3rd Putnam 1940

Problem B2

C1, C2 are cylindrical surfaces with radii r1, r2 respectively. The axes of the two surfaces intersect at right angles and r1 > r2. Let S be the area of C1 which is enclosed within C2. Prove that S = 8r22A = 8r12C - 8(r12 - r22)B, where A = ∫01 (1 - x2)1/2(1 - k2x2)-1/2 dx, B = ∫01 (1 - x2)-1/2(1 - k2x2)-1/2 dx, and C = ∫01 (1 - x2)-1/2(1 - k2x2)1/2 dx, and k = r2/r1.




It is hard to see the point of the second half [ 8r22A = 8r12C - 8(r12 - r22)B ], which is trivial. Multiply top and bottom of the integrand in A by (1 - x2)1/2, so that we get (1 - x2) on the top. Then note that k2(1 - x2) = 1 - k2x2 + (k2 - 1).

For the first half, the part of C1 enclosed inside C2 comprises two bent ovals, one at each end. It is tempting to think that when rolled flat, each piece is an ellipse. If that were true, then the problem would be trivial - the semi-axes are r2 and r1sin-1k, so the total area would be 2π r1r2sin-1k. But conics only remain conics when projected onto flat surfaces and here we are projecting onto a curved surface.

So we have to use integration. Take C1 to be x2 + y2 = r12, and C2 to be y2 + z2 = r22. We calculate the area of the quarter of the piece with x > 0 having y, z > 0. We may divide it into strips parallel to the z-axis. Take the angle θ to be the axial angle - the angle by which the strip has to be rotated about the z-axis from the y = 0 position (keeping in the surface C1). [Points on the strip have the same x and y coordinates, but varying z-coordinates.] The strip has y-coordinate r1sin θ and hence length (r22 - r12sin2θ)1/2 and width r1dθ. Thus the area of the quarter-oval is ∫0sin-1k (r22 - r12sin2θ)1/2 r1 dθ. Put t = 1/k sin θ, and the integral becomes r2201 (1 - t2)1/2 (1 - k2t2)-1/2 dt. Hence the complete oval is 4 times this, and the total area 8 times.

Comments. The original question was badly worded, because at first sight it appeared to be about volumes, not areas, which may have confused some. It is important to get a clear picture of the geometry - the easiest way to do this is to roll a piece of paper into a cylinder.



3rd Putnam 1940

© John Scholes
15 Sep 1999