### 11th Putnam 1951

Problem B7

In 4-space let S be the 3-sphere radius r: w2 + x2 + y2 + z2 = r2. What is the 3-dimensional volume of S? What is the 4-dimensional volume of its interior?

Solution

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 √(r2 - k2). So the 4-dimensional volume = ∫-rr 4π/3 (r2 - x2)3/2 dx = 4π r4/3 ∫-11 (1 - t2)3/2 dt.

Let K = (1 - t2)3/2 dt. We find that A = t(1 - t2)3/2 differentiates to K - 3t2(1 - t2)1/2. B = t(1 - t2)1/2 differentiates to (1 - t2)1/2 - t2(1 - t2)-1/2. C = sin-1t differentiates to (1 - t2)-1/2. Hence B + C differentiates to 2(1 - t2)1/2, so 2A + 3(B + C) differentiates to 8K. So ∫-11 (1 - t2)3/2 dt = ( terms involving powers of (1 - t2)1/2 + 3/8 sin-1t )|-11 = 3 π /8. Hence the 4-dimensional volume is π2 r4/2.

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 π2 r3 .