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Centroid formula homework help

  1. Jul 21, 2006 #1
    Ok, I think I've figured this out but the book gives me different answer for my x value:

    [tex]y=4-x^2[/tex] [tex]y=0[/tex]

    [tex]A= \int_0^2(4-x^2)dx[/tex] A=16/3

    [tex]x bar= \frac{3}{16}\int_0^2 x(4-x^2)dx[/tex]

    and this comes out to 3/4..The book says 0 for the X-value...where did I go wrong?
     
  2. jcsd
  3. Jul 21, 2006 #2

    0rthodontist

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    Your region goes from -2 to 2 on the x-axis, not 0 to 2.
     
  4. Jul 21, 2006 #3
    ohhhhhh ya..good poinT.
     
  5. Jul 21, 2006 #4
    but wont that give me A=0?
     
  6. Jul 21, 2006 #5
    Well...

    [tex]
    A = \int\limits_{ - 2}^2 {\left( {4 - x^2 } \right)dx} = \left[ {4x - \frac{{x^3 }}{3}} \right]_{ - 2}^2 = \left( {8 - \frac{8}{3}} \right) - \left( { - 8 + \frac{8}{3}} \right) \ne 0
    [/tex]

    BTW since you're finding the centroid of a 2D solid shouldn't you use a double integral? Or is there some sort of formula that you're already using.
     
  7. Jul 22, 2006 #6
    Alternatively, you can deduce the x-coordinate of the centroid by symmetry.
     
  8. Jul 22, 2006 #7
    oh..ok..No there is a formula..Find A..and then the X and Y coordinates
     
  9. Jul 23, 2006 #8

    HallsofIvy

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    Yes, there is a formula: but that doesn't mean you can't use your intelligence!

    The "centroid" of a figure is the geometric center. Since you figure is symmetric about the y-axis, obviously the centroid must be on that axis: the x coordinate of the centroid is 0.
     
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