# Why does the volume of a hypersphere decrease in higher dimensions? [duplicate]

This question already has an answer here:

First let us define an $n$-ball as the euclidean sphere in $\mathbb{R}^n$ including its interior and its surface where $n$ refers to the number of coordinates needed to describe the object (the geometer's notation), NOT the topologist's notation which refers to the dimension of the manifold. All $n$-balls considered here are of radius 1 centered at the origin. So an $n$-ball is the set of points

$$\{x=(x_1,x_2,...,x_n)\in \mathbb{R}^n : \sum_{i=1}^n x_i^2 \leq 1\}.$$

It is well known that the volume of the unit $n$-ball is given by

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

and before anyone points it out, we will consider all volumes to be unit-less so that they can be compared with each other. So for example

$$\pi=V(2)<V(3)=\frac{4}{3}\pi.$$

Now, my question is why does $V(n)\rightarrow 0$ as $n\rightarrow\infty$? This behavior is independent of the radius. Depending on the radius, $V(n)$ may (or may not) increase at first, hit a peak at some $n$, and then monotonically decrease and converge to zero. Considering $n\in\mathbb{N}$ to take on only discrete values, for the unit $n$-ball, the max volume is achieved at $n=5$.

I have read all I could find (here/wikipedia and elsewhere) and I do see that the $n$-ball occupies a smaller and smaller portion of the circumscribing cube $[-1,1]^n$ and I have seen the arguments that the diameter should be used as the fundamental quantity instead of the radius. This way, the volume monotonically decreases to zero for all $n$ at least for the unit ball. I also see the analytic reason why the function converges to zero. The gamma function in the denominator grows much faster than the numerator and eventually the whole fraction converges to zero even if the radius is a googol.

But my question is, intuitively speaking (as if intuition is a good guide in higher dimensions), $V(n)$ should be monotonically increasing or at the very least non-decreasing. The way I see it, a 2-ball is contained in a 3-ball and a 2-ball can be rotated in $\mathbb{R}^3$ to create a 3-ball. Similarly, you can rotate any $(n-1)$-ball in $\mathbb{R}^n$ around the appropriate axis to create an $n$-ball. I know that any $(n-1)$-ball in $\mathbb{R}^n$ has Lebesgue measure zero but inside any $n$-ball, an $(n-1)$-ball can be rotated around any of the $n$ axes so $V(n)$ should be larger than $V(n-1)$. This specific "argument" hasn't been addressed on any of the previous questions that I could find.

Any geometric interpretation of what's happening? Bonus points for something intuitive and/or 1-2-3 dimensional examples to show me the fallacy of my argument. Thanks!

Addendum: I have seen this thread (and even this one and many others on stack exchange and math overflow) and like I said, they don't address my argument presented in this question. They do mention other arguments such as comparison with the circumscribing cube or analytically looking at fraction to see why it goes to zero. But those I already know...namely by reading these very threads.

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## marked as duplicate by T. Bongers, L.G., Daniel Rust, Dominic Michaelis, azimutAug 22 '13 at 10:18

This has been discussed before. –  Potato Aug 22 '13 at 9:13
Imagine a difference between a ball and a cylinder, why the former has smaller volume than the latter? For a hypersphere similar situation happens with each dimension. –  dtldarek Aug 22 '13 at 9:15
I order to compare the volumes of an $n$-ball with a $(n+1)$-ball, you should make the $n$-ball into a $(n+1)$-cylinder by extruding it along an interval of length $1$. Now the cylinder has the same $(n+1)$-dimensional volume as the $n$-dimensional volumen of the $n$-ball. –  Christoph Aug 22 '13 at 9:19
Check my answer added here (because of closure of this question). You are right that $n$-ball produces $n+1$-ball with rotations, but the thing is that the unit cube $[0,1]^n$ also produces $[0,1]^{n+1}$ with translations and the volume doesn't grow, because that's how it works. See more in the answer. –  savick01 Aug 22 '13 at 10:35
In my opinion it is already a wrong point of view to compare measures of objects of different dimension. In "real life" you don't compare lengths with areas or volumes: Take a disc of area 1 and form the cylinder of height $h<1$ over it. Its volume is smaller than $1$. But is this really surprising? Compare this situation with the "rotation argument" in the original posting. –  Hagen Aug 22 '13 at 11:01

I think this is one of the best examples which show that we must not rely on our (low dimensional geometric) intuition too much. Kudryavtsev's "Mathematical analysis" textbook contains another one:

The $n$-dimensional unit cube belongs to the $n$-dimensional unit ball only for $n=1,2,3,4$.

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Can you explain what "belongs to" means in this context? –  Christoph Aug 22 '13 at 9:21
"is a subset of". Here "unit cube" means $\{ (x_1, \ldots, x_n) \in \mathbb{R}^n \colon |x_k| \le 1/2,\ k=1,\ldots,n \}$. –  njguliyev Aug 22 '13 at 9:25

The upper limit of a sphere is to be contained in a volume of $(2\pi r^2)^n$ for $2n$, but the denominator increases as the factorial, because it's a kind of prism on a base, so the denominator is here $2 n$ for going from $n-1$ to $n$.

Since ultimately the increase is $2\pi$ over two dimensions, but decreases by $2n$ over 2 dimensions, the value of $2n > 2\pi$ happens at n=3.5 or 4 (for N=7 and N=8). Thereafter, the denominator is expanding faster than the numerator, without limit.

Although only loosely related, consider the packing of cubes, which fills all space, against the packing of spheres. The maximum packing of spheres supposes that all of the holes are simplex in shape. This means that if you have a packing of spheres of unit radius, there are no more than $n+1$ sphere-sectors in a simplex of edge $2$.

It's not easy to calculate the portion of the sphere in the sector, but it comes out as $s/n!$, where $1 < s < \sqrt{n/4}$. The volume of a simplex of edge $2$ is $\sqrt{2^n/(n+1)}/n!$, so the spheres occupy ever-diminishing parts of space even at the tightest packing, ie $s \sqrt{(n+1)/2^n}$ of space.

The famous $E_8$ lattice, for example, the densest packing in eight dimensions, the spheres occupy $1/3.98$ of space.

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