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Homework Help: Triple Integral in Cylindrical Coordinates

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the mass and center of mass of the solid S bounded by the paraboloid [tex]z=4x^2+4y^2[/tex] and the plane [tex]z=a\;\;(a>0)[/tex] if S has constant density K.


    2. Relevant equations
    In cylindrical coordinates, [tex]x^2+y^2=r^2[/tex].


    3. The attempt at a solution
    In order to find the mass, I tried
    [tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K\;drdzd\theta[/tex]
    This didn't seem to work, though.
     
  2. jcsd
  3. Nov 10, 2008 #2

    tiny-tim

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    Hi daveyman! :smile:

    Hint: the volume element isn't drdzdθ. :wink:
     
  4. Nov 10, 2008 #3
    Okay, so this is the mass:

    [tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}\text{Kr}\;\;drdzd\theta = \frac{1}{8} a^2 K \pi[/tex]
     
    Last edited: Nov 10, 2008
  5. Nov 10, 2008 #4
    Now that I have the mass (m), I want to find the r component of the center of mass. I tried

    [tex]\frac{1}{m}\int _0^{2 \pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K r^2drdzd\theta [/tex]

    This is incorrect. What am I doing wrong?
     
    Last edited: Nov 10, 2008
  6. Nov 10, 2008 #5

    tiny-tim

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    That should come out as 0, which is the r component of the centre of mass. :wink:
     
  7. Nov 10, 2008 #6
    Mathematica is telling me that this integral is not equal to zero. Have I made a mistake?
    I've included the output as an attachment.
     

    Attached Files:

  8. Nov 10, 2008 #7

    tiny-tim

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    ah … just realised … you can't do centre of mass that way in cylindrical coordinates … you'd have to convert to x and y coordinates first.

    (but the z coordinate will be ok :smile:)
     
  9. Nov 10, 2008 #8
    Right, that makes sense. So how can I solve for the r and theta components in cylindrical coordinates? I know qualitatively that they should be zero, but how do I show this mathematically?
     
  10. Nov 10, 2008 #9

    tiny-tim

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    I think you're perfectly entitled to say that it's obvious. :smile:

    (btw, if r is zero, then θ is meaningless :wink:)
     
  11. Nov 10, 2008 #10
    Very true. I think that's it. Thank you!
     
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