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The centroid of a hemispherical surface

  1. Nov 16, 2005 #1
    I have to find the centroid of a hemispherical surface with radius r.
    I want to use Pappus's theorem.S=2*Pi*x*s
    The surface S, is 2*Pi*r^2 and the length of the curve s, is Pi*r so the distance from the centroid ,which lies on the y axis by symmetry, to the x axis is S/(2*Pi*Pi*r) = r/Pi

    This is not the answer by text book gives for the y coordinate of the centroid. What am I doing wrong? Please help me.
     
  2. jcsd
  3. Nov 17, 2005 #2

    Fermat

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    Homework Helper

    x is the centroid of the generating surface which is a semicircle. And the centroid of a semicircle (a 2D object) is different to that of a hemispherical surface (a 3D object). I'm afraid you can't use Pappus.
     
    Last edited: Nov 18, 2005
  4. Nov 17, 2005 #3

    Astronuc

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    Staff: Mentor

    Try this instead - http://mathworld.wolfram.com/Hemisphere.html

    I believe the hemispherical is a semi-circle rotated by pi, or a quarter circle rotated by 2pi about an axis corresponding to a radius and corresponding to either end of the arc.
     
  5. Nov 18, 2005 #4
    I understand I can't use Pappus's theorem.
    So I try to cut the hemispherical shell into horizontal shells up the y axis. Each shell should have the surface 2*Pi*R where R=sqrt(r^2-y^2)
    So dS = 2*Pi*R dy
    But there is something wrong here, When I integrate from 0 to r I don't get the right answer for the surface of the hemispherical surface.
    Could you help me some more, please.
     
  6. Nov 18, 2005 #5
    I got it right. I had to calculate the width of the shell.And then integrate.
     
  7. Feb 15, 2011 #6
    How can I calculate the centroid of the hemisphere surface using only simple integrals and rectangular coordinates?
     
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