1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Triple Integral in Cylindrical Coordinates
Loading...