The unit ball in R^{n} is defined as the set of points (x_{1},…,x_{n}) such that

x_{1}^{2} + … + x_{n}^{2}

What is the volume of the unit ball in various dimensions?

Let's investigate! The 1-dimensional volume (i.e., length) of the 1-dimensional ball (the interval [-1,1]) is 2.

The 2-dimensional volume (i.e., area) of the unit disc in the plane is pi.

The 3-dimensional volume of the unit ball in R^{3}is 4/3 Pi.

The “volume” of the unit ball in R^{4}is (Pi/2) * Pi.

So apparently, as the dimension increases, so does the volume of the unit ball. What does this volume tend to as the dimension tends to infinity?

Intuitively, one may think that in higher and higher dimensions there's more and more “room” in the unit ball, allowing its volume to become larger and larger. Does the volume become infinite, or does it approach a sufficiently large constant as the dimension increases?

The answer is surprising and shows how our intuition is often misleading. Using multivariable calculus one can calculate the volume of the unit ball in R^{n} to be

V(n) = Pi^{n/2} / Gamma(n/2 + 1),

where Gamma is the Gamma function that generalizes the factorial function (i.e., Gamma(z+1) = z!). For n even, say n=2k, the volume of the unit ball is thus given by

V(n) = Pi^{k}/k!.

Since k! tends to infinity faster than Pi^{k}, it follows that V(n) tends to 0 as n tends to infinity!

In higher dimensions you can fit less and less stuff into the unit ball. Of course, by stuff we mean n-dimensional stuff, since the unit ball in R^{n} always contains all the lower dimensional unit balls!

**Presentation Suggestions:**

Try computing the volume of the unit ball in R^{3} and R^{4} using multivariable calculus. Then using a computer algebra package plot V(n) using the formula above. What dimension seems to have the maximal volume? Now plot V(n)^{1/n}. Explain. Explore these same ideas with the surface area. See also Surface Area of a sphere and High Dimensional Spheres in Cubes.

**The Math Behind the Fact:**

One may work with the formula for V(n) by applying Stirling's Formula, which approximates Gamma(x+1) by x^{x} e^{-x} (2 Pi x)^{1/2} for large x, to see why the surprising fact above is true.

Another heuristic is the following probabilistic argument. Pick n points independently and identically distributed (i.i.d.) from a uniform distribution in [-1,1], and form an n-tuple out of these numbers. The resulting vector represents a point picked randomly out of the unit box B=[-1,1]^{n}, so the probability that such a point is in the unit n-ball is the ratio R(n) of the volume V(n) to the volume of the unit box, which is 2^{n}.

Notice that if there are just two coordinates of this point that are greater than 1/Sqrt[2], then the point cannot be in the unit n-ball. As n grows, we choose more and more coordinates i.i.d. from the uniform distribution, and the smaller the probability is that just zero or one of those n coordinates are bigger than 1/Sqrt[2]. A little thought reveals that for large n, this probability decreases by about 1/Sqrt[2] for each new coordinate that is chosen. This shows that the ratio R(n) tends to 0 as n goes to infinity.

However, we hope to show that V(n)=2^{n}R(n) tends to 0 as n goes to infinity. A refinement of the above argument will do the trick: if there are just five coordinates of this point that are greater than 1/Sqrt[5], then the point cannot be in the unit n-ball. For large n, as each new coordinate chosen, the probability than less than five coordinates are bigger than 1/Sqrt[5] drops by about 1/Sqrt[5]. So V(n) changes by about a factor 1/Sqrt[5] as n is incremented, for large n. On the other hand, the factor 2^{n} changes by a factor of 2 as n is incremented, for large n. Hence 2^{n} changes by a factor of 2/Sqrt[5] for large enough n, so whatever this quantity is, it eventually gets smaller and smaller.

**How to Cite this Page:**

Su, Francis E., et al. “Volume of a Ball in N Dimensions.” *Math Fun Facts*. <https://www.math.hmc.edu/funfacts>.

**Fun Fact suggested by: **

Jon Jacobsen