The unit ball in Rn is defined as
the set of points
x12 + ... + xn2
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
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 R3
is 4/3 Pi.
The "volume" of the unit ball in R4 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
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 Rn to be
V(n) = Pin/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) = Pik/k!.
Since k! tends to infinity faster than Pik,
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 Rn always contains all the lower
dimensional unit balls!
Try computing the volume of the unit ball in R3 and R4 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
xx 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 2n.
Notice that if there are just two coordinates of
this point that are greater than 1/Sqrt, 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.
A little thought reveals that for large n, this probability decreases by about 1/Sqrt for each new coordinate that is chosen. This shows that the ratio R(n) tends to 0
as n goes to infinity.
we hope to show that V(n)=2nR(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, 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 drops by about 1/Sqrt.
So V(n) changes by about a factor 1/Sqrt
as n is incremented, for large n.
On the other hand, the factor 2n changes by
a factor of 2 as n is incremented, for large n.
Hence 2n changes by a factor of 2/Sqrt 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.