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Centroid of a solid of revolution

  1. Oct 31, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    The curve 3x2+2y2-12y=32 is rotated about the x-axis and forms a solid hemisphere.

    Verify that the weight is 8cm from the bottom of the hemisphere.


    2. Relevant equations



    3. The attempt at a solution

    Now, I can only do a little bit in centroids but that is for just plane surfaces. I do not think I know how to do it when it is rotated.

    I thought it would be similar to the formula for centroids like

    [tex]\overline{x}=\frac{\int x dA}{\int dA}[/tex]

    would be similar to

    [tex]\overline{x}=\frac{\int x dV}{\int dV}[/tex]

    when rotated. I need some help in doing it.


    EDIT:

    I considered a cylindrical element of radius x and width dy.
    So that the volume of this small element is [itex]dV= \pi x^2 dy[/itex]

    Then I should have to take the first moment of volume about the x-axis? (like the first moment of area)
    But I am not sure how to take this moment of volume
     
    Last edited: Oct 31, 2008
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  3. Nov 1, 2008 #2

    HallsofIvy

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    This makes no sense at all. Rotating 3x2+ 2y2- 12y= 32 around the x-axis does NOT form a "solid hemisphere" and there is no "weight".

    No, it wouldn't be. If a figure is rotated around the x-axis then it would be [itex]\pi y^2 dx[/itex].

     
  4. Nov 1, 2008 #3

    rock.freak667

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    sorry...I meant rotated around the y-axis.
     
  5. Nov 1, 2008 #4

    HallsofIvy

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    Rotating that around the y-axis still does not give a "solid hemisphere", it gives an ellipsoid.

    It shoule be obvious, by symmetry, that the x and z components of the centroid are 0. Yes, the y component of the centroid is given by
    [tex]\overline{y}= \frac{\int y dV}{\int dV}[/tex]

    I would recommend doing the integration in cylindrical coordinates except altered to use polar coordinates in the xz-plane rather than the xy-plane. That is, [itex]x= r cos(\theta)[/itex], [itex]y= y[/itex], [itex]z= r sin(\theta)[/itex] and [itex]dV= r sin(\theta)drd\theta dy[/itex]. Alternatively, just replace "y" by "z" in the original equation and find the "z" component of the centroid.
     
  6. Nov 1, 2008 #5

    rock.freak667

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    I have never done double/triple integrals as dv is in terms of three differentials. But the question just showed me a picture of a kind of semi-circle and said that was the curve.
     
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