volume of the $n$-sphereVolumeOfTheNSphere2003-07-23 03:56:132006-10-18 01:40:37Derivationsphere% this is the default PlanetMath preamble. as your knowledge
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\def\R{\mathbb{R}}The volume contained inside $S^n$, the $n$-sphere (or hypersphere), is
given by the integral
\[
V(n) = \int_{\sum_{i=1}^{n+1}x_i^2\le1} d^{n+1} x.
\]
Going to polar coordinates ($r^2=\sum_{i=1}^{n+1}x_i^2$) this becomes
\[
V(n) = \int_{S^n} d\Omega \int_0^1 r^{n}\, dr.
\]
The first integral is the integral over all solid angles subtended by the
sphere and is equal to its area
$A(n)=\frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}$,
where $\Gamma(x)$ is the gamma function.
The second integral is elementary and evaluates to
$\int_0^1 r^{n}\, dr = 1/(n+1)$.
Finally, the volume is
\[
V(n) = \frac{\pi^{\frac{n+1}{2}}}{\frac{n+1}{2}\Gamma\left(\frac{n+1}{2}\right)}
= \frac{\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+3}{2}\right)}.
\]
If the sphere has radius $R$ instead of $1$, then the correct volume is
$V(n)R^{n+1}$.
Note that this formula works for $n\ge0$. The first few cases are
\begin{itemize}
\item[$n=0$] $\Gamma(3/2)=\sqrt{\pi}/2$, hence $V(0)=2$ (this is the
length of the interval $[-1,1]$ in $\R$);
\item[$n=1$] $\Gamma(2)=1$, hence $V(1) = \pi$ (this is the familiar
result for the area of the unit circle);
\item[$n=2$] $\Gamma(5/2)=3\sqrt{\pi}/4$, hence $V(2) = 4\pi/3$
(this is the familiar result for the volume of the \PMlinkescapetext{unit} sphere);
\item[$n=3$] $\Gamma(3)=2$, hence $V(3) = \pi^2/2$.
\end{itemize}