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Area of ring = Circumference*dr

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data
    My professor put on the board today, that the area of a ring (used in discussion of moments of inertia) was = the circumference of the ring *dr = 2*pi*r*dr.

    This may sound trivial, but I cannot seem to work out in my head how this related to the formula I know for the area of a ring = pi*(r^2_outer-r^2_inner). I see the units are ok in either, but the factor of 2 is confusing me.

  2. jcsd
  3. Feb 8, 2012 #2


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    Call the inner radius r and the outer radius r+dr. Then the area is, as you said:
    A = pi((r+dr)^2 - r^2) = pi(r^2 + 2r dr +dr^2 - r^2) = pi (2r dr +dr^2)

    Now as dr gets very small, the dr^2 term becomes vanishingly small compared to the 2 r dr term, so we can ignore it, giving:

    A = 2 pi r dr

    which is what your professor said.
  4. Feb 8, 2012 #3
    Thank you very much.
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