area of the $n$-sphereAreaOfTheNSphere2003-07-23 02:20:182006-10-18 01:23:52Derivationsphere% this is the default PlanetMath preamble. as your knowledge
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\def\R{\mathbb{R}}The area of $S^n$ the unit $n$-sphere (or hypersphere) is the same as the total
solid angle it subtends at the origin. To calculate it, consider the following
integral
\[
I(n) = \int_{\R^{n+1}} e^{-\sum_{i=1}^{n+1} x_i^2}\, d^{n+1} x.
\]
Switching to polar coordinates we let $r^2=\sum_{i=1}^{n+1} x_i^2$ and the
integral becomes
\[
I(n) = \int_{S^n} d\Omega \int_{0}^{\infty} r^{n} e^{-r^2}\, dr.
\]
The first integral is the integral over all solid angles and is exactly what we
want to evaluate. Let us denote it by $A(n)$. With the change of variable
$t=r^2$, the second integral can be evaluated in terms of the gamma function
$\Gamma(x)$:
\[
I(n)/A(n) = \frac{1}{2}\int_0^\infty t^{\frac{n-1}{2}} e^{-t}\, dt
= \frac{1}{2}\Gamma\left(\frac{n+1}{2}\right).
\]
We can also evaluate $I(n)$ directly in Cartesian coordinates:
\[
I(n) = \left[ \int_{-\infty}^\infty e^{-x^2}\, dx \right]^{n+1}
= \pi^{\frac{n+1}{2}},
\]
where we have used the standard Gaussian integral $\int_{-\infty}^\infty
e^{-x^2}\, dx = \sqrt{\pi}$.
Finally, we can solve for the area
\[ A(n) = \frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}. \]
If the radius of the sphere is $R$ and not $1$, the correct area is
$A(n)R^{n}$.
Note that this formula works only for $n\ge0$. The first few special cases
are
\begin{itemize}
\item[$n=0$] $\Gamma(1/2)=\sqrt{\pi}$, hence $A(0)=2$ (in this case, the
area just counts the number of points in $S^0=\{+1,-1\}$);
\item[$n=1$] $\Gamma(1)=1$, hence $A(1)=2\pi$ (this is the familiar result
for the circumference of the unit circle);
\item[$n=2$] $\Gamma(3/2)=\sqrt{\pi}/2$, hence $A(2)=4\pi$ (this is the
familiar result for the area of the unit sphere);
\item[$n=3$] $\Gamma(2)=1$, hence $A(3)=2\pi^2$;
\item[$n=4$] $\Gamma(5/2)=3\sqrt{\pi}/4$, hence $A(4)=8\pi^2/3$.
\end{itemize}