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Double Integral Help

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A lamina occupies the region inside the circle x^2 + y^2 = 2y but outside the circle x^2 + y^2 = 1. Find the center of mass if the density at any point is inversely proportional to its distance from the origin.

    2. Relevant equations

    Xcm = double integral of y*f(x,y)
    Ycm = double integral of x*f(x,y)

    3. The attempt at a solution

    I know my integrand will be k(constant)/r
    and I know how to solve for centre of mass, my only trouble is setting up the double integral. I think these are circles centred at 2 and 0, and when I draw them in cartesian coords, my region R is the space in between them that looks like a crescent moon. I don't know how to set up my double integral, because I can't express the first circle in terms of y, since there are two terms of them. Please help?
  2. jcsd
  3. Feb 1, 2009 #2


    Staff: Mentor

    The first circle is centered at (0, 1), which you can see by completing the square. The second circle is centered at (0, 0). The radius of both circles is 1.

    Because of the symmetry of the lamina and of the density function, the center of mass will be on the y-axis, so you only need to find the mass of the region and the moment about the x-axis to find Ycm. The double integral looks like this:
    [tex]Mx =\int \int_A x \frac{k}{\sqrt{x^2 + y^2}} dA[/tex]

    Now you need to describe the region A. Whether you have vertical strips that sweep left to right, or horizontal strips that sweep from bottom to top, things are complicated by the fact that the left and right boundaries change with y and the lower and upper boundaries change with x.

    This problem is really ripe for changing to polar coordinates, it seems to me, since the region A has a somewhat simpler description in polar coordinates, and your density function is just k/r. The lower circle is just r = 1. The upper circle is a bit more complicated, but you can convert with the formulas
    x = r cos(theta)
    y = r sin(theta)
    r^2 = x^2 + y^2

    The only remaining matter is that dA = dx dy = dy dx in Cartesian coordinates, but dA = r dr dtheta in polar coordinates.

    Hope that helps.
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