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Calculus, center of mass

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the center of mass of the 2-dimensional plate which occupies the region inside the circle x^2 + y^2 = 2y, but outside the circle x^2 + y^2 = 1, and for which the density is inversely proportional to its distance from the origin.

    2. Relevant equations

    m = integral (p(x,y)) dA

    3. The attempt at a solution

    -> m = integral k/r
    x^2 + (y-1)^2 = 1

    It's not the integration itself I'm having trouble with just finding the boundaries of integration as they are between the two circles, any help/hints would be appreciated.
  2. jcsd
  3. May 13, 2010 #2
    From looking at a graph of the two circles, it appears that the two intersection points occur between the upper half of the circle x2 + y2 = 1 and the lower half of the circle x2 + (y-1)2 = 1. That is, between the functions

    [tex]y = \sqrt{1-x^2} \qquad \text{ and } \qquad y = 1-\sqrt{1-x^2}.[/tex]​

    From here you can find the intersection points.
  4. May 13, 2010 #3


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    Science Advisor

    First, do you see, from symmetry, that [itex]\overline{x}= 0[/itex]?

    The region you seek lies outside the unit circle but inside the circle with center at (0, 1) and radius 1. Those intersect where [itex]x^2+ (y-1)^2= 1= x^2+ y^2[/itex] or [itex]y^2- 2y+ 1= y^2[/itex] so [itex]y= 1/2[/itex]. Then [itex]x= \pm\sqrt{3}/2[/itex].

    There is a slight complication in that the circle actually extends to x=-1 and x= 1 so it is probably best to do this in three separate integrals.

    1) x from -1 to [itex]-\sqrt{3}/2[/itex], y from [itex]1- \sqrt{1- x^2}[/itex] to [itex]1+ \sqrt{1- x^2}[/itex] (from solving [itex]x^2+ (y- 1)^2= 1[/itex] for y).

    2) x from [itex]-\sqrt{3}/2[/itex], y from [itex]\sqrt{1- x^2}[/itex] (the lower circle) to [itex]1+ \sqrt{1- x^2}[/itex]

    3) x from [itex]\sqrt{3}/2[/itex] to 1, y from [itex]1- \sqrt{1- x^2}[/itex] to [itex]1+ \sqrt{1- x^2}[/itex].
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