The volume of a $d$ dimensional hypersphere of radius $r$ is given by:

$$V(r,d)=\frac{(\pi r^2)^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$$

What intrigues me about this, is that $V\to 0$ as $d\to\infty$ for any fixed $r$. How can this be? For fixed $r$, I would have thought adding a dimension would make the volume bigger, but apparently it does not. Anyone got a good explanation?

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    $\begingroup$ Think about the square of side length $1/2$. What is its volume in $\mathbb{R}^d$? $\endgroup$
    – JavaMan
    Sep 23, 2011 at 18:53
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    $\begingroup$ Imagine a square of side length $\frac{1}{2}$: the area is $\frac{1}{4}$. A cube of edge length $\frac{1}{2}$ has volume $\frac{1}{8}$. A hypercube of edge length $\frac{1}{2}$ has hypervolume $\frac{1}{16}$; etc. For a cube, whether going to the "next dimension" increases the total hypervolume or decreases it (as a scalar) depends on whether the edge side was less than $1$ or more than $1$. That suggests to me that "adding a dimension" does not, in and of itself, necessarily mean "making the volume bigger"... $\endgroup$ Sep 23, 2011 at 18:55
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    $\begingroup$ What about a sphere of radius $10^{10}$? The volume eventually goes to zero :-) $\endgroup$ Sep 23, 2011 at 19:03
  • $\begingroup$ @probabilityislogic: The point is that simply "adding a dimension makes things bigger" doesn't work: it's not just about the dimension, there are other factors at play. $\endgroup$ Sep 23, 2011 at 19:26
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    $\begingroup$ Shouldn't the title include the phrases hypershpere and n-->inf? $\endgroup$
    – ripper234
    Oct 4, 2011 at 23:19

7 Answers 7


I suppose you could say that adding a dimension "makes the volume bigger" for the hypersphere, but it does so even more for the unit you measure the volume with, namely the unit cube. So the numerical value of the volume does go towards zero.

Really, of course, it is apples to oranges because volumes of different dimensions are not commensurable -- it makes no sense to compare the area of the unit disk with the volume of the unit sphere.

All we can say is that in higher dimensions, a hypersphere is a successively worse approximation to a hypercube (of side length twice the radius). They coincide in dimension one, and it goes downward from there.

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    $\begingroup$ This is a very nice explanation. Thanks :) $\endgroup$
    – Srivatsan
    Sep 23, 2011 at 21:27
  • $\begingroup$ This is nice and concise. Here is somebody's blog post saying the same thing with more words. $\endgroup$
    – yasmar
    Sep 24, 2011 at 10:27
  • $\begingroup$ You might also want to read this similar post on MSE math.stackexchange.com/questions/15656/… $\endgroup$
    – Tpofofn
    Sep 24, 2011 at 15:38
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    $\begingroup$ I don't think OP was bewildered over the volume of n-dimension to volumes of n+1 dimensions. Yes, that would be like "apples to oranges" as you say (or apples to apple trees). However, this downplays the intuitive significance that the ratio of volume of an n-sphere (greater than ~5 dimensions) to its corresponding n-cube decreases to virtual nothingness with the dimension. $\endgroup$
    – aiwyn
    Apr 22, 2019 at 19:02
  • $\begingroup$ It does not go straight downwards from $n=1$, since for $n=1$ you get $2r$ and for $n=2$ you get $\pi r^2$ which is larger, certainly for $r\ge 1$ and it keeps increasing up to $n=5$ for $r\ge 1$ and further for larger $r$ $\endgroup$
    – Henry
    Mar 24, 2020 at 22:53

The reason is because the length of the diagonal cube goes to infinity.

The cube in some sense does exactly what we expect. If it's side lengths are $1$, it will have the same volume in any dimension. So lets take a cube centered at the origin with side lengths $r$. Then what is the smallest sphere which contains this cube? It would need to have radius $r\sqrt{d}$, so that radius of the sphere required goes to infinity.

Perhaps this gives some intuition.

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    $\begingroup$ This is the clearest answer for me. $\endgroup$
    – Sklivvz
    Sep 24, 2011 at 7:56
  • $\begingroup$ I think you mean the cube has side lengths $2r$. $\endgroup$
    – Xiang Yu
    Nov 23, 2018 at 17:30

This was thoroughly discussed on MO. I quote from the top answer:

The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is of size $\frac{1}{\sqrt{n}}\ll 1$. This can be turned into a simple geometric argument (as suggested by fedja) using the fact that an $n$-element set has $2^n$ subsets:

At least $n/2$ of the coordinates of a point in the unit ball are at most $\sqrt{\frac{2}{n}}$ in absolute value, and the rest are at most $1$ in absolute value. Thus, the unit ball can be covered by at most $2^n$ bricks (right-angled parallelepipeds) of volume $$\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$ each corresponding to a subset for the small coordinates. Hence, the volume of the unit ball is at most $$2^n \cdot \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = \left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, so the volume of the ball of any fixed radius also goes to $0$.

