Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    15 Dec '08 05:33
    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
  2. Subscriber deriver69
    Keeps
    15 Dec '08 13:41
    I would simplify the process and use the points (0,0), (h,0) and (h,r)

    the line is then y=r/h*x and then work out the integral of pi*(r/h*x)^2 between 0 and h.
  3. Standard member TheMaster37
    Kupikupopo!
    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....
  4. Subscriber joe shmo On Vacation
    Strange Egg
    15 Dec '08 15:36 / 1 edit
    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....
    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

    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!
  5. Subscriber joe shmo On Vacation
    Strange Egg
    16 Dec '08 02:45
    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....
    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?
  6. Standard member TheMaster37
    Kupikupopo!
    17 Dec '08 13:41
    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?
    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.
  7. Subscriber joe shmo On Vacation
    Strange Egg
    17 Dec '08 14:57 / 1 edit
    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.
    whoops, I missed that small detail..haha

    Thanks again