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In this section, you will find the volume and the surface area of a ball in n-dimensional space.
You know that a circle of radius r has area pi r2 and circumference 2 pi r. These two quantities are related. Indeed, a disk may be divided into a number of thin concentric annuli; an annulus of radius r and width dr has area approximately 2 pi r dr; and adding these areas by integrating gives the area of the disk. Thus pi r2 is the integral of 2 pi r, or, equivalently, 2 pi r is the derivative of pi r2.
The same principle relates the volume and the surface area of a ball in three-dimensional space. Namely, to find the volume of an onion, we can peel off a succession of spherical layers, and the sum of the volumes of the layers is the volume of the onion. Thus, the integral of the expression for the surface area of a ball of radius r should give the expression for the volume of a ball; and the derivative of the expression for the volume of a ball should give the expression for the surface area. Indeed, a ball of radius r in three-dimensional space has volume 4 pi r3/3 and surface area 4 pi r2.
Exercise
The same relationship holds in n-dimensional space. Namely, if V(r) denotes the volume of a ball of radius r in n-dimensional space, and S(r) denotes the corresponding surface area, then the derivative V'(r)=S(r).
It is a standard problem to compute the formulas for the volume V(r) and surface area S(r) of the n-ball. (See, for example, James Stewart's book Calculus, third edition, Brooks/Cole, 1995, page 966, problem 12.) However, the easy way to do the computation is less well known than the problem.
As just observed, it is enough to find either V(r) or S(r), as the other quantity can be obtained by differentiating or integrating. It turns out to be easier to find the surface area S(r).
Exercise
Use a scaling argument to show that S(r) equals a constant times rn-1, and the value of the constant equals S(1).
Now the trick is to compute an auxiliary integral in two different ways, and to equate the results. First integrate exp(-x12-x22-...-xn2) over the whole n-dimensional space as an iterated integral in Cartesian coordinates. (The improper integral is convergent.) You should find that this integral is the nth power of a one-dimensional integral, and if you do not recognize the one-dimensional integral, then you can use Maple to evaluate it.
Next integrate exp(-x12-x22-...-xn2) over the
whole space by viewing the space as made up of spherical
shells. You should get S(1) times a one-dimensional
integral with respect to the radius r. If you tell Maple to assume(n, positive), then Maple can compute the
one-dimensional integral in terms of the Gamma function.
Exercise
By following the hints above, show that the volume V(r) of a ball of radius r in n-dimensional space equals 2 pin/2 rn/(n Gamma(n/2)). Also verify that this formula agrees with the formulas that you know in dimensions 2 and 3.
The shape of n-dimensional space changes in a surprising
way when n becomes large. Let h(n) denote the volume of
the unit ball in dimension n. Try running the Maple loop
"for n from 2 to 20 do evalf(h(n)) od;" to see how this
volume changes with the dimension. You should find that the
volume increases until dimension 5 and then decreases
rapidly. In fact, if you execute the Maple command
limit(h(k), k=infinity); you will find that the
volume of the unit ball tends to zero as the dimension tends to infinity.
That the volume of the unit ball is maximal in dimension 5
is often cited as a curious fact. However, the question is the
wrong one: volumes of balls in different dimensions are not
comparable, for the same reason that volume and surface area
are not comparable. The volume of a ball in dimension 2 is
measured in square meters, while the volume of a ball in
dimension 3 is measured in cubic meters. The right quantity
to consider when varying the dimension n is
h(n)1/n. If you modify the above do loop,
replacing h(n) by h(n)1/n, you will find that this
quantity decreases steadily toward zero as n increases;
there is no longer a mysterious peak at n=5.
Exercise
Perform a similar analysis on the behavior of the surface area of the unit ball in n-dimensional space as n tends to infinity.
The Math 696 course
pages were last modified April 5, 2005. | | | Peeling an onion |