Very neat argument --- but I still don't believe it. Why should it not work for a single cylinder? Just to placate those who haven't heard of principal values, or get distracted by minor matters like compactness (?), I'll give this large finite length and assert that the ratio S/V approaches 2 rather than 3, cones notwithstanding. On 1/3/11, Veit Elser <ve10@cornell.edu> wrote:
Minor simplification:
Each element of the surface dS is tangent to a plane whose distance from the origin is 1, so dV = (1/3)1 dS.
(We need at least two cylinders with non-colinear axes for X to be compact.)
Veit
On Jan 3, 2011, at 12:22 PM, Eugene Salamin wrote:
From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Sun, January 2, 2011 11:54:06 PM Subject: [math-fun] Cylinder puzzle
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin.
Let X denote the intersection of all the C_k's.
Prove that the surface area of X is exactly three times its volume.
--Dan _______________________________________________ Here is a proof for finitely many cylinders. Consider a cone of solid angle dΩ with vertex at the origin. Except for a set of measure zero, the cone intersects the boundary of X on a single cylinder. If r is the distance from the origin to the boundary point on the cone axis, then the cone has volume dV = r^3 dΩ/3, and intercepts surface area dS = r^2 dΩ sec i, where the inclination angle i is that between the cone axis and the normal to the surface element. For cylinders of unit radius r = sec i, so dS = 3 dV.
-- Gene
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