Centroid of the region bounded by the curve...need help

  1. centroid of the region bounded by the curve....need help!!!

    Find the centroid of the region bounded by the curve x=2-y^2 and the y-axis:

    my work shown:

    therefore if A= 2 times the integral of sqrt(2-x) dx

    is the M_x equal to the integral of (2-x) dx from 0 to 2?

    and the M_y equal to the integral of (2)(x)(sqrt(2-x) dx from 0 to 2?

    therefore x-coordinate of the centroid is M_y/A

    and the y-coordinate of the centroid is M_x/A

    therefore centroid is [(M/y/A),(M_x/A)]

    is this correct?

    then the x-coordinate of the centroid is (M_y / A)

    i've been told the centroid of the y-coord. is zero... .however i dont' believe that is correct.. how do i determine the centroid and are M_x and M_y values correct... because if they are ... isn't the centroid simply x--> M_y/A and y--> M_x/A... please help me with this problem!!!

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Re: centroid of the region bounded by the curve....need help!!!

    Why not work out your integrals on y instead of x, -2<=y<=+2?

    Then just work the problem out and see what you get without worrying in advance what it is supposed to be.
     
  4. Re: centroid of the region bounded by the curve....need help!!!

    i'm confused with what u mean.. do u mean integrate the equation in terms of x instead of y to determine the centroid.. how do i go about doing that...

    isn't my A(y) value correct, i just need help with my values for M_y, M_x assuming my A(y) is correct therefore i could find the centroid as M_y/A,M_x/A for co-ord. of (x,y)

    please help
     
  5. Re: centroid of the region bounded by the curve....need help!!!

    The given boundary function is x = f(y), so you can calculate the area and the moments as integrals that involve things like
    A = int(x) dy from -2 to 2, etc.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook