- 15 Dec '08 05:33Hello all,

Im trying to derive the formula for volume of a right circular cone fo no particular reason.

hope this is clear,

starting with a right triangle on the x- axis the points are

(a,0);(b,0);(a,z)

the base of the triangle (b-a), and the height (z-0)=z

using the disk method, the radius of the disk is:

z/(a-b)*(x-a) + z,

the equation of a line with a negative slope, passing through the point (a,z), and an x intercept (b,0).

so the volume should equal ( not sure how to format this, see key below)

INT( ) , Integral

a_b , Limits of integration, from a to b

INT(a_b) Pi[z/(a-b)*(x-a) + z]^2 dx

after expanding and evaluating the integral from a to b, I come to:

(Pi*z^2/(a-b)^2)(1/3*b^3 -1/3*a^3 -b^2*a + b*a^2)

but after plugging in some ramdom values for a,b, and z, I come up with 3 times the volume obtained using the formula 1/3*Pi*r^2*h

so I have given up on further simplification for the time being!!

If anyone is willing to put the time in to point out my error I would greatly appriciate it

thanks

Eric - 15 Dec '08 13:54 / 1 edit"INT(a_b) Pi[z/(a-b)*(x-a) + z]^2 dx"

This is correct, after that you go wrong. Writing out the square:

Pi*INT(a_b) [z^2 - (2z^2/(b-a))*(x-a) + (z^2/(b-a)^2)*(x-a)^2] dx

Calculating the integraal:

Pi*[ z^2 * x - (z^2/(b-a))*(x-a)^2 + (z^2/3(b-a)^2)*(x-a)^3] with boundaries a and b.

Evaluating in the endpoints:

Pi/3 * z^2 * (b-a) wich is exactly what you're after.

I would have chosen the points (0,0), (0,z) and (a,0) b.t.w.... - 15 Dec '08 15:36 / 1 edit

Thanks alot, I used Mathcad to do all the expansion to expidite the process and avoid any simlpe mistakes...go figure. Apparently I need to learn how to use the program better..lol*Originally posted by TheMaster37***"INT(a_b) Pi[z/(a-b)*(x-a) + z]^2 dx"**

This is correct, after that you go wrong. Writing out the square:

Pi*INT(a_b) [z^2 - (2z^2/(b-a))*(x-a) + (z^2/(b-a)^2)*(x-a)^2] dx

Calculating the integraal:

Pi*[ z^2 * x - (z^2/(b-a))*(x-a)^2 + (z^2/3(b-a)^2)*(x-a)^3] with boundaries a and b.

Evaluating in the endpoints:

Pi/3 * z^2 * (b-a) wich is ...[text shortened]... exactly what you're after.

I would have chosen the points (0,0), (0,z) and (a,0) b.t.w....

And I see that the points you and deriver whould choose are easier to work with. I'm new to all this, I tend to miss simplification shortcuts! - 16 Dec '08 02:45

Actually, I do have a question. In your expansion of the integrand where does the negative sign come from in front of the middle term of*Originally posted by TheMaster37***"INT(a_b) Pi[z/(a-b)*(x-a) + z]^2 dx"**

This is correct, after that you go wrong. Writing out the square:

Pi*INT(a_b) [z^2 - (2z^2/(b-a))*(x-a) + (z^2/(b-a)^2)*(x-a)^2] dx

Calculating the integraal:

Pi*[ z^2 * x - (z^2/(b-a))*(x-a)^2 + (z^2/3(b-a)^2)*(x-a)^3] with boundaries a and b.

Evaluating in the endpoints:

Pi/3 * z^2 * (b-a) wich is ...[text shortened]... exactly what you're after.

I would have chosen the points (0,0), (0,z) and (a,0) b.t.w....

Pi*INT(a_b) [z^2 - (2z^2/(b-a))*(x-a) + (z^2/(b-a)^2)*(x-a)^2] dx

Does it have something to do with the negative slope? I thought that was accounted for in the transposing of the addition from (a-b) to (b-a)......Im sure your correct, I just don't understand? - 17 Dec '08 13:41

You are right, it does come from the negative slope. I haven't made the transposition wich results in a minus sign there.*Originally posted by joe shmo***Actually, I do have a question. In your expansion of the integrand where does the negative sign come from in front of the middle term of**

Pi*INT(a_b) [z^2 - (2z^2/(b-a))*(x-a) + (z^2/(b-a)^2)*(x-a)^2] dx

Does it have something to do with the negative slope? I thought that was accounted for in the transposing of the addition from (a-b) to (b-a)......Im sure your correct, I just don't understand?

Had I made the transposition (like you have) then there would have been a plus sign in that place. - 17 Dec '08 14:57 / 1 edit

whoops, I missed that small detail..haha*Originally posted by TheMaster37***You are right, it does come from the negative slope. I haven't made the transposition wich results in a minus sign there.**

Had I made the transposition (like you have) then there would have been a plus sign in that place.

Thanks again