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Volume of cathedral dome (Using volumes of revolution, disk method)

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A cathedral dome is designed with three semi circular supports of radius r so that each horiontal cross section is a regular hexagon. Show that the volume of the dome is r^3 * sqrt(3)

    an accompanying figure - http://imgur.com/3fSqh
    2. Relevant equations

    Vdisk=∫pi*(f(x))^2 dx
     
  2. jcsd
  3. Feb 10, 2012 #2

    LCKurtz

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    That dome is not a solid of revolution since the cross sections are not circles. What you need to do is figure out the area of the hexagonal cross section at height h and do the volume by integrating cross section areas for h from 0 to r.
     
  4. Feb 11, 2012 #3
    Okay, that helps a little bit. So I would derive an expression for the area of the base, then take the integral from 0 to h? (since h = r correct)?
     
  5. Feb 11, 2012 #4

    LCKurtz

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    No, that isn't correct. You don't start with the area of the base. You want the area of the upper hexagon in the picture, which will depend on h. h is the distance from the base to the plane of that hexagon, in other words the variable distance between the two red dots.
     
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