hgb>This is very cool! Do you have a reference to an article for amateurs? A couple of years ago, the Car Talk radio show and website featured the puzzle of how to quarter-fill a horizontal cylindrical tank, so a lot of math teachers may have been embarrassed already. Unfortunately, I didn't see their solution, and it's no longer obvious on their site. I also haven't seen the area problem phrased as a "parabolic" (epsilon = 1) case of Kepler's, but it's fairly obvious. An efficient area solution might be useful as the lead term of an expansion at epsilon = 1 of the high eccentricity Kepler case. Watson gives the solution in terms of Bessel functions as a standard Fourier expansion involving a slick integration by parts which turns Kepler's equation into one of the usual definite integral definitions of the Bessel function. I'll send it along if there's any interest. Twice I said

b) although the equation of motion of an elliptical orbit requires higher transcendental functions,[...] (The hard part is theta(t), which apparently even transcends Lambert_W.)

But just barely. t = time ~ M, the "mean anomaly", and the problem is to solve E - eps sin(E) = M for E, the "eccentric anomaly". In a (comfortable) neighborhood of M = 0, Kepler's eqn is (nonobviously) equivalent to | i M - eps cos(E) | |W(- e eps)| = eps.

Also, do the transcendental functions show up for "even" numbers of dimensions?

For dimensions 0..5, thickness x, the volume formula gives sqrt(x) 2 asin(-------) sqrt(2) v(0) = ---------------, v(1) = x, %pi sqrt(x) sqrt(2) asin(-------) (x - 2) (x - 1) sqrt(2) v(2) = - sqrt(2) (--------------- - ---------------------) sqrt(x), x sqrt(x) 2 sqrt(1 - -) 2 x 2 v(3) = %pi (1 - -) x , v(4) = sqrt(2) %pi sqrt(x) 3 sqrt(x) 2 3 sqrt(2) asin(-------) (x - 2) (x - 1) (2 x - 4 x - 3) sqrt(2) (-------------------------------- + -----------------------)/6, x sqrt(x) 2 sqrt(1 - -) 2 x 3 (1 - -) x 2 3 5 2 %pi x (1 - -----------) 4 v(5) = ---------------------------, 3 and the hypergeometric three-term relations give the recurrence v(n) = 2 %pi (2 %pi v(n - 4) (x - 2) x 2 - v(n - 2) ((n - 2) (x - 2 x - 1) - 1))/((n - 1) n), which proves transcendence iff n even. --rwg SELF-DEVOURING GIRDLE-OF-VENUS STREAMLINES RAIMENTLESS ETERNALISMS