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    $\begingroup$ I'm not seeing why the bricks each have volume at most $\left(2\sqrt{\frac{2}{n}}\right)^{n/2}.$ Shouldn't this be multiplied by a factor of $2^{n/2}$ since the other coordinates have absolute value at most $1$? This doesn't change the limit obviously, but some enlightenment would be nice. $\endgroup$
    – cats
    Feb 14, 2013 at 8:30

Let $X_j$ be a sequence of independent random variables with uniform distribution in the interval $[-r,r]$ (i.e. you are picking an infinite sequence of random numbers from the interval). The probability that $(X_1, \ldots, X_d)$ lies in your hypersphere, i.e. that $R_d = X_1^2 + \ldots + X_d^2 \le r^2$, is $V(r,d)/(2r)^d$ (which by scaling doesn't depend on $r$, so let's call it $P(d)$). Of course $P(d) \to 0$ as $d \to \infty$, but the point is that it goes to 0 faster than an exponential in $d$. Indeed, by the theory of large deviations, for any $t > 0$ we should have $P(R_d \leq t d) \approx e^{-d I(t)}$ as $d \to \infty$ for some function $I(t)$, where $I(t) \to \infty$ as $t \to 0+$.

  • $\begingroup$ Came into this thread to give this proof, but I see someone beat me to it. $\endgroup$ Sep 23, 2011 at 20:44

Let me develop on the idea presented by Christoph in the comments of this post.

The first thing is that the volume depends on the radius. $n$-ball or cube of radius $r$ has $r^n$ times bigger volume than that of radius $1$. One should expect then, that at least for big radii the volume of the ball will increase (as it is in the case of the unit cube $[-1,1]^n$ of volume $2^n$).

However, the thing is that somebody once decided (could they decide otherwise?) that Fubini theorem would be nice and $n$-dimensional volume of $[0,1]^n$ will be the same as $(n+1)$-dimensional volume $[0,1]^n\times [0,1]$ -- even though of course one is much smaller than the other!

That's the key point - volume is invariant under cartesian product with the unit interval - like it or not.

So, as Christoph Pegel says it is reasonable to compare volume of the ball $B^n$ with volume of the cylinder $B^n\times [-1,1]$. The second is of course $2$ times bigger but it is a matter of radius as already discussed.

Note that if we compare $B^{n+1}$ with $B^n\times [-1,1]$, we notice that only the zero section is the same. At level $t$ the ball is smaller with radius $\sqrt{1-t^2}$. That means that its volume is $(1-t^2)^{n/2}$ times smaller! This function converges to zero with $n$ (and so does its integral) and thus there's no wonder that it kills any geometric growth (where multiplicative factor $>1$ is constant).

  • $\begingroup$ In my opinion the key point is that there are two different measures involved. Namely the Hausdorff measures $\mathcal H^n$ and $\mathcal H^{n+1}$. The $n$-ball has positive $\mathcal H^n$-volume, while its $\mathcal H^{n+1}$-volume vanishes, since there is really no $(n+1)$-dimensional interior in an $n$-ball. To compare the volumes $\mathcal H^{n}(B^n)$ and $\mathcal H^{n+1}(B^{n+1})$ under a common measure, we make use of $\mathcal H^{n}(A)=\mathcal H^{n+1}(A\times[0,1])$. $\endgroup$
    – Christoph
    Aug 22, 2013 at 11:08
  • $\begingroup$ Could you explain why are you working with a cylinder of height two instead of height one? Shouldn't we compare $B^{n+1}$ with $B^n \times [\textrm{something with length one}]$? Otherwise we do double the volume. Everything else makes sense to me. I just don't understand "but it is a matter of radius as already discussed". $\endgroup$ Aug 23, 2013 at 8:53
  • $\begingroup$ @FixedPoint Just because "height" (diameter) of the unit-ball is $2$. If our default ball has radius $1/2$, then I would work with some unit interval. The difference between those cases is a factor proportional to the radius and dependence radius ~ volume was discussed in the first paragraph. $\endgroup$
    – savick01
    Aug 23, 2013 at 9:36

Consider the distance from the center of the sphere of radius $R$ in $\mathbb{R}^n$ to the corner of its enclosing cube: $R\sqrt{n}$. This is like covering the surface of the sphere with a layer of tall corners.

For $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$, define $$ Q(x)=x\frac{|x|}{\max_i|x_i|}\tag{1} $$ $Q$ maps a sphere to its enclosing cube. As described above, $\frac{|Q(x)|}{|x|}$ reaches a maximum of $\sqrt{n}$ in the corners. To be exact, $$ \frac{|Q(x)|^2}{|x|^2}=\sum_j\frac{|x_j|^2}{\max_i|x_i|^2}\tag{2} $$ By considering each $x_i$ to be normally distributed and using homogeneity of $(2)$, we get that the mean of $\frac{|Q(x)|^2}{|x|^2}$ over the sphere is $\frac{n+2}{3}$. Thus, the rms average of $\frac{|Q(x)|}{|x|}$ is $\sqrt{\frac{n+2}{3}}$. In this way, the volume of the cube should be approximated by $\sqrt{\frac{n+2}{3}}^{\;n}$ times the volume of the sphere. This grows faster than any $R^n$.


Imagine you have n circles fitting perfectly in a square. Pick a point at random inside each of these squares. What's the probability that the randomly chosen point is inside each of the circles? It doesn't matter. As long as the area of the target (a circle) is less than the area of the square the probability of hitting all the targets will approach 0. This is (roughly) analogous to taking a 2n dimensional sphere and projecting it onto n planes. A point in the 2n sphere will fit into all n of the projected circles. This shows that the volume of the n-sphere relative to the box that contains it approaches 0.

Forget volume or area or anything with a physical interpretation. You are picking n positive numbers according to some probability distribution (in our case its distance from a fixed point). The larger n is the less likely their sum is to be less than any fixed constant.


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