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Thread: calculate moments of inertia

  1. #1
    Jul 2010

    calculate moments of inertia

    Hi everyone,I got a question here and I thought I have solved it,and I've double checked my answer again and again,but I still get different answer compare to solution.I really feel solution is wrong,can anyone help me please?

    Consider the solid region that is bounded by the top half of the hyperboloid of two sheets $\displaystyle z^2-x^2-y^2=1$ and the plane $\displaystyle z=3$.
    If the mass per unit volume is $\displaystyle z$,find the three moments of inertia $\displaystyle I_x$,$\displaystyle I_y$,and $\displaystyle I_z$.

    I got same answer for $\displaystyle I_z$,so please don't worry for it.My problem is just $\displaystyle I_x$ and $\displaystyle I_y$,and for sure they are equal.
    I used spherical coordinate to calculate.Also the formula $\displaystyle \int \int \int_D (y^2+z^2)\mu(x,y,z) dxdydz$. I did times by Jacobian as well.
    My final answer is $\displaystyle 80\pi$,but solution is $\displaystyle 40\pi$.
    Can anyone please help me? Thanks a lot.
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  2. #2
    MHF Contributor

    Apr 2005
    Well, of course, they are equal, by symmetry. I would recommend using cylindrical coordinates rather than spherical coordinates. In cylindrical coordinates, the boundaries are given by $\displaystyle z^2- r^2= 1$ and $\displaystyle z= 3$.

    The mass is given by $\displaystyle M= \int_{\theta= 0}^{2\pi}\int_{r= 0}^{2\sqrt{2}}z drd\theta dz$.

    $\displaystyle I_x$ is given by $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^{2\sqrt{2}}r^2 cos^2(\theta) drd\theta dz$

    (($\displaystyle x^2= y^2+ z^2$ only on the lower boundary of the solid.)
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