Rod (Solid Cylinder)
If $L \gg R$ then we typically refer to it as a rod. If $L \ll R$ then we typically refer to it as a disc. If $L \sim R$ then you can decide what to call it…a log maybe? For our purposes, we will draw it proportionally like a disc because it’s easier to see the geometry.
Parameterizing the Volume
We can do this with just standard cylindrical coordinates
\[\b{r}(s, \phi, z) = s \; \u{s} + z \; \u{z} \\[10pt] V = \{ \b{r}(s, \phi, z) \ : \ 0 \leq s \leq R \quad 0 \leq \phi < 2\pi \quad -L/2 \leq z \leq L/2 \}\] \[\frac{\partial \b{r}}{\partial s} ds = ds \; \u{s} \qquad \frac{\partial \b{r}}{\partial \phi} d\phi = s \; d\phi \; \u{\phi} \qquad \frac{\partial \b{r}}{\partial z} dz = dz \; \u{z}\]Therefore
\[dV = \left \lvert \begin{array}{ccc} ds & 0 & 0 \\ 0 & s \; d\phi & 0 \\ 0 & 0 & dz \end{array} \right \rvert = s \; ds \; d\phi \; dz\]Mass
Moment of Inertia About Central Axis
From the diagram, we see that $r_{axis} = s$.
\[\begin{align} I &= \int r_{axis}^2 \;dm \\[10pt] &= \rho \int r_{axis}^2 \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{-L/2}^{L/2} r_{axis}^2 \; s \; ds \; d\phi \; dz \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{-L/2}^{L/2} s^2 \; s \; ds \; d\phi \; dz \\[10pt] &= \rho \left ( \int_{0}^{R} s^3 \; ds \right ) \left ( \int_{0}^{2\pi} d\phi \right ) \left ( \int_{-L/2}^{L/2} dz \right ) \\[10pt] &= \rho \left ( \tfrac{1}{4} R^4 \right ) \left ( 2\pi \right ) \left ( L \right ) \\[10pt] &= \rho \cdot \tfrac{1}{2} \pi R^4 L \\[10pt] &= \tfrac{1}{2} M R^2 \end{align}\]Moment of Inertia About Central Diameter
Without loss of generality, we can assume the axis of rotation lies on the $x$-axis. As an exercise, evaluate the integral if the axis of rotation is on the $y$-axis.
We use cylindrical coordinates to evaluate this integral. Since it is a solid cylindrical, all parameters $s$, $\phi$, and $z$ vary. Therefore, $dV = s \; ds \; d \phi \; d z$. Either using the Euclidean distance formula or a little bit of geometry and Pythagoras, the diagram shows that $r_{axis} = \sqrt{(s \sin \phi)^2 + z^2}$.
\[\begin{align} I &= \int r_{axis}^2 \;dm \\[10pt] &= \rho \int r_{axis}^2 \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{-L/2}^{L/2} r_{axis}^2 \; s \; ds \; d\phi \; dz \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{-L/2}^{L/2} ((s \sin \phi)^2 + z^2) \; s \; ds \; d\phi \; dz \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{-L/2}^{L/2} (s^3 \sin^2 \phi + sz^2) \; ds \; d\phi \; dz \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} (L s^3 \sin^2 \phi + \tfrac{1}{12} L^3 s) \; ds \; d\phi \\[10pt] &= \rho \int_{0}^{R} (\pi L s^3 + \tfrac{1}{6} \pi L^3 s) \; ds \\[10pt] &= \rho \cdot \left ( \tfrac{1}{4} \pi L R^4 + \tfrac{1}{12} \pi L^3 R^2 \right ) \\[10pt] &= \rho \cdot \pi R^2 L \left ( \frac{1}{4} \pi R^2 + \tfrac{1}{12} L^2 \right ) \\[10pt] &= \tfrac{1}{4} M R^2 + \tfrac{1}{12} M L^2 \end{align}\]A Thin Rod or Thin Disc About the Central Diameter
Sometimes called a slender rod as well.
If $L \gg R$, then sometimes we neglect the contribution due to the radius. Essentially, this reduces to a line segment. Therefore, the moment of inertia is simply
\[I = \frac{1}{12} M L^2\]Likewise, if $L \ll R$, then sometimes we neglect the contribution due to the length. Essentially, this reduces to a flat circular disc. In topology, we would call this a 2D ball. Therefore, the moment of inertia is simply
\[I = \frac{1}{4} M R^2\]