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Moment of Inertia for Ellipsoid

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    a)Evaluate ∫∫∫E dV, where E is the solid enclosed by the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2 =1. Use the transformation x=au, y=bv, z=cw.

    b)If the solid in the above has density k find the moment of inertia about the z-axis.

    2. Relevant equations
    ∅=phi


    3. The attempt at a solution
    I got a correct 4/3∏abc

    i thought b) would be 0 based on the following.

    Jacobian is abc and the solid enclosed would be u^2+v^2+w^2≤1

    x^2*y=(a^2*u^2)(bv)

    V=∫∫∫(a^2bu^2v)(abc)dudvdw
    =∫0→2∏ ∫0→∏ ∫0→1 (a^3b^2c)(ρ^2sin^2∅cos^2θ)(ρsin∅sinθ)ρ^2sin∅dρd∅dθ
    =a^3b^2c∫∫∫(ρ^5sin^4∅cos^2θsinθ)dρd∅dθ
    =a^3b^2c∫0→2∏ cos^2θsinθdθ∫0→∏ sin^4∅d∅ ∫0→1 ρ^5dρ
    =0 since the 1st integral is 0
     
  2. jcsd
  3. Nov 4, 2011 #2

    LCKurtz

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    The integrand for a second moment is always positive. Your problem is in the last step above. The moment arm is the distance squared from the z axis: r2=x2+y2, not x2y.
     
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