Hello 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
"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....
Originally posted by TheMaster37Thanks 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..lol๐๐
"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!
Originally posted by TheMaster37Actually, 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
"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?
Originally posted by joe shmoYou are right, it does come from the negative slope. I haven't made the transposition wich results in a minus sign there.
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.
Originally posted by TheMaster37whoops, I missed that small detail..haha
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 again๐