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Finding bounds of a centroid problem

  1. Apr 8, 2006 #1
    I am having a problem finding the upper and lower (x,y) bounds for this problem.

    Find the centroid of r = 1 + cos(theta) which lies in the 1st quadrant.

    I come up with (2,0) and (1,0) or the axis intercept points. Is this the correct way to go about it?


    m=((∫)[0]^2 ) (∫)[0]^1
     
  2. jcsd
  3. Apr 8, 2006 #2

    arildno

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    It is simplest to calculate the coordinates of the centroid with the use of polar representation.
    As a help, the area of the region is:
    [tex]\int_{0}^{\frac{\pi}{2}}\int_{0}^{1+\cos\theta}rdrd\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{3}{2}+2\cos\theta+\frac{\cos{2\theta}}{2}{d\theta}=\frac{3\pi}{8}+1[/tex]
    And, most importantly, remember the relations:
    [tex]x=r\cos\theta,y=r\sin\theta[/tex]
     
    Last edited: Apr 9, 2006
  4. Apr 9, 2006 #3
    Thanks so much for the assistance!
     
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