# How to interpret the power of $\pi$ in the expression of the volume of an n-dimensional ball?

I'm now learning multiple integrals. I reached this formula through computation of multiple integrals: $$V_{2n}(R)=\frac{\pi^n}{n!~}\cdot R^{2n}, \quad V_{2n+1}(R)=\frac{2^{n+1}\pi^n}{(2n+1)!!~}\cdot R^{2n+1}.$$ Here $V_m(R)$ denotes the volume of a $m$-dimensional ball of radius $R$.

It looks quite strange that $V_{m+2}(R)$ has one more $\pi$ than $V_m(R)$, while the pattern of degree of $\pi$ are not the same between odd dimensions and even dimensions.

Question: How to interpret the power of $\pi$ occur in the formula above?

Geometric interpretations could be nice, since higher dimensional spaces is difficult for me to imagine.

Thank you !

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Take a look at this. I am not sure if the power of $\pi$ is important here since there are different multipliers in the front. – Patrick Li Jan 6 '13 at 6:48
I'd say it has to do with the rank of the groups $SO(2n)$ and $SO(2n+1)$ both of which is $n$. This roughly says that there are $n$ "independent rotations" in both $2n$ and $2n + 1$ dimensional space since for each one you need to pick a plane where the rotation occurs. The rotation in this one plane then corresponds to a circle, which is where the $\pi$ comes from. But take this with a lot more than a grain of salt! :) – Marek Jan 6 '13 at 8:30
@Marek Thank you very very much! I'm really interested in your comment, also looking forward to a more detailed answer:) – rhenskyyy Jan 9 '13 at 9:30

The formula is actually quite symmetric for even and odd dimensions. The Jacobian in $n$-dimensions is: $$J_n = r^{n-1}\prod_{k=1}^{n-2}\sin^k \phi_{n-k-1}$$ And the resultant volume is also symmetric in odd and even powers of $n$: $$V_n \sim \frac{\pi^{n/2}}{\Gamma(n/2+1)}$$ So in essence the difference in odd and even volumes comes from the insistence to convert a perfectly friendly gamma function into factorials, thus leading to an "extra" factor of $\sqrt{\pi}$. Of course you may intuitively expect there to be whole powers of $\pi$ since that is what we are used to in low dimensions, but I think we should just get used to the fact that the familiar $\pi$ is just a reduction of the "real $\pi$" which involves the Gamma function.
I don't agree that we should treat the odd and even cases uniformly. There is no sense in which the factor jumps by $\sqrt\pi$ in every dimension. Rather, it only jumps by $\pi$ every two dimensions. There is just a whole world of difference between the odd and even case. E.g. you can't comb the hair on $S^2$ (Hairy ball theorem) but there's no problem doing it on $S^3$. This essentially boils down to Euler characteristic which is $\chi(S^n) = 1 + (-1)^n$. Similarly, observe that antipodal map ($x \mapsto -x$) is not a proper rotation only for odd $n$. – Marek Jan 9 '13 at 11:27