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Science Forum

  1. 14 Jun '10 21:13
    Anyone remember how to integrate to find surface area. Ugh.

    For example, a 150 ml erlemeyer, bottom diameter of 6 cm and top diameter of 2.5 cm.

    Outside circum. of 2*pie*r integrated from 1.25 to 3? I know I am missing something here.
  2. 14 Jun '10 21:36
    Originally posted by mlprior
    Anyone remember how to integrate to find surface area. Ugh.

    For example, a 150 ml erlemeyer, bottom diameter of 6 cm and top diameter of 2.5 cm.

    Outside circum. of 2*pie*r integrated from 1.25 to 3? I know I am missing something here.
    I think I found a solution:

    http://mathworld.wolfram.com/ConicalFrustum.html
  3. Subscriber joe shmo
    Strange Egg
    15 Jun '10 03:54 / 1 edit
    Originally posted by mlprior
    I think I found a solution:

    http://mathworld.wolfram.com/ConicalFrustum.html
    for your example I get

    119*Pi/16

    start by finding 2 points.

    I used; A(0,3) & B(7/4,5/4)

    Find the slope, which I get to be -1

    Next find equation of line

    y = 3-x

    then since lateral surface area of a right circular cylinder = 2*Pi*r*h

    r being f(x) and h = dx

    we have

    2*pi*Int{0, 7/4}[3-x]dx

    = 2*pi*[ 3x-1/2x^2 |{0,7/4}]

    = 119*pi/16 [mm^2]

    EDIT: but according to wolfram im missing a factor of sqrt(2)!?( bummer)
  4. Subscriber joe shmo
    Strange Egg
    15 Jun '10 15:43
    Originally posted by joe shmo
    for your example I get

    119*Pi/16

    start by finding 2 points.

    I used; A(0,3) & B(7/4,5/4)

    Find the slope, which I get to be -1

    Next find equation of line

    y = 3-x

    then since lateral surface area of a right circular cylinder = 2*Pi*r*h

    r being f(x) and h = dx

    we have

    2*pi*Int{0, 7/4}[3-x]dx

    = 2*pi*[ 3x-1/2x^2 |{0,7/4}]

    = 119*pi/16 [mm^2]

    EDIT: but according to wolfram im missing a factor of sqrt(2)!?( bummer)
    I found where I went wrong. Apparently the formula for a surface of revolution is based off the frustrum of a cone ( the very thing you seek to find ) and not the right circular cylinder.

    S = 2*pi*[Int{a,b}r(x)*Sqrt( 1 + (r'(x))^2 )]

    can anybody explain why it would need to be done this way?

    Eric