Trouble calculating a volume

*To*: mathgroup at smc.vnet.net*Subject*: [mg33806] Trouble calculating a volume*From*: "Jay D. Martin" <jdm111 at psu.edu>*Date*: Tue, 16 Apr 2002 03:52:15 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Respected Colleagues: I am attempting to calculate the volume of the frustum of a cone. I have been able to calculate the volume of a cone using a triple integral as follows: Integrate[1, {x, 0, h}, {y, (-r)*(x/h), r*(x/h)}, {z, -Sqrt[(r*(x/h))^2 - y^2], Sqrt[(r*(x/h))^2 - y^2]}] This integral does evaluate to the correct answer (1/3)*h*Pi*r^2 where r is the base diameter and h is the height. In order to calculate the volume of the frustum of a cone, I have used the following equation : Simplify[Integrate[1, {x, 0, h}, {y, -(r1 + ((r2 - r1)*x)/h), r1 + ((r2 - r1)*x)/h}, {z, -Sqrt[(r1 + ((r2 - r1)*x)/h)^2 - y^2], Sqrt[(r1 + ((r2 - r1)*x)/h)^2 - y^2]}]] where r1 and r2 are the top and bottom radii of the section and h is the height. This integral evaluates to include Logarithms and imaginary numbers (I*Sqrt[h^2*r2]*(r1^2 + r1*r2 + r2^2)* (Log[-2*I*h*r2] - Log[2*I*h*r2]))/(3*Sqrt[r2]) when the solution should be (1/3)*h*Pi*(r1^2+r1*r2+r2^2) Which, if you substitute r1=0 and r2=r you get the volume of a cone. I am using Mathematica Version 4.1.2.0 running on Windows 2000 Thank you for any help Jay Martin PS: what I am ultimately after are the moments of inertia of the section That is why I am using the triple integral in terms of x, y, and z