Q:

How to derive the moment of inertia of a thin hoop about its central diameter?

For lack of a better image, I am searching for the moment of inertia of this

where$\ r_1 = r_2$ (negligible thickness), and where the object would be rotating around its central diameter, which is perpendicular to the z-axis. Basically, either$\ I_x$ or$\ I_y$.
I could easily use the perpendicular axis theorem to find that$\ I_z = I_x + I_y = 2I_x = 2I_y$ and solve for my desired moment (x- or y-axis), which would give me $\frac{1}{2}MR^2$. Easy.
However, I have perused multiple sources and even attempted on my own to arrive at such an expression, and all of them (including mine) have included a second term including the width of the hoop, or in the case of my illustration, the height of the cylinder:$$I_x=I_y=\frac{1}{2}MR^2 + \frac{1}{12}MW^2$$
To make matters more confusing, my attempts at deriving left me with an identical expression to the one above, except$\ W$ is to the third power. I tried splitting the hoop into many rings and using the parallel axis theorem to account for the fact that every ring except the central one is a distance$\ z$ from the axis of rotation. It's likely that my issues are due to either poor math or a misconception of how I should even set this up based on$\ I=\int r^2dm$, but I would like to hear how others would go about solving this.

A:

The inertia $I$ is actually a tensor whose components are
$$
I_{ij} = \int{\rm d}^3{\bf x}~\rho({\bf x}) [{\bf x}\cdot{\bf x}\delta_{ij} - x_ix_j] \tag{1}
$$
So, for example the component $I_{11}$ can be calculated as
$$
I_{11} = \int{\rm d}^3{\bf x}~\rho({\bf x}) [x^2 + y^2 + z^2 -x^2 ] = \int{\rm d}^3{\bf x}~\rho({\bf x}) [y^2 + z^2] \tag{2}
$$
To calculate this we need the density, which for this problem is just
$$
\rho({\bf x}) = \rho(r,\phi,z) = \frac{M}{2\pi R h}\delta(r-R) \tag{3}
$$
Replacing (3) in (2) you get
\begin{eqnarray}
I_x &\stackrel{\rm def.}{=}& I_{11} = \int {\rm d}r {\rm d}\phi {\rm d}z ~r \left[  \frac{M}{2\pi R h}\delta(r-R) \right] (y^2 + z^2), ~~y=r\sin\phi \\
&=& \frac{M}{2\pi R h}\left\{ 
\int {\rm d}r {\rm d}\phi {\rm d}z ~r \delta(r-R)(r^2\sin^2\phi) +
\int {\rm d}r {\rm d}\phi {\rm d}z ~r \delta(r-R)(z^2)
\right\} \\
&=& \frac{M}{2\pi R h}\left\{ R^3h \int_0^{2\pi}{\rm d}\phi~\sin^2\phi + 2\pi R \int_{-h/2}^{h/2}{\rm d}z~z^2 \right\}  \\
&=&\frac{M}{2\pi R h}\left\{ \pi R^3h  + 2\pi R \frac{h^3}{12}\right\} \\
&=& \frac{M}{2}\left(R^2 + \frac{h^2}{6}\right) \tag{4}
\end{eqnarray}
with a similar expression for $I_y$

