Volume of a Hypersphere⎯C.E. Mungan, Spring 2010 Problem: Find the volume Vn of an n-dimensional hypersphere of radius R. The three lowest values of n are well known. In one dimension, we have a line segment extending a distance R in each direction, so that its length is V1 = 2R . The case of n = 2 corresponds to a circle, whose 2 3 area is V2 = ! R . Finally, n = 3 corresponds to a sphere of volume V3 = 4! R / 3. Derive a compact formula for the general case. n Method #1: (Courtesy of Bob Sciamanda.) We can write the answer as Vn (R) = R !n , where !n " Vn (1) is the volume of a hypersphere of unit radius, since R is the only quantity in the problem with dimensions of length. The volume of any closed solid is z f V3 = ! A(z)dz (1) zi where A(z) is the cross-sectional area of a slab of thickness dz cut through the solid like a loaf of bread, and we integrate from zi to z f along any arbitrary axis z. It is convenient to choose z = Rcos! to be the polar axis, where we integrate upward from !i = " to ! f = 0 , and where the area of a slice through the sphere is V2 (r) with r = Rsin! . Making these substitutions in Eq. (1) gives " 3 3 V3(R) = R V2 (1)# sin ! d! . (2) 0 The reader is invited to perform this integral (by using the identity sin2 ! = 1" cos2 ! ) and check that it correctly gives 4/3, as implied in the problem statement above. Generalizing Eq. (2) to n dimensions gives $ n !n = !n"1 % sin # d# . (3) 0 This equation will correctly reproduce the one-dimensional value !1 = 2 if we identify !0 " 1. (That identity can be interpreted as stating that a zero-dimensional sphere is composed of one point.) The problem is now reduced to performing this definite integral and then finding a nonrecursive formula for !n . The integral is recognized as a beta function, which can be recast in terms of gamma functions as % n + 1( % 1( # +' * +' * n % n + 1 1( & 2 ) & 2) In ! sin " d" = B' , * = . (4) $ & 2 2) % n + 2( 0 + &' 2 )* Recall that the gamma function is a generalization of the factorial function, !(n + 1) = n!(n) , where !(1) = 1 and !(1 / 2) = " . Now notice that $ n + 1' $ 1' $ n' $ 1' # # # # %& 2 () %& 2() %& 2() %& 2() 2* In ! In 1 = ! = . (5) " $ n ' $ n + 1' n # + 1 # %& 2 () %& 2 () Substituting this result into Eq. (3) results in a near-miraculous simplification, 2# ! = ! I = ! I I = ! n n"1 n n"2 n n"1 n"2 n $2# 2# 2# ! 1 if n even (6) & n n " 2 2 = % 2# 2# 2# & ! 2 if n odd & ' n n " 2 3 where I made use of the facts that !0 = 1 and !1 = 2 to terminate the even and odd recursions, respectively. Both of these cases can be written in the single compact way, " n/2 !n = (7) $ n ' # + 1 %& 2 () which completes the exercise. The first six values are tabulated below. Notice the recurring patterns in which factors of π appear. Also note that υn is a maximum for n = 5 and decreases monotonically to zero for larger values of n. " n % n I ! + 1 ! V n #$ 2 &' n n 0 ! 1 1 1 1 1 2 2 ! 2 2R 1 2 2 2 ! 1 ! ! R 4 3 4 4 3 3 3 4 ! 3 ! 3 ! R 3 1 2 1 2 4 4 8 ! 2 2 ! 2 ! R 16 15 8 2 8 2 5 5 15 8 ! 15 ! 15 ! R 5 1 3 1 3 6 6 16 ! 6 6 ! 6 ! R The surface area of a unit hypersphere, an, can be obtained by multiplying Eq. (7) by the derivative with respect to R of Rn and then setting R to 1. In other words, multiply Eq. (7) by n to get 2! n/2 an = . (8) # n& " $% 2'( Again this area has a maximum, namely at n = 7 , and thereafter it decreases to zero. Method #2: (Courtesy of Matt Springer.) The differential of the volume of a hypersphere of radius r ! R is n n$1 Vn = !nr " dVn = !n # nr dr . (9) But the usual integral of a Gaussian is " 2 # e!x dx = $ . (10) !" Multiply this integral by itself n times, subscripting each dummy variable x by a different index i so that we can keep track of them, * * * # n & n 2 n/2 ! exp% !" xi ( ) dxi = , . (11) + + + $ ' !* !* !* i=1 i=1 However, the summation is simply equal to r2 in n dimensions, and the product of differentials is just the n-dimensional volume element. Writing the volume element in spherical coordinates and performing all of the angular integrals, that element becomes Eq. (9), so that Eq. (11) reduces to # !r2 n!1 n/2 $ e "nnr dr = % . (12) 0 2 Pull the constants !n and n out of the integral and change variables to u ! r to get # 1 ! n e"uun/2"1 du = % n/2 . (13) 2 n $ 0 The remaining integral defines !(n / 2). Finally, noting that " n % n " n% ! + 1 = ! , (14) #$ 2 &' 2 #$ 2&' we can rearrange Eq. (13) to get the solution in Eq. (7). .