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Surface of revolutions

  1. May 4, 2012 #1
    When I learned solids of revolutions, we divided the solid into infinitely thin "disks" and then summed up their individual volumes to get the volume of the surface.

    Now in surface of revolutions, we are divided the solid into infinitely thin fustrums and summing up their individual surface area to get the total surface area of the solid.

    My question is: Why can't we just use disk to find the surface of revolution by summing up the surface areas of infinitely thin disks? Why use a fustrum for the surface of revolution but not for the solid of revolution?

  2. jcsd
  3. May 5, 2012 #2


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  4. May 5, 2012 #3
    It seems that surface of revolution only works with fustrum but not with disks.
    I have done the calculations, and because the formula for surface of a disk is different from that for a fustrum, the integrals come out different.

    Interesting, for solids of revolutions, it doesn't matter whether you use disks or fustrums.

    Or maybe I am making a mistake.

  5. May 5, 2012 #4


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    Hi Bipolarity! :smile:
    Because the error in the volume is πx(dx)2tanθ / πx2dx, which -> 0 as dx -> 0, so i's ok to use a disc instead of a frustrum.

    (And the error in the area is secθ - 1)
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