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Choosing dr question

  1. Jun 4, 2009 #1
    in this solution:

    http://i39.tinypic.com/2e4w7f4.gif

    they chose thivkness to be dr

    althogh r is the radius of a sub cilinder.

    ??
     
  2. jcsd
  3. Jun 4, 2009 #2

    Mark44

    Staff: Mentor

    Read the paragraph that starts "To analyze the problem, we divide the silicon into concentric elements..." (emphasis added)
     
  4. Jun 4, 2009 #3
    i dont understand this thing of taking a small ring
    i can try and take a small 3d part
    breaking it into 2d rings
    and each ring breaks into small parts
    so we start from defining this small part:
    our radius of a ring is r
    so the length is rd\theta
    the width is dr
    so the area of one 2d ring is
    [tex]
    \int_{a}^{b}\int_{0}^{2\pi}rd\theta dr=\pi(b^2-a^2)
    [/tex]
    our formula for calculating R is [tex] R=\frac{L}{\sigma A}=\frac{L}{\sigma \pi(b^2-a^2)}[/tex]
    but they got other result

    where is the mistake in my way?
     
    Last edited: Jun 4, 2009
  5. Jun 4, 2009 #4

    Mark44

    Staff: Mentor

    Your mistake is that the area you are thinking about is different from the one in the article. You are thinking about a ring, or annulus, and the article is talking about the surface area of a concentric shell of silicon that runs the length of the coax conductor.

    The formula you show is correct for the area of an annulus (a ring), but not for the the surface area of a long cylindrical shell, just like what they show in figure a. Is that clear?
     
  6. Jun 4, 2009 #5
    but in the end it doesnt matter how we calculate the area
    i should get the same result
     
  7. Jun 4, 2009 #6

    Mark44

    Staff: Mentor

    It does matter, since you and the article are calculating different areas. You are calculating the area of a 2D shape that is a circle with a circular hole in it. The text is calculating the surface area of 2D shape wrapped around a long 3D cylinder. The two formulas are completely different.
     
  8. Jun 4, 2009 #7
    "area of 2D shape wrapped around a long 3D cylinder"
    i cant imagine that

    can you explain that term??
     
  9. Jun 4, 2009 #8

    Mark44

    Staff: Mentor

    OK, how about this. You know what a log (made of wood) is, right? That's a 3D object. Now wrap a sheet around the log. The sheet is a 2D object wrapped around a 3D object.
     
  10. Jun 5, 2009 #9
    so they are making an integral (sum of sheets).
    we calculate the resistivity of one sheet and make an integral for every sheet
    every time the r is changing
    why do i need dr??
    R=\frac{L}{\sigma A}=\int_{a}^{b}\frac{L}{\sigma 2\pi r}
     
  11. Jun 5, 2009 #10

    Cyosis

    User Avatar
    Homework Helper

    Staying with the log analogy. A log has tree-rings, each with a different radius. You need to find the resistance of all those rings together.
     
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