Question

proof for moment of inertia of a rectangular lamina

Answer 1

See diagram.

Aerial Density of the lamina = M / ab  ,  M = mass of lamina, a and b are dimensions of the lamina.

Let us take an axis passing through the central line of rectangle and perpendicular to the base "a".

Take a strip of width dx and length b. Its mass = dm = b *dx *M/ab = Mdx/a

Moment of Inertial = I = 
$=\int\limits^{x=a/2}_{x=-a/2} {x^2} \, dm= \int\limits^{x=a/2}_{x=-a/2} {x^2\ \frac{M}{a}} \, dx\\\\ I=\frac{M}{3a} [x^3 ]_{-a/2}^{a/2}=\frac{M}{3a} *2* \frac{a^3}{8}\\\\ I=\frac{1}{12}M a^2$

If moment of inertia is found about the central axis along the other dimension,
          I = M b²/12