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The Centroid

  1. Mar 26, 2009 #1
    I think I'm right when I say that the centroid of a 2D shape is found by the intersection of the lines that separate that shape into two shapes of equal areas.
    Is that correct? I don't want to (for now) think about it in terms of moments and integrals, because frankly, it's a little confusing.
    It would make sense if that's the case.

    Then lastly, say we have a parabola, just a standard y = x^2. It's centroid will be found on the y axis, but the exact value is only able to be determined by calculus. I think I'm correct when I say that a section of parabola (made with a horizontal cut) will be similar to the parabola before the cut. Also as a result of the similarity, a parabola will take up the same amount of space for a given rectangle with vertexes on (0, 0) and (x, y).

    Using this definition, we can say that the centroid of a parabola that extends to x = 10, y = 100 is found by this method:

    [tex]x = 10[/tex]
    [tex]y = x^2 = 100[/tex]
    [tex]A_{rectangle}= xy = x^3 = 1000[/tex]

    [tex]A_{underparabola} = \int_{0}^{10} x^2 dx = 333.3333...[/tex]

    [tex]A_{rectangle} - A_{underparabola} = A_{parabola}[/tex]
    [tex]1000 - 333.333.... = 666.666...[/tex]

    [tex]A_{parabola} / A_{rectangle} = P:A = 0.666...[/tex]

    P:A is the ratio this parabola takes of it's envelope rectangle (should be constant for all values of [tex]\infty > x > 0 [/tex]
    Now the area of the parabola is halved (which gives the area of the lower, parabola shaped section:

    [tex]666.666... / 2 = 333.333....[/tex]
    [tex]333.333... / P:A = A_{smallrectangle} = 500[/tex]

    [tex]500 = xy = x^3[/tex]
    [tex]x = \sqrt[3]{500}[/tex]
    [tex]y = x^2[/tex]
    [tex]y = 500^{2/3}[/tex]

    y ~ 63
    Centroid = (0, 63)

    I'm pretty sure this is correct, but can someone who's a bit more senior confirm for me?
    Thanks.
     
  2. jcsd
  3. Mar 26, 2009 #2

    Doc Al

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    Staff: Mentor

    No. The lines must divide the shape into parts of equal moment about the line.
    Oh well...

    Think of it as finding the center of mass of the object.
     
  4. Mar 26, 2009 #3
    For the parabola example then, how would you do it?
     
  5. Mar 26, 2009 #4

    Doc Al

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    Staff: Mentor

    See: http://mathworld.wolfram.com/ParabolicSegment.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
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