From: Pertti Lounesto Subject: Re: Volume and surface Date: Fri, 20 Aug 1999 17:45:20 +0200 Newsgroups: sci.math Keywords: can of paint with finite volume but infinite surface area "Clifford J. Nelson" wrote: > I thought: given a can of paint to be applied with a given thickness on an > object, what shape (meaning polyhedron) should be chosen for the object to > minimize the volume. Irregular tetrahedron? I learned this in Calculus I: Is there a can of paint, filled with paint, but with so little paint inside that one cannot paint the surface of the can? The answer is. "Yes". Here is the solution: Consider the graph y = 1/x, x > 1. Rotate the graph around the x-axis to form the can. The volume of the can is V = int_1^infty pi*(1/x)^2 dx = pi*(-1/infty-(-1/1)) = pi. The surface area of the can is more than S = 2*int_1^infty 1/x dx = 2*(log(infty)-log(1)) = infty. In fact, S is the area of a cardboard, between the line x = 1 and the curves y = +-1/x, dipped into the can. Thus, the can filled with paint, contains so little paint that it cannot even paint a card within the can. ============================================================================== From: Clive Tooth Subject: Re: Gabriel's Horn Date: Fri, 19 Mar 1999 23:09:42 +0000 Newsgroups: sci.math Andrew Theisen wrote: > Help! Please! > I am looking for a proof showing that Gabriel's Horn has both a finite > volume and infinite surface area. Thank you in advance. Gabriel's Horn is the surface of revolution obtained by rotating the curve y=1/x, for x>=1, about the x axis. The volume is obtained by integrating pi y^2 dx from 1 to infinity. The surface area is obtained by integrating 2pi y sqrt(1+(dy/dx)^2) dx from 1 to infinity. The first integral is easy. As for the second, note that sqrt(1+(dy/dx)^2)>1. Post again if you need more help. -- Clive Tooth http://www.pisquaredoversix.force9.co.uk/ End of document