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Finding centroids

  1. Aug 15, 2006 #1
    I'm doing a little self tutoring here, but I'm having trouble finding centroids for geometric shapes. I have a book that gives the formula for x and y coordinates (something like x = int(x dA)/int(dA) and similar for y). I still don't understand it though. Can someone explain it in more detail or link me to somewhere where they do that? A simple example would help, something like finding the centroid of a 3x2 rectangle with a corner at the origin using the calculus formulas. I think once I understand a simple case I'll be able to do more complicated ones.

    Thanks!
     
  2. jcsd
  3. Aug 15, 2006 #2

    Hurkyl

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    As far as understanding goes:

    [tex]
    \bar{x} = \frac{\int_R x \, dA}{\int_R \, dA}
    [/tex]

    is just the expression for an average. The centroid is just the average position of the shape. (So, it's x-coordinate is the average x-coordinate over the shape)


    As far as calculuation goes, just do it! Make R the rectangle you described and compute the integrals! (Of course, for this particular shape, there are easier ways to figure out the average position)
     
  4. Aug 16, 2006 #3
    This might seem very stupid, but I don't know how to evaluate those integrals. When I'm finding the x coordinate, do I use the boundaries on the x axis? How do I evaluate an integral with an x inside and a dA? They're two different variables.
     
  5. Aug 16, 2006 #4

    Hurkyl

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    Oh, you've never done area integrals before? Now I understand your problem!

    Generally, you evaluate them by splitting them into two integrals. You can think of it as chopping your region up into lines. (or, if you're really clever, other interesting shapes) The outer integral is on the parameter that selects which line, and the inner integral integrates over that line.

    For example, if we chop your 3x2 integral up into vertical lines, then we have:

    [tex]
    \int_R x \, dA = \int_0^3 \int_0^2 x \, dy \, dx
    [/tex]

    (what do you get if you chop it up into horizontal lines?)

    Because it's a rectangle aligned with the coordinate axes, the integral is particularly easy. Usually, the bounds for inner integral are functions of the outer variable.
     
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