A solid in Euclidean 3-space extends from z = -h/2 to z = +h/2 and the area of the section z = k is a polynomial in k of degree at most 3. Show that the volume of the solid is h(B + 4M + T)/6, where B is the area of the bottom (z = -h/2), M is the area of the middle section (z = 0), and T is the area of the top (z = h/2). Derive the formulae for the volumes of a cone and a sphere.

**Solution**

*Easy*

Let the polynomial be az^{3} + bz^{2} + cz + d. Then the volume is ∫_{-h/2}^{h/2} (az^{3} + bz^{2} + cz + d) dz = bh^{3}/12 + dh. But B + T = bh^{2}/2 + 2d, M = d, so h(B + 4M + T)/6 = bh^{3}/12 + dh, which proves the formula. For a sphere radius R, we have h = 2R, B + T = 0 and M = πR^{2}, so the formula gives 4/3 πR^{3}, as usual. For a cone height h, base area A, we have B = A, T = 0, M = A/4, so the volume is hA/3, as usual.

The original text of the problems and the official solutions are available in: A M Gleason, R E Greenwood & L M Kelly, The William Lowell Putnam Mathematical Competition, Problems and Solutions, 1938-1964, MAA 1980.

Out of print, but available in some university libraries.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999

Last corrected/updated 11 Nov 03