Lebesgue Volume of a closed Euclidean ball

Question

I am reading a paper, and it defines $$\mathbf{B}_\delta(\mathbf{x}):= \{\mathbf{x}: \|\mathbf{x}\|\leq \delta\}$$ as the closed Euclidean ball of radius $\delta$ and centered at $\mathbf{x}$. And it says its Lebesgue volume $\mbox{vol}(\mathbf{B}_\delta(\mathbf{x}))$ satisfies $$\mbox{vol}(\mathbf{B}_\delta(\mathbf{x}))=\frac{\pi^{\frac{p}{2}}}{\Gamma(\frac{p}{2}+1)}\delta^p, \ \ \ \ \forall \mathbf{x}\in \mathbb{R}^p$$

Can anyone please remind me how to obtain this volume? I have not seen this before.

Thanks!

Answer 1

I have a short lecture note about this question at this link

Unfortunately it is in italian but may be it can be still useful

Answer 2

It is geometrically obvious that the ball $B^n(r)\subset{\mathbb R}^n$ has volume $\kappa_n\>r^n$ for some constant $\kappa_n$. We know that $\kappa_1=2$ and $\kappa_2=\pi$. We now have to establish a recursion for the $\kappa_n$ $(n\geq3)$.

Write the points of ${\mathbb R}^n$ in the form $(x,y,{\bf z})$ with $x$, $y\in{\mathbb R}$ and ${\bf z}\in{\mathbb R}^{n-2}$. We write $x^2+y^2=:r^2$, and then have $$B^n(1)=\bigl\{(x,y,{\bf z})\in{\mathbb R}^n\bigm| x^2+y^2+|{\bf z}|^2\leq1\bigr\}=\bigl\{(x,y,{\bf z})\in{\mathbb R}^n\bigm| 0\leq r\leq1, \>|{\bf z}|\leq\sqrt{1-r^2}\bigr\}\ .\tag{1}$$ Write $D$ for the unit disc in the $(x,y)$-plane. According to $(1)$ and Fubini's theorem we then can write $$\eqalign{\kappa_n&={\rm vol}(B^n(1))=\int_{B^n(1)}1\>{\rm d}(x,y,{\bf z})=\int_D\int_{B^{n-2}(\sqrt{1-r^2})}1\>{\rm d}({\bf z})\>{\rm d}(x,y)\cr &=\int_D \kappa_{n-2}(1-r^2)^{(n-2)/2}\>{\rm d}(x,y)=\kappa_{n-2}\>2\pi\int_0^1(1-r^2)^{(n-2)/2}\>r\>dr\ .\cr}$$ The last integral is elementary, and leads to the recursion $$\kappa_n={2\pi\over n}\kappa_{n-2}\qquad(n\geq3)\ ,$$ from which your formula follows.