we obtain
\begin{equation}\tag{1}
I_n=\prod_{j=1}^n \int_{-\infty}^\infty e^{- \frac{x_j^2}{2}}\, dx_j=(2\pi)^{\frac{n}{2}}.
\end{equation}
Second, applying formula (Sph), we find that
\begin{equation}\tag{2}
\begin{split}
I_n &=\int_0^\infty \left(\int_{S(r)} e^{-\frac{\lvert x \rvert^2}{2}}\, dS \right)\, dr\\
&=\int_0^\infty \left(\int_{S(r)} e^{-\frac{r^2}{2}}\, dS\right)\, dr \qquad\left(\text{because }\lvert x \rvert=r\text{ on }S(r)\right) \\
&=\int_0^\infty e^{-\frac{r^2}{2}}\left(\int_{S(r)} \, dS\right)\, dr \\
&=\int_0^\infty \lvert S(r)\rvert e^{-\frac{r^2}{2}}\, dr.
\end{split}
\end{equation}
Now we observe that 
\begin{equation}
\lvert S(r)\rvert= r^{n-1}\lvert S(1)\rvert;
\end{equation}
this is due to the fact that $S(r)$ can be obtained by dilating $S(1)$ by a factor of $r$, which scales its $n-1$-dimensional area by a factor of $r^{n-1}$. Inserting this into (2) we arrive at 
\begin{equation}\tag{3}
\begin{split}
I_n&= \lvert S(1)\rvert \int_0^\infty r^{n-1} e^{-\frac{r^2}{2}}\, dr \\ 
&=\lvert S(1)\rvert \int_0^\infty (2t)^{\frac{n}{2}-1} e^{-t}\, dt \qquad \left(\text{change variables }t=\frac{r^2}{2}\right)\\
&=2^{\frac{n}{2}-1}\lvert S(1) \rvert \Gamma\left(\frac{n}{2}\right).
\end{split}
\end{equation} 
Equating (1) and (3) we get 
\begin{equation}
\lvert S(1)\rvert =2\frac{\pi^{\frac{n}{2}} }{\Gamma\left(\frac{n}{2}\right)},
\end{equation}
so we can conclude that 
\begin{equation}
\begin{array}{cc}
\lvert S(R)\rvert= \frac{2 (\pi)^{\frac{n}{2}} }{\Gamma\left(\frac{n}{2}\right)}R^{n-1}, & 
\text{vol}(B(R))=\frac{2 (\pi)^{\frac{n}{2}} }{  n  \Gamma\left(\frac{n}{2}\right)} R^{n}. 
\end{array}
\end{equation}
P.S.: This result coincides with Christian Blatter's one, because of the following property of the Gamma function: 
\begin{equation}
\Gamma\left(\frac{n}{2}+1\right) = \frac{n}{2}\Gamma\left(\frac{n}{2}\right).
\end{equation}
A: Denote the volume of the $n$-dimensional unit ball $B(n,1)$ by $\kappa_n$.  The ball of radius $r\geq0$  then has volume ${\rm vol}_n\bigl(B(n,r)\bigr)=\kappa_n r^n$. 
In order to obtain a recursion formula for the $\kappa_n$ we write the points of ${\mathbb R}^n$ in the form
$(x,y,{\bf z})$ with $(x,y)\in{\mathbb R}^2$, $\>{\bf z}\in{\mathbb R}^{n-2}$.
Fubini's theorem, applied to the product ${\mathbb R}^2\times{\mathbb R}^{n-2}$, then gives
$$\kappa_n={\rm vol}_n\bigl(B(n,1)\bigr)=\int\nolimits_{B(n,1)}1\ {\rm d}(x,y,{\bf z})=
\int\nolimits_{B(2,1)}\left(\int\nolimits_{B(n-2,\sqrt{1-x^2-y^2})}1\ {\rm d}({\bf z})\right)\ {\rm d}(x,y)\ .$$
Here we have used that $(x,y)$ is restricted to $x^2+z^2\leq1$, and that for given $(x,y)$ the variable ${\bf z}$ is restricted to $|{\bf z}|^2\leq 1-x^2-y^2$. It follows that in the inner integral the volume of $B(n-2,\sqrt{1-x^2-y^2})$ is computed.
Introducing polar coordinates in the $(x,y)$-plane we therefore obtain
$$\kappa_n=2\pi\int_0^1 \kappa_{n-2}(1-r^2)^{(n-2)/2}\>r\ dr\ ,$$
and this leads to
$$\kappa_n={2\pi\over n}\kappa_{n-2}\ .$$
Together with $\kappa_1=2$, $\kappa_2=\pi$ it is now easy to verify that
$$\kappa_n={\pi^{n/2}\over \Gamma\left({n\over2}+1\right)}\qquad(n\geq1)\ .$$
