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Solid of revolution knowing only area?

  1. Jan 28, 2013 #1
    Suppose I have a region R whose boundary extremely complicated. While it would take me hundreds of years to approximate the boundary with formulae, I can easily estimate the area of R within a desired precision. I want to find the volume of the solid of revolution of R .

    My intuition told me that I should be able to integrate some function solely dependent on the area with respect to θ from 0 to 2π .

    However, using what knowledge I already have, this is not the case. For instance, the volume of the solid obtained by rotating a right triangle whose base is r and height is h about the axis coninciding with the leg of length h is given by the well-known formula [tex]\frac{1}{3}\pi r^{2}h=\pi r(\frac{1}{2}rh)\cdot\frac{2}{3}=\frac{2A\pi r}{3} [/tex]. The volume of a solid of a similar solid formed by a rectangle of base r and height 1/2 h is given by the well-known fromula [tex]\pi r^{2}(\frac{1}{2}h)=\pi r(\frac{1}{2}rh)=A\pi r [/tex]. The solids are both generated by regions of equal area, yet the volumes are different by a factor of 2/3 .

    My question is what parameters are involved in this process? Clearly, I am missing at least one.
     
  2. jcsd
  3. Jan 28, 2013 #2

    mfb

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    It is not sufficient to know the area - you have to know how far away this area is from the rotation axis "on average" (as integral).
     
  4. Jan 28, 2013 #3
    How would the centroid of the region figure into my volume? I am picturing a small portion of the solid subtending an angle of dθ.
     
  5. Jan 28, 2013 #4

    mfb

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    That is not enough. If you can calculate the average x (which leads to the barycenter) and the average x^2 (in physics, this would be related to the inertial moment) for your area, this might work. x is the distance to the rotation axis here.
     
  6. Jan 28, 2013 #5

    rollingstein

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    Pappus' 2nd centroid theorem

    The second theorem states that the volume V of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d traveled by its geometric centroid.

    http://en.wikipedia.org/wiki/Pappus's_centroid_theorem
     
  7. Jan 28, 2013 #6
    Bingo. Thanks.
     
  8. Jan 29, 2013 #7
    There's no proof provided in that page, but you can prove it by doing the "natural" process. That is, imagine the solid as a layered cake whose layers are all Δy tall. Take one particular slice subtending an angle of Δθ. Then, you estimate the volume of a slice by adding the volumes of Δθ-slices of cylinders of height Δy and radius x (dependent on y, so it's different at each layer). Taking the limit as Δy→0, you get an integral that you can simplify using the properties of the x-coordinate of the center of mass. Then, just add up all the slices.
     
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