1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

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


    User Avatar
    Science Advisor

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted