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Center of Mass (Triple Integral)

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    T is the solid bounded by the cylinder y^2+z^2=4 and the planes x=0 and x=3. The mass density at a point P of T is directly proportional to the distance between P and the yz-plane.

    Find the center of mass of the solid T.

    2. Relevant equations

    y^2+z^2=4

    x=0

    x=3

    3. The attempt at a solution

    I drew the solid and got a cylinder extending from x=0 (yz-plane) all the way to x=3 with a radius of 2.

    I also attempted to set up an integral but I think my main problem is figuring out what the density to integrate is.

    I set up my integral as the integral from x=0 to x=3, the integral from y= -2 to y=2, and the integral from
    z= -√(4-y^2) to z=√(4-y^2) dz dy dx.

    Is that correct? I don't know how to go about determining my p(x,y,x) aka my density.


    Edit: Initially I tried to use x as my density but I couldn't integrate that so I tried y and then z but none of them worked out.
     
    Last edited: Apr 14, 2013
  2. jcsd
  3. Apr 15, 2013 #2
    No one knows? :confused:
     
  4. Apr 15, 2013 #3

    tiny-tim

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    Science Advisor
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    Hi Reefy! :smile:

    (try using the X2 button just above the Reply box :wink:)
    Yes :smile:

    (but in this case the density is constant over each circular slice, so you could just use πr2, and integrate only over x :wink:)
    You can't integrate ∫ x dx ? :confused:

    (or ∫∫∫ x dxdydz)
     
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