3
$\begingroup$

We were shown in class this next calculation: (Here, $V_n(RB^n)$ is the volume of an $n$ dimensional ball of radius $R$, likewise $S_{n-1}$ is the surface area of the $n$ dimensional sphere in $\mathbb{R}^n$. $rS^{n-1}$ denotes the $n$ dimensional sphere of radius $r$ and integrating $d\textbf{S}$ means a surface integral.) $V_n(RB^n)=\int_{RB^n}1dx=\int_0^R\int_{rS^{n-1}}1d\textbf{S}dr=\int_0^R\int_{S^{n-1}}r^{n-1}d\textbf{S}dr=$$=\int_0^Rr^{n-1}\int_{S^{n-1}}1d\textbf{S}dr=\int_0^Rr^{n-1}S_{n-1}dr=\frac{R^n}{n}S_{n-1}$ and finally $V_n=\frac{1}{n}S_{n-1}$ since $V_n(RB^n)=R^nV_n$. My problem is with the 3rd equality. The first is obvious and the second is the coarea formula. I assume the third equality is a result of a change of variables, but since this is taking place in $\mathbb{R}^n$ I'd expect the change of variables to be $x\mapsto rx$ which gives the Jacobian of $r^n$ - not the $r^{n-1}$ we see after the third equality.

It'd be easier for me to assume the teacher had a mistake here, had she not used this result later on in her lectures... So my question is, was she wrong in the change of variables there or am I missing something about surface integrals?

  • 1
    This wikipedia article on [n-Sphere](http://en.wikipedia.org/wiki/N-sphere) is highly relevant. Check hyperspherical volume element section.2011-10-07

2 Answers 2

3

The third equality comes from the fact that the map $f:\quad{\mathbb R}^n\to{\mathbb R}^n, \quad u\mapsto x:=r\>u$ ($r$ is constant here) multiplies volume elements by its Jacobian $r^n$ but multiplies $(n-1)$-dimensional surface elements by $r^{n-1}$.

While we are at it: The set $B^n:=\{x\in{\mathbb R}^n\ |\ |x|<1\}$ is the ($n$-dimensional) unit ball in ${\mathbb R}^n$. On the other hand its boundary $\{x\in{\mathbb R}^n\ |\ |x|=1\}$ is not the $n$-dimensional unit sphere, but the $(n-1)$-dimensional unit sphere $S^{n-1}$. The latter has an $(n-1)$-dimensional surface area which you might call $\omega(S^{n-1})$ or similar, but certainly not $S_n$. Therefore the proper way to write your formula would be ${\rm vol}(R\>B^n)={R^n\over n}\omega(S^{n-1})\ .$

  • 0
    @Donjim: Think of a small patch, say $1$"$\times1$" on a ball $100$" in radius (in $\mathbb{R}^3$). If that ball were blown up to $200$" in radius, the patch would be $2$"$\times2$" and have $4\times$ the area. If you were thinking of a chunk of the ball $1$"$\times1$"$\times1$" just under the surface, then that would become $2$"$\times2$"$\times2$" and have $8\times$ the volume.2011-10-07
1

As an example, if we let $n=3$, the area of a two-sphere is proportional to $r^2$, not $r^3$. You haven't changed the $dr$ integral, which still goes from $0$ to $R$.

  • 0
    Indeed, the surface area of the sphere in $\mathbb{R}^3$ is $4\pi r^2$. (+1)2011-10-07