# calculate moments of inertia

• October 8th 2010, 06:44 PM
tsang
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 $z^2-x^2-y^2=1$ and the plane $z=3$.
If the mass per unit volume is $z$,find the three moments of inertia $I_x$, $I_y$,and $I_z$.

I got same answer for $I_z$,so please don't worry for it.My problem is just $I_x$ and $I_y$,and for sure they are equal.
I used spherical coordinate to calculate.Also the formula $\int \int \int_D (y^2+z^2)\mu(x,y,z) dxdydz$. I did times by Jacobian as well.
My final answer is $80\pi$,but solution is $40\pi$.
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 $z^2- r^2= 1$ and $z= 3$.
The mass is given by $M= \int_{\theta= 0}^{2\pi}\int_{r= 0}^{2\sqrt{2}}z drd\theta dz$.
$I_x$ is given by $\int_{\theta= 0}^{2\pi}\int_{r= 0}^{2\sqrt{2}}r^2 cos^2(\theta) drd\theta dz$
(( $x^2= y^2+ z^2$ only on the lower boundary of the solid.)