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MathGroup Archive 2002

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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/



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