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Find the centroid

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the centroid bounded by the given curves.
    y = x^3 , x+y = 2, y = 0
    So if you graph this you will see that if you did it with respect to x you would have the break the graph at the point (1,1). Or at least I think you would have to. So in my book they give the equations


    2. Relevant equations

    x bar = (1/A)∫ x(f(x) - g(x)) dx and y bar = (1/A)∫ (1/2) ( f(x)^2 - g(x)^2 ) dx

    3. The attempt at a solution


    So I was thinking what if I just turned did it with respect to Y.


    So would it be OK if I did.

    y bar = (1/A)∫ y(f(y) - g(y)) dy and x bar = (1/A)∫ (1/2) ( f(x)^2 - g(x)^2 ) dx

    Would this be OK to try? I just want to make sure I have it straight before I plug and chug.
    Thanks,
    J
     
  2. jcsd
  3. Oct 17, 2013 #2

    verty

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    Homework Helper

    I think this is similar to a center of mass calculation, you can read more here.
     
  4. Oct 17, 2013 #3

    Mark44

    Staff: Mentor

    If I'm understanding what you're asking, no, you can't do this. The geometry of the two integrands is very different, so you can't just arbitrarily switch from integrating w.r.t x to integrating w.r.t. y.

    For ##\bar{x}##, the typical integration element is the moment for a vertical slice, and looks like x * (yupper - ylower)Δx.

    For ##\bar{y}##, the typical integration element is the moment for a horizontal slice, and looks like y * (xright - xleft)Δy.
     
  5. Oct 17, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The definition of "centroid" of a region is [itex]\bar{x}= \frac{1}{A}\int x dA[/itex] and [itex]\bar{y}= \frac{1}{A}\int ydA[/itex] where A is the area of the region and dA is the differential of area.

    In particular, if the region lies between y= f(x) and y= g(x), x between a and b, with f(x)< g(x) for all x in [a, b], then [itex]\int xdA= \int_{x= a}^b\int_{y= (x)}^{g(x)} x dy dx= \int_a^b x(f(x)- g(x))dx[/itex] and [itex]\int_a^b \int_{f(x)}^{g(x)} y dydx= \int_a^b \frac{1}{2}\left[y^2\right]_{f(x)}^{g(x)} dx[/itex]
    [itex]= \frac{1}{2}\int_a^b g^2(x)- f^2(x) dx[/itex], just as you say.

    But if the region lies between x= f(y) and x= g(y), y between a and b, with f(y)< g(y) for all y in [a, b], you can also do [itex]\int dA[/itex] reversing the orders of x and y. That would give [itex]\int xdA= \int_{y= a}^b\int_{x= f(y)}^{g(y)} x dxdy= \frac{1}{2}\int_a^b g^2(y)- f^2(y) dy[/itex] and [itex]\int ydA= \int_{y= a}^b \int_{x= f(y)}^{g(y)} y dxdy= \int_a^b y(f(y)- g(y))dy[/itex].

    In other words, all you are doing is "swapping" x and y. And, of course, it really doesn't matter which axis you call "x" and which "y", as long as you don't get them confused.
     
    Last edited by a moderator: Oct 17, 2013
  6. Oct 17, 2013 #5
    Wait I'm confused. Mark44 said no and Hallsofivy yes? Or is HallsofIvy saying something different?
     
  7. Oct 17, 2013 #6
    Different I should of had dy
     
  8. Oct 17, 2013 #7
    Going back to my original post then.
    Find the centroid bounded by the given curves.
    y = x^3 , x+y = 2, y = 0
    A = 3/4
    So I did x = 1/A (1/2) ∫ (2-y)^2 -(y)^1/3 dy

    = 2/3 ∫ 4 - 4y +y^2 - y^(2/3) dy
    = (2/3) [ 4y -2y^2 + y^3/3 - (3/5)y^5/3]
    evaluate from [0,1] and you get 52/45 = 1.155


    And y = (4/3) ∫ y(2-y - y^1/3) dy = (4/3) ∫ 2y -y^2 -^4/3 dy
    = (4/3)[y^2 - y^3 /3 - 93/7)^7/3 from [0,1]

    = 20/ 63 = .317460

    So centroid is (52/45 , 20/ 63)

    Yes no ?
     
  9. Oct 18, 2013 #8
    I think you made a mistake

    [itex]\int ydA= \int_{y= a}^b \int_{x= f(y)}^{g(y)} y dxdy= \int_a^b y(f(y)- g(y))dy[/itex].

    Should if be g(y) - f(y) ? Since f(y) <g(y) ?
     
  10. Oct 18, 2013 #9

    Mark44

    Staff: Mentor

    Yes. The last integral should have an integrand of y(g(y) - f(y)). The upper limit of integration is g(y) and the lower limit is f(y).
     
  11. Oct 18, 2013 #10

    Mark44

    Staff: Mentor

    That makes a difference. Also, I'm not sure I understood what you were asking, so we might have interpreted what you wrote in different ways.
     
  12. Oct 18, 2013 #11
    Ah. It;s Ok. I usually write like a 6 year old. Doesn't help ...ha!
     
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