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Products of Inertia

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data



    http://i45.tinypic.com/hwcsy0.png



    3. The attempt at a solution

    I'm not sure how to find the rest; Ixy, Iyz and Izx...
    Usually for integrals such as moments of inertia you will be able to reduce it to only one variable. However, there are 2 variables here; xy, yz and zx. How do i reduce it to only one?
     
    Last edited by a moderator: Feb 3, 2013
  2. jcsd
  3. Feb 2, 2013 #2

    vela

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    You don't. You evaluate the triple integrals.
     
  4. Feb 3, 2013 #3

    Are my Ixx, Iyy and Izz are correct?

    Oh, so it's dM = dx dy dz,

    then the range for dx is from -√(a2-y2-z2) to √(a2-y2-z2)

    for dy it's from -√(a2-z2) to √(a2-z2)

    for dz it's from 0 to a

    Are my ranges of integration right?
     
  5. Feb 3, 2013 #4

    vela

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    No, they're not. You can't say ##y^2+z^2=a^2-x^2## because that's true only on the spherical part of the surface. Inside the hemisphere, it doesn't hold.

    That's dV, not dM. You should write dM = k dV = k dx dy dz.

    Yes.
     
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