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Area of a fence over a curve

  1. Mar 5, 2010 #1
    Think of a rectangular flag. When it is moved by the wind, we may think of the inferior border of the flag as drawing a curve in some plane. But, obviously, the area of the waving flag is the same of the rectangular flag.

    So, if I draw a curve of length L in a plane, and I set up a fence of constant height H over said curve, I suppose its area is LH. But what is the rigorous justification?? Line integrals??
  2. jcsd
  3. Mar 5, 2010 #2


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    The simplest way to do it is to write the lower curve as y= f(x) and the upper curve as y= f(x)+ L. That way, to find the area you are integrating (f(x)+ L)- f(x)= L over the length of the curve. That will be a constant no matter how you bend the curve.
  4. Mar 5, 2010 #3
    Thanks, but if the length of the curve at the base is L and the constant heigth is H, the area of this waving flag, according to your procedure, would be

    (Integral) [f(x) + H]dx - (Integral) f(x)dx = (Integral) Hdx.

    And that is not LH, which is the obvious area of a flag of length L and heigth H when it is "at rest".
  5. Mar 5, 2010 #4
    Probably I have been unclear with my statement.

    I draw a smooth curve in the plane xy. I know its length, it is L. I "build" a fence over it, reaching the same height H for every point of the curve.

    It is pretty obvious that this fence has area LH, but what is the rigorous justification? Do I need to go to line integrals to have it?
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