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