For point masses, the moment of inertia about a chosen axis is

$$I=\sum_i m_i r_{\perp i}^2,$$

where $r_{\perp i}$ is each mass's perpendicular distance from the axis. For a continuous body, the same idea becomes an integral over small mass elements:

$$\boxed{I=\int r_\perp^2,dm}.$$

The main task is to express the mass element $dm$ in coordinates that match the geometry.

For a one-dimensional distribution with linear mass density $\lambda$,

$$dm=\lambda,ds.$$

For a thin surface with surface mass density $\sigma$,

$$dm=\sigma,dA.$$

For a three-dimensional body with volume mass density $\rho_m$,

$$dm=\rho_m,dV.$$

For a uniform body these densities are constant and can be found from total mass divided by total length, area, or volume.

Uniform rod about its center

Consider a thin uniform rod of length $L$ and mass $M$, rotating about an axis through its center and perpendicular to the rod. Put the rod along the $x$-axis. Its linear density is

$$\lambda=\frac{M}{L},$$

so

$$dm=\lambda,dx.$$

A mass element at coordinate $x$ is a perpendicular distance $|x|$ from the axis. Therefore

$$I=\int_{-L/2}^{L/2}x^2\lambda,dx =\frac{M}{L}\int_{-L/2}^{L/2}x^2,dx.$$

Evaluating the integral,

$$I=\frac{M}{L}\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} =\boxed{\frac{1}{12}ML^2}.$$

The result scales as $ML^2$, as dimensional reasoning suggests, but the numerical factor comes from the mass distribution.

Uniform disk about its symmetry axis

For a thin uniform disk of radius $R$ and mass $M$, use concentric rings of radius $r$ and thickness $dr$. The surface density is

$$\sigma=\frac{M}{\pi R^2}.$$

A ring has area

$$dA=2\pi r,dr,$$

so its mass is

$$dm=\sigma 2\pi r,dr.$$

Every point on that ring is distance $r$ from the symmetry axis, hence

$$I=\int_0^R r^2,dm =2\pi\sigma\int_0^R r^3,dr.$$

Thus

$$I=2\pi\frac{M}{\pi R^2}\frac{R^4}{4} =\boxed{\frac12MR^2}.$$

Choosing the mass element

Different decompositions can describe the same body, but a good choice makes $r_\perp$ simple and groups together material at the same distance from the axis. Rod elements work naturally for slender bodies; concentric rings are natural for disks and cylinders; shells or slices can be useful for other geometries.

The integral is not a new definition unrelated to the point-mass formula. It is the continuous limit of the same weighted sum: mass farther from the axis contributes more strongly because its distance is squared.