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The quarter disk in the first quadrant bounded by x^2+y^2=4

  1. Nov 16, 2015 #1
    • members are reminded that homework-type questions belong in our homework forums, and we require that you show your attempt at the solution
    Find the coordinate of center of mass.
    Given: The quarter disk in the first quadrant bounded by x^2+y^2=4
    I tried to solve this problem but can't figure out how to do it.
    so y integration limit is: 0 <= y <= sqrt(4-x^2))
    x limit of integration: 0 <= x <= 2

    and then after the dy integral I got something really messy
    integral from 0 to 2(sqrt(4-x^2)+x^2sqrt(4-x^2)+(sqrt(4-x^2)^3/3)dx)
    Where did I do wrong?
     
  2. jcsd
  3. Nov 16, 2015 #2

    Svein

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    You could look at it another way: The quarter disk is given by R=2, φ∈[0, π/2].
     
  4. Nov 16, 2015 #3

    HallsofIvy

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    The area of the quarter circle with radius 2 is, of course, [itex](1/4)\pi (4)= \pi[/itex]. Because of symmetry the x and y coordinates of the centroid (I would not say "center of mass" since we are not given a density) are the same:
    [tex]\overline{x}= \overline{y}= \frac{\int xdA}{A}= \frac{\int ydA}{A}[/tex].

    Personally, I would do the integration in polar coordinates, as Svein suggests:
    [tex]\int xdA= \int_{r= 0^2}\int_{\theta= 0}^{\pi/2} (r cos(\theta) rd\theta dr= \left(\int_0^2 r^2dr\right)\left(\int_0^{\pi/2} cos(\theta)d\theta\right)[/tex]

    But since you specifically ask about the integration in xy- coordinates,
    [tex]\int x dA= \int_0^2\int_0^{\sqrt{4- x^2}} x dydx= \int_0^2x\sqrt{4- x^2}dx[/tex]
    and now use the substitution [itex]u= 4- x^2[/itex].

    I can't say what you did wrong because you don't show what you did!
     
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