Hello Pierre,
You're as bad as Coxeter! I notice Coxeter even uses 'volume' to refer to nD hypervolume... and uses 'sphere' to refer to an nD hypersphere or nD hyperball...
I can't claim to be a Calculus wiz... in fact, I never took Calculus III, which is where I believe material of this sort, at least up to 3 dimensions, is usually handled. Can you demonstrate how you derived the formula inside your integral, and show how to compute the hypervolume of a nD hyperball with it?
Coxeter does speak to this subject in
Regular Polytopes, H.S.M. Coxeter, pp. 125-126. Also, if one combines Coxeter's treatment with Weisstein's work given here...
Hypersphere...
It seems Coxeter and Weisstein are both using some method analoguous to the taking of "shells" integration. That is, they find the "surface" content of the nD hypersphere, and then I believe use this as a basis to form an integral for summing all the (n-1)D hyperspherical "shells" from the center of the hypersphere to, or through, its (n-1)D surface.
Coxeter and Weisstein both state that the hypervolume of the nD hypersphere, or ball, is given by the integral:
\[
V_n = \int\limits_0^R {S_n r^{n - 1} dr = \frac{{S_n R^n }}{n}} \;\left( 1 \right)
\]
where S
n is the (n-1)D hyper-surface "content" of the nD unit hyperball.
Both do some pretty involved manipulation with the exponential function and derive this integration formula for the gamma, or generalized factorial, function:
\[
\Gamma \left( m \right) = 2\int\limits_0^\infty {e^{ - r^2 } r^{2m - 1} dr}
\]
Now, with utilizing
\(\Gamma \left( {\frac{1}{2}} \right) = \pi ^{\frac{1}{2}} \) one can derive:
\[
S_n = \frac{{2\pi ^{\frac{n}{2}} }}{{\Gamma \left( {\frac{1}{2}n} \right)}}
\]
At this point, Coxeter comes up with a recurrence formula that can be used to derive S
n. However, Weisstein comes up what I think is a more useful rule for determination of S
n, n integral, for n even and n odd only utilizing the single and
double factorial functions:
\[
S_n = \left\{ \begin{array}{l}
\frac{{2^{\frac{{n + 1}}{2}} \pi ^{\frac{{n - 1}}{2}} }}{{\left( {n - 2} \right)!!}}\;\;\;\;for\;n\;odd \\
\frac{{2\pi ^{\frac{n}{2}} }}{{\left( {\frac{1}{2}n - 1} \right)!}}\;\;for\;n\;even, \\
\end{array} \right.
\]
You can calculate the double factorial for an odd integer with the single factorial function with this formula:
\[
\left( {2n - 1} \right)!! = \frac{{\left( {2n} \right)!}}{{2^n n!}}
\]
Then, the hypervolume of a nD hyperball is found by taking the value found for S
n, found by the rule immediately above, and using it in formula (1).
So, for a 4D unit hyperball:
\[
\begin{array}{l}
S_n = \frac{{2\pi ^{\frac{n}{2}} }}{{\left( {\frac{1}{2}n - 1} \right)!}} = \frac{{2\pi ^{\frac{4}{2}} }}{{\left( {\frac{1}{2}4 - 1} \right)!}} = \frac{{2\pi ^2 }}{1} = 2\pi ^2 \\
\\
V_n = \frac{{S_n R^n }}{n} = \frac{{2\pi ^2 \cdot 1^4 }}{4} = \frac{1}{2}\pi ^2 \\
\end{array}
\]