11th Putnam 1951 Problem B7 In 4-space let S be the 3-sphere radius r: w 2 + x 2 + y 2 + z 2 = r 2 . What is the 3-dimensional volume of S? What is the 4-dimensional volume of its interior? Solution
: 2π 2 r 3 , 1/2 π 2 r 4 . A hyperplane perpendicular to the x-axis will cut S in a sphere. If the hyperplane is at a distance k from the origin, then the radius of the sphere will be √(r 2 - k 2 ). So the 4-dimensional volume = ∫ -r r 4π/3 (r 2 - x 2 ) 3/2 dx = 4π r 4 /3 ∫ -1 1 (1 - t 2 ) 3/2 dt. Let K = (1 - t 2 ) 3/2 dt. We find that A = t(1 - t 2 ) 3/2 differentiates to K - 3t 2 (1 - t 2 ) 1/2 . B = t(1 - t 2 ) 1/2 differentiates to (1 - t 2 ) 1/2 - t 2 (1 - t 2 ) -1/2 . C = sin -1 t differentiates to (1 - t 2 ) -1/2 . Hence B + C differentiates to 2(1 - t 2 ) 1/2 , so 2A + 3(B + C) differentiates to 8K. So ∫ -1 1 (1 - t 2 ) 3/2 dt = ( terms involving powers of (1 - t 2 ) 1/2 + 3/8 sin -1 t )| -1 1 = 3 π /8. Hence the 4-dimensional volume is π 2 r 4 /2. The volume of the 3-dimensional surface is the derivative wrt r of the 4-dimensional volume (because the 4-dimensional volume for r + δr is the 4-dimensional volume for r + the 3-dimensional volume of the bounding sphere times δr). Hence the 3-dimensional volume is 2 π 2 r 3 . 11th Putnam 1951 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002