Gene, is there a rigorous proof of that using just Eucidean solid geometry, without calculus?
Yes. For a region R on the surface of the sphere, we define a volume V_R by taking the union of every line segment {OX : X in R}, where O is the centre of the sphere. Now we will compute this volume using solid geometry. Assume wlog the sphere is the sphere centred at the origin of radius r, and R is the region bounded between planes z = a and z = b, with a < b. Then the volume V_R can be computed as follows: V_R = pi*a*(r^2 - a^2)/3 - pi*b*(r^2 - b^2)/3 + Q where Q is the volume of the `slice' of sphere bounded between the planes z = a and z = b. To calculate Q, we appeal to Cavalieri's principle: https://cp4space.files.wordpress.com/2014/06/cavalieri.png Hence Q = pi*r^2*(b-a) - pi*b^3/3 + pi*a^3/3. Consequently, V_R = (2/3)*pi*r^2*(b-a). This is a function of b-a, so we are done. :) Sincerely, Adam P. Goucher
--Dan
----- That z is uniform in [-1,+1] is a consequence of the solid geometry theorem that the area of a sphere between two parallel planes depends only on the separation between the planes, and is independent of what part of the sphere lies between the planes. ----- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun