Circle
This is one of the simplest moment of inertia derivations. Counter-intuitively, I am actually going to be extremely rigorous in this post, explaining every step in detail. I think this will greatly aid in future posts, which have more complex geometries.
Parameterizing the Curve
Consider a circle with radius $R$ centered at the origin on the $xy$-plane.
We can parameterize this circle using cylindrical coordinates.
\[\b{r}(\phi) = R \cos \phi \; \u{x} + R \sin \phi \; \u{y} = R \; \u{s} \\[10pt] \ell = \{ \b{r}(\phi) \ : \ 0 \leq \phi < 2\pi \}\]Notice that the direction of $\u{s}$ implicitly depends on $\phi$. Therefore it would be incorrect to write $\frac{\partial}{\partial \phi} R \u{s} = \u{s}$. Instead, we use Cartesian coordinates for the derivatives, since the unit vectors are always constant.
\[d \b{\ell} = \frac{\partial \b{r}}{\partial \phi} d\phi = \left ( - R \sin \phi \; \u{x} + R \cos \phi \; \u{y} \right ) \; d\phi = R \; d\phi \; \u{\phi}\]So actually, we have shown that $\frac{\partial}{\partial \phi} \u{s} = \u{\phi}$, which is interesting. Finally, we write the differential of the curve.
\[d\ell = \abs{d \b{\ell} } = R \; d\phi\]We are now ready to integrate!
Mass
Since we are always assuming a uniform mass density, mass is going to always be equivalent to arc length, surface area, or volume.
In most cases, we will know the value of the mass from middle school geometry. For example, in this case, we know the circumference of a circle of $2 \pi R$. However, I am always going to derive the value of the mass rigorously. 1) It is just interesting to see where these formulas come from, and 2) it will verify that we have, in fact, correctly parameterized the object.
\[\begin{align} M &= \int dm \\[10pt] &= \lambda \int d\ell \\[10pt] &= \lambda \int_{0}^{2 \pi} R \; d \phi \\[10pt] &= \lambda R \left ( \int_{0}^{2 \pi} d \phi \right ) \\[10pt] &= \lambda R \left ( 2 \pi \right ) \\[10pt] &= \lambda \cdot 2 \pi R \end{align}\]Moment of Inertia About Central Axis
Now, to compute the moment of inertia is just a matter of finding an expression for $r_{axis}$. From the diagram, we can see it is just the constant $R$. Now, we just go into integration autopilot.
\[\begin{align} I &= \int r_{axis}^2 dm \\[10pt] &= \lambda \int r_{axis}^2 d\ell \\[10pt] &= \lambda \int_{0}^{2 \pi} R^2 \cdot R \; d \phi \\[10pt] &= \lambda R^3 \left ( \int_{0}^{2 \pi} d \phi \right ) \\[10pt] &= \lambda R^3 \left ( 2 \pi \right ) \\[10pt] &= \lambda \cdot 2 \pi R^3 \\[10pt] &= M R^2 \end{align}\]Moment of Inertia About Central Diameter
We do likewise here. In this case, $r_{axis}$ is not constant, but rather depends on $\phi$.
\[\begin{align} I &= \int r_{axis}^2 dm \\[10pt] &= \lambda \int r_{axis}^2 d\ell \\[10pt] &= \lambda \int_{0}^{2 \pi} (R \ \sin \phi )^2 \cdot R \; d \phi \\[10pt] &= \lambda R^3 \left ( \int_{0}^{2 \pi} \sin^2 \phi \; d \phi \right ) \\[10pt] &= \lambda R^3 \left ( \pi \right ) \\[10pt] &= \lambda \cdot \pi R^3 \\[10pt] &= \tfrac{1}{2} M R^2 \end{align}\]Inertia Tensor
Given the above orientation to the $xyz$-axes, the inertia tensor of a circle is
\[I = \begin{bmatrix} \frac{1}{2} M R^2 & 0 & 0 \\ 0 & \frac{1}{2} M R^2 & 0 \\ 0 & 0 & M R^2 \end{bmatrix} = \tfrac{1}{2} M R^2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\]As an exercise, verify that the products of inertia are, in fact, $0$.
Also notice that $I_{zz} = I_{xx} + I_{yy}$. This will always be true if the curve/surface lies in the $xy$ plane. As an additional exercise, prove why this is the case (hint: just look at their definitions). This provides an additional sanity check when we have more complicated results.