How do I evaluate the Lebesgue measure of a ball?

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 $$\kappa_n={\pi^{n/2}\over \Gamma\left({n\over2}+1\right)}\qquad(n\geq1)\ .$$

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}