1
$\begingroup$

What is the volume of spheres in higher dimensions?

$\endgroup$

closed as off-topic by Did, H. R., Claude Leibovici, Daniel, Peter Woolfitt Nov 27 '15 at 15:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, H. R., Claude Leibovici, Daniel, Peter Woolfitt
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ @user1754 How do you define the spheres? by center point and radius or by n-points on its surface? $\endgroup$ – user2468 Mar 16 '11 at 17:47
15
$\begingroup$

Did you try the obvious?

$\endgroup$
8
$\begingroup$

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} $

so

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

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

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

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

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

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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

As posted earlier here.

$\endgroup$
  • 2
    $\begingroup$ just a note to the uninitiated, this is the volume of a unit sphere. $\endgroup$ – Please Delete Account Mar 23 '11 at 22:01
-5
$\begingroup$

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 $vol(sph) = 2^N * c2p(n)$

$\endgroup$
  • $\begingroup$ bolts, nuts, drills are not spheres? $\endgroup$ – levitopher Jun 1 '13 at 1:56
  • $\begingroup$ @levitopher they are sold by their diameter, as are ball bearings. The point is circles, cylinders and spheres are sold by thei diameters. $\endgroup$ – wendy.krieger Jun 1 '13 at 4:54
  • 3
    $\begingroup$ 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. $\endgroup$ – levitopher Jun 1 '13 at 23:24
  • $\begingroup$ On the other hand, the formula is against the diameter, not the radius. Some mathematical folk just don't get it. $\endgroup$ – wendy.krieger Jun 2 '13 at 3:51