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Math Help - Where is the error in this iintegration?

  1. #1
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    Where is the error in this iintegration?

    The classic derivation of the volume of a cone by the shell method in integration starts with the cone right-side up. And of course there are other ways of finding the volume of a cone by integration. I know, so please do not send me one of these derivations. I just want to see the mistake in the following derivation.
    Take an inverted cone, so that the tip is at the origin, the maximum radius is the constant R, and the maximum height is the constant H, and h and r stand for the variables of radius and height. Then
    Let "Int" = The integral from r= 0 to r = R
    Volume= Int 2*pi* r*h dr
    Int 2pi*r (H/R) r dr
    2pi (H/R Int r^2)dr
    2pi (H/R) r^3/3 from 0 to R
    2pi*H(R^2)/3
    Which is a factor of 2 too much. Where is the error?

    Thanks.
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  2. #2
    Grand Panjandrum
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    Re: Where is the error in this iintegration?

    Quote Originally Posted by nomadreid View Post
    The classic derivation of the volume of a cone by the shell method in integration starts with the cone right-side up. And of course there are other ways of finding the volume of a cone by integration. I know, so please do not send me one of these derivations. I just want to see the mistake in the following derivation.
    Take an inverted cone, so that the tip is at the origin, the maximum radius is the constant R, and the maximum height is the constant H, and h and r stand for the variables of radius and height. Then
    Let "Int" = The integral from r= 0 to r = R
    Volume= Int 2*pi* r*h dr
    Int 2pi*r (H/R) r dr
    2pi (H/R Int r^2)dr
    2pi (H/R) r^3/3 from 0 to R
    2pi*H(R^2)/3
    Which is a factor of 2 too much. Where is the error?

    Thanks.
    What is the surface area of a cone?

    CB
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  3. #3
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    Re: Where is the error in this iintegration?

    I presume you mean that, instead of using infinitesimal cylinders, I should be using infinitesimal truncated cones?
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  4. #4
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    Re: Where is the error in this iintegration?

    Ah, never mind, I found the error. And it wasn't anything to do with the surface area of cones.
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