# Technical question regarding using the coarea formula to calculate the relation between the $n$-ball's volume and the $(n-1)$-sphere volume

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?

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This wikipedia article on n-Sphere is highly relevant. Check hyperspherical volume element section. – Sasha Oct 7 '11 at 14:07

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 $${\rm vol}(R\>B^n)={R^n\over n}\omega(S^{n-1})\ .$$

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yeah I noticed the mistake in notation and just edited that in. I know that the boundary of the $n$-ball is a $(n-1)$ manifold but I thought it's still a subset of $\mathbb{R}^n$ and thus the appropriate change of variables would have Jacobian $r^n$. In the calculation above the change of variables takes place before the surface element is "resolved" in order to obtain $\int_{S^{n-1}}d\textbf{S}$ which is replaced by $S_{n-1}$. – Donjim Oct 7 '11 at 16:02
Could you please explain why would the transformation you've mentioned for a change of variables would affect $(n-1)$ surfaces differently than $n$ surfaces? I mean, the change of variables formula is invariant to the dimension of the surface: $\int_{T({\Omega})}f = \int_{\Omega} (f\circ T) | J_T|$ – Donjim Oct 7 '11 at 16:12
@Donjim: just because an $n-1$ manifold is imbedded in an $n$ dimensional space doesn't mean its area becomes $n$ dimensional. The map from the unit sphere in $\mathbb{R}^n$ taking $u\to ru$ only multiplies the $n-1$ dimensions of the sphere, not the $n$ dimensions of the ball contained in the sphere. That is why the volume element is $\mathrm{d}\sigma\;\mathrm{d}r$, you get $n-1$ dimensions in the $\mathrm{d}\sigma$ and one more in $\mathrm{d}r$. – robjohn Oct 7 '11 at 16:35
@robjohn - Basically what you're saying is that because this is a $(n-1)$ manifold I have a parametrization of it and then I would apply the change of variables in $(n-1)$ dimensional space which would result in a surface element of $r^{n-1}$. Am I correct in understanding here? – Donjim Oct 7 '11 at 17:03
@Donjim: it transforms as its dimension when scaled, and $S^{n-1}$ has $n-1$ dimensions. The extra dimension of volume is perpendicular to the surface: its thickness. – robjohn Oct 7 '11 at 17:36

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 $R$.

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Indeed, the surface area of the sphere in $\mathbb{R}^3$ is $4\pi r^2$. (+1) – robjohn Oct 7 '11 at 14:46