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88 CHAPTER 3. Multiple Integrals

Example 3.30. In this example we compute the volume ωn of the n-dimensional unit ball.

By the Fubini theorem,

                                                  ??
                                       ż 1 ż 1´x21                  ż 1´x12´¨¨¨´x2n´1

                                ωn =              ? ¨¨¨ ?                                              dxn ¨ ¨ ¨ dx1 .

                                          ´1 ´ 1´x12                   ´ 1´x21´¨¨¨´xn2´1

                                      ??
                                        1´x21              1´x12´¨¨¨´xn2´1
                                   ż                  ż                                                                                         n´1
                                       ?          ¨¨¨     ?                                                                                       2
Note that the integral                                                              dxn     ¨      ¨¨  dx2   is  in  fact    ωn´1(1  ´  x12  )       ,  the

                                   ´ 1´x21            ´ 1´x21´¨¨¨´xn2´1

volume of (n ´ 1)-dimensional ball of radius a1 ´ x12; thus

                                      ż1                                                               ż  π
                               ωn =                                                                       2
                                                                x2     n´1                                   cosn                                    (3.8)
                                                ωn´1(1    ´         )    2   dx     =  2 ωn´1                      θ  dθ  .

                                          ´1 0

Integrating by parts,

ż     π                     ż  π                                                             ˇθ=   π                  ż   π
      2                        2                                                            θˇ     2               1)     2
         cosn                     cosn´1          d(sin θ)      cosn´1              sin                   (n                 cosn´2  θ sin2
               θ   dθ    =                   θ               =                  θ           ˇθ=0       +      ´                                 θ  dθ

      00                                                                                                              0

                                      ż   π
                                   1)
                            (n               2  cosn´2 θ(1 ´        cos2

                         =      ´                                               θ) dθ

                                          0

which implies that                        żπ      cosn θ dθ     =      n  ´     1   żπ      cosn´2 θ dθ .
As a consequence,                              2                          n              2
                                            0                                         0

                                             $    (n  ´  1)(n   ´      3) ¨ ¨ ¨ 2   ż  π
                                                                                       2
                                                                                            cos θ dθ             if n is odd ,
                   ż  π                      ’        n(n ´ 2) ¨ ¨ ¨ 3 0                                         if n is even ,
                      2                      ’
                         cosn                &

                                θ  dθ  =              (n  ´ 1)(n ´ 3) ¨ ¨ ¨ 1                   π
                                                          n(n ´ 2) ¨ ¨ ¨ 2
                   0’                                                                       ż2     dθ
                                    ’
                                    %                                                         0

and   the  recursive        formula       (3.8)      implies    that            ωn  =       2 ωn´2 π .       Further         computations          shows

that                                                                                           n

                                                  $       (2π)  n´1                         if n is odd ,
                                                                  2
                                                  ’   n(n ´ 2) ¨ ¨ ¨ 3 ω1
                                                  ’
                                                  ’
                                                  &

                                       ωn =                     n´2

                                                  ’      (2π) 2        ¨  ¨  4  ω2          if n is even .
                                                  ’   n(n ´ 2) ¨
                                                  ’
                                                  %

                                                                      ż8

Let Γ be the Gamma function defined by Γ(t) = xt´1e´x dx for t ą 0. Then Γ(x + 1) =

xΓ(x)    for  all  x  ą  0,    Γ(1)    =  1     and    (  1  )  =   ?π.               0     fact          that   ω1   =   2  and  ω2  =  π,     we      can
                                                      Γ
                                                          2                     By the

express ωn as                                                ωn = Γ(πn+n22 2 ) .
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