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Centroid of a uniform shape, using area

  1. Jul 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the coordinates of the centroid of the uniform area.
    20150728_223532_zpsyzygiicc.jpg

    2. Relevant equations
    equations for centroid coordinates at the top of my paper.

    3. The attempt at a solution
    20150728_223514_zpslatnbvec.jpg
     
  2. jcsd
  3. Jul 28, 2015 #2
    How does k go from the denominator to the numerator of your integrals as you are evaluating them?
     
  4. Jul 28, 2015 #3
    (2/3k)(58.1)
    =(.667k)(58.1)
    =38.7k
    where k is a constant. And i treated the other integral similarly.
     
  5. Jul 28, 2015 #4

    Nathanael

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    In one of your steps you basically said ##\sqrt{\frac{y}{k}} =\frac{1}{k}\sqrt{y}##

    Also Dr. Courtney's point still stands... it's not 2/3k it's 2/(3k)
     
  6. Jul 28, 2015 #5
    Ok thanks Dr. Courtney and Nathanael. I now have 38/k. In my final answer the k's cancel out and im left with the same 7.5/38.7.
     
  7. Jul 28, 2015 #6

    Nathanael

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    38.7/k is still not right. It should be 38.7/√k

    When you find the x-coordinate of the centroid, you should be integrating with respect to x.
    The "dA" in the formula ##\bar x=\frac{1}{A}\int xdA## is the area of the thin strip between x and x+dx.
     
  8. Jul 28, 2015 #7
    Wouldnt it be easier to integrate with respect to y, because of the shaded region is above the curve.
     
  9. Jul 28, 2015 #8

    Nathanael

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    It's not about what is easier, it's simply wrong.

    When you integrate with respect to y, you are taking horizontal strips of area, right? Well when you find the x-coordinate of the centroid you want to take vertical strips of area. The reason for this is that you want to take strips of area which all have the same x-value, and then multiply them by that x-value. You just can't do this when you integrate w.r.t. y.
     
  10. Jul 28, 2015 #9
    ##\bar x=\frac{1}{A}\int xdA##
    = ##1/2 \int kx^3dx## does this look ok?
     
  11. Jul 28, 2015 #10

    Nathanael

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    kx2dx is the area under the curve. Try to figure out a way to find the area above the curve.

    Also where did the 1/A go? And where did this 1/2 come from? (Are you saying the area is 2?)
     
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