Re: Trouble calculating a volume
- To: mathgroup at smc.vnet.net
- Subject: [mg33830] Re: [mg33806] Trouble calculating a volume
- From: BobHanlon at aol.com
- Date: Fri, 19 Apr 2002 02:28:07 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 4/16/02 7:10:06 AM, jdm111 at psu.edu writes:
>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
>
Use the assumptions option
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]}],
Element[{h,r1,r2}, Reals]&&h>=0 && r2>0]
(1/3)*h*Pi*(r1^2 + r2*r1 + r2^2)
Bob Hanlon
Chantilly, VA USA