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What is the volume of spheres in higher dimensions?

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@user1754 How do you define the spheres? by center point and radius or by n-points on its surface? –  user2468 Mar 16 '11 at 17:47
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6 Answers

Did you try the obvious?

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Let $\kappa_n$ be the volume of the unit $n$-sphere $B_n$ and write the points of $B_n$ in the form $(x,y,{\bf z})$ with ${\bf z}\in{\mathbb R}^{n-2}$. For given $(x,y)\in B_2$ one has $|{\bf z}|^2\leq 1-r^2$, where $r:=\sqrt{x^2+y^2}$. These ${\bf z}$ fill an $(n-2)$-sphere of radius $\sqrt{1-r^2}$, and the $(n-2)$-dimensional volume of this sphere amounts to $\kappa_{n-2}(1-r^2)^{(n-2)/2}$. Therefore we get $$\kappa_n=\int_{B_2}\kappa_{n-2}(1-r^2)^{(n-2)/2}{\rm d}(x,y)=2\pi \kappa_{n-2}\int_0^1 (1-r^2)^{(n-2)/2}\>r\>dr={2\pi\over n}\>\kappa_{n-2}\ .$$ By means of this this recursion formula and using the known values $\kappa_1=2$, $\kappa_2=\pi$ one easily obtains the $\Gamma$-formula quoted in other answers.

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Start with an unrelated integral

$$I = \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty e^{(-x_1^2+x_1^2+\cdots x_n^2) } dx_1 dx_2 \cdots dx_n = \pi^{n/2}$$

This should be easy to easy if you notice that this can be factored as a product of n integrals each in the variable $x_i$ and that $\int_{-\infty}^\infty e^{-x^2}dx =\sqrt{\pi}$

Now try to calculate the same integrate by making a change to polar coordinates. These new coordinates will have one $r$ coordinate and $n-1$ angular coordinates $\{\theta_i\}$

$$I = \int e^{(-x_1^2+x_1^2+\cdots x_n^2) } dV_n =$$

To express the volume element in the new coordinates you can either use inference from lower dimensions (or if you're a physicist, a dimensional argument) and say that $$V = C_n r^n \Rightarrow dV_n = nC_n r^{n-1}dr$$

or you can explicitly compute the Jacobian to show that $|J(r,\theta_1,\cdots,\theta_n)|= r^{n-1} \cos^{n-2}\theta_1\cos^{n-3}\theta_2 \cdots \cos \theta_{n-2} $


$$I = \int_0^\infty e^{-r^2} nC_nr^{n-1} dr = C_n\frac{n}{2} \Gamma (\frac{n}{2}) = C_n\left(\frac{n}{2}\right)! $$

$$ C_n = \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!} $$

$$V_n = \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!} r^n$$

on a sidenote, you also get the surface area from $dV_n = S_n(r) dr$

$$S_n(r)= \frac{(2\pi)^{n/2}}{\Gamma (\frac{n}{2})} r^{n-1}$$

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$$V_n = \frac{\pi^{n/2}}{\Gamma(n/2+1)}$$

As posted earlier here.

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just a note to the uninitiated, this is the volume of a unit sphere. –  Please Delete Account Mar 23 '11 at 22:01
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Perhaps something like this:

Imagine a 1D curve that follows the curved edge of a semi-circle, lets call a constant number determined by this curve $c$ and the curve $v$.

For each additional dimention in the $n$ dimensional sphere generalization the internal space would increase relative to $c$?

For example a sphere could be thought of as neighbouring circles of varying sizes, the sizes would vary through the sphere to follow $v$.

To put it a more generic way:

An $n$ sphere could be thought of as neighbouring $n-1$ spheres of varying sizes, the sizes would vary through the $n$ sphere to follow $v$.

$space_1 = length$

$space_n = space_{n-1}*c$

Therefore, possibly, the internal space of the n-sphere could be defined as:

$units^n = c^n$


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In the real world, spheres are measured by their diameter, not their radius. Were one to buy tyres, or plates, or ironmongery (bolts, nuts, drills, that sort of thing), and holes of all sorts, are measured by their diameter.

Moreover, there are named units corresponding to the area of an inch (diameter) circle, and the volume of an inch diameter sphere - circular inch, and spheric inch. This is because the three regular solids that are the Nth power of a line, all define valid products, and coherent units.

Thus, we have the conversion factor in rexx style.

   C2P(N) = PPI( 1/FACT(N, 2),  N // 2)
       PPI(a, b) = a * (PI/2)^2
       FACT(a, b) =  a * (a-b) * (a-2b) ... while a-nb >0

In mathematical terms, one could write $c2p(n) = (\pi/2)^{n//2} / (n-1)!!$, and the volume of a sphere is then $vol(sph) = 2^N * c2p(n)$

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bolts, nuts, drills are not spheres? –  levitopher Jun 1 '13 at 1:56
@levitopher they are sold by their diameter, as are ball bearings. The point is circles, cylinders and spheres are sold by thei diameters. –  wendy.krieger Jun 1 '13 at 4:54
But the question was about volumes in higher dimensions, nothing to do with how you buy a specific circular or spherical object. Your answer might be fine but the whole unit question is not related. –  levitopher Jun 1 '13 at 23:24
On the other hand, the formula is against the diameter, not the radius. Some mathematical folk just don't get it. –  wendy.krieger Jun 2 '13 at 3:51
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