Moment of Inertia of a Solid Sphere

Result

$I = \frac{2}{5}MR^2 \quad \text{(about any diameter)}$

Derivation

Divide the sphere into thin discs perpendicular to the rotation axis ($z$-axis).

At height $z$ from the centre, disc radius: $r = \sqrt{R^2 - z^2}$

Disc MI about $z$-axis: $dI = \frac{1}{2}(dm)r^2 = \frac{1}{2}r^2 \cdot \rho\pi r^2 \, dz = \frac{1}{2}\rho\pi r^4 \, dz$

$I = \int_{-R}^{R} \frac{1}{2}\rho\pi(R^2-z^2)^2 \, dz = \frac{\rho\pi}{2}\int_{-R}^{R}(R^4 - 2R^2z^2 + z^4) \, dz$

$= \frac{\rho\pi}{2}\left[R^4(2R) - 2R^2\frac{2R^3}{3} + \frac{2R^5}{5}\right] = \frac{\rho\pi}{2} \cdot 2R^5\left(1 - \frac{2}{3} + \frac{1}{5}\right)$

$= \rho\pi R^5 \cdot \frac{8}{15}$

With $M = \frac{4}{3}\pi R^3 \rho$, so $\rho = \frac{3M}{4\pi R^3}$:

$I = \frac{3M}{4\pi R^3} \cdot \pi R^5 \cdot \frac{8}{15} = \frac{2MR^2}{5}$

$\boxed{I = \frac{2}{5}MR^2}$

About a tangent

Using parallel axis theorem ($d = R$):

$I_{tangent} = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$

Solid sphere vs hollow sphere

The hollow sphere has larger MI — its mass is entirely at the rim, far from the axis. The solid sphere has mass throughout its volume, with inner layers close to the axis reducing the average $r^2$.