Re: Trouble calculating a volume

*To*: mathgroup at smc.vnet.net*Subject*: [mg33840] Re: [mg33806] Trouble calculating a volume*From*: Rob Pratt <rpratt at email.unc.edu>*Date*: Fri, 19 Apr 2002 02:28:56 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

On Tue, 16 Apr 2002, Jay D. Martin wrote: > 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 Using Simplify with assumptions {h >= 0, r2 > 0} yields your desired answer. Rob Pratt Department of Operations Research The University of North Carolina at Chapel Hill rpratt at email.unc.edu http://www.unc.edu/~rpratt/