Consider a cylindrical shell of radius $r$ and mass per unit volume $p$ rotating about the axis of the cylinder. Find the moment of inertia $I_{zz}$ (Do not evaluate integral)
I started of by saying:
Note, $I_{zz}=\int \int \int_{V} p (r^2-z^2)$ $dV$
Using cylindrical cooridnates $x= r cos\theta$, $y=rsin\theta$ and $z=z$, and so
$dV=rdzd\theta$. Note that here, $r$ is constant, so
$I_{zz}=\int_{0}^{2\pi}\int_{0}^Lp(r^2-z^2)rdz d\theta$
However, I was told that this is incorrect. I'm not sure why this is incorrect. May someone elaborate?
I'm assuming the $z$ axis and the axis of the cylinder coincide. Then, $$I_{z} = \int_{r_{1}}^{r_{2}}\int_{0}^{2\pi}\int_{0}^{L} \rho r^{2} rdr d\theta dz$$ where the cylinder has height $L$ and the shell has inner radius $r_{1}$ and outer radius $r_{2}$. Note that this integral treats the cylinder as a "superposition" of rings.
The integrand does not depend on $\theta$ and $z$ and $\rho$ is assumed to be constant, so $I_{z}$ can be simplified: $$I_{z} = 2\pi \rho L \int_{r_{1}}^{r_{2}}r^{3}dr.$$ Your term $(r^{2}-z^{2})$ seems incorrect to me. It is almost as if the axis of the cylinder was displaced from the $z$ axis but, even in this case, we would have $(r-z)^{2}$ instead of $r^{2}-z^{2}$.