C_{1}, C_{2} are cylindrical surfaces with radii r_{1}, r_{2} respectively. The axes of the two surfaces intersect at right angles and r_{1} > r_{2}. Let S be the area of C_{1} which is enclosed within C_{2}. Prove that S = 8r_{2}^{2}A = 8r_{1}^{2}C - 8(r_{1}^{2} - r_{2}^{2})B, where A = ∫_{0}^{1} (1 - x^{2})^{1/2}(1 - k^{2}x^{2})^{-1/2} dx, B = ∫_{0}^{1} (1 - x^{2})^{-1/2}(1 - k^{2}x^{2})^{-1/2} dx, and C = ∫_{0}^{1} (1 - x^{2})^{-1/2}(1 - k^{2}x^{2})^{1/2} dx, and k = r_{2}/r_{1}.

**Solution**

Straightforward.

It is hard to see the point of the second half [ 8r_{2}^{2}A = 8r_{1}^{2}C - 8(r_{1}^{2} - r_{2}^{2})B ], which is trivial. Multiply top and bottom of the integrand in A by (1 - x^{2})^{1/2}, so that we get (1 - x^{2}) on the top. Then note that k^{2}(1 - x^{2}) = 1 - k^{2}x^{2} + (k^{2} - 1).

For the first half, the part of C_{1} enclosed inside C_{2} 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 r_{2} and r_{1}sin^{-1}k, so the total area would be 2π r_{1}r_{2}sin^{-1}k. 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 C_{1} to be x^{2} + y^{2} = r_{1}^{2}, and C_{2} to be y^{2} + z^{2} = r_{2}^{2}. 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 C_{1}). [Points on the strip have the same x and y coordinates, but varying z-coordinates.] The strip has y-coordinate r_{1}sin θ and hence length (r_{2}^{2} - r_{1}^{2}sin^{2}θ)^{1/2} and width r_{1}dθ. Thus the area of the quarter-oval is ∫_{0}^{sin-1k} (r_{2}^{2} - r_{1}^{2}sin^{2}θ)^{1/2} r_{1} dθ. Put t = 1/k sin θ, and the integral becomes r_{2}^{2} ∫_{0}^{1} (1 - t^{2})^{1/2} (1 - k^{2}t^{2})^{-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.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999