# ball (mathematical)

Mathematicians, unlike the rest of the human race, draw a
sharp distinction between a sphere and a
ball. A sphere (in mathematics) is only a surface,
whereas a ball is everything inside, and possibly including, that surface
– the filling of the sphere. An **open ball** consists
of all the points that are less than a given distance (the radius) away
from a given point (the center); a **closed ball**, consists
of all the points that are less than or equal to the radius.

Mathematical balls can also exist in any number of dimensions.
A one-dimensional ball of radius *r* is just a line segment. It consists
of all the points on a line between -*r* and *r*, or, in the case
of a one-dimensional unit ball (a ball with a radius of 1), between -1 and
1. A 1-d unit ball thus has a length, or "1-d volume," of 2. A 2-d unit
ball, which is the filling of a unit circle, has an area, or 2-d volume,
of π. The volume of a unit ball in 3-d is 4/3 π. In 4-d it is π^{2}/2.
Apparently, as the number of dimensions increases, so does the volume of
the unit ball. What does this volume tend to as the dimension tends to infinity?
Intuitively, it might seem 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 a technique called multivariable
calculus the volume of the unit ball in n dimensions, *V*(*n*),
can be shown to be π^{n/2} / Γ(*n*/2 + 1),
where Γ is the gamma function that generalizes the factorial function
(i.e., Γ(*z* + 1) = *z*!). For *n* even, say *n* = 2*k*, the volume of the unit ball is thus given by *V*(*n*)
= π^{k} / *k*!.