Moment of inertia of rectangular lamina For calculating the moment of inertia of a rectangular lamina, we take thin strips of rectangles and then integrate considering two axes. There we take a strip at a distance $x$ from the center of gravity and this distance is the perpendicular distance to the thin rectangle. But moment of inertia is defined for point masses. I already raised a question on MOI before but that was for ring and I got my answer on that. But here how do we calculate the MOI of that thin strip using the definition of MOI from point masses?
 A: You can use the same idea as in the linked answer to show that the moment of inertia of any object whose mass is at distance exactly $R$ from the axis is $MR^2$. Therefore it holds not only for a ring of radius $R$, but for any subset of a ring such as a union of circular arcs of radius $R$.
Let $a$ and $b$ be the sides of our rectangular lamina with $a > b$ and let $M$ be its mass. We can assume that $a$ is the horizontal side and $b$ the vertical side. I'm assuming that the axis is perpendicular to the rectangle and goes through its center. Notice that you can decompose the rectangle as a union of objects of the above type. Namely, for any $r \in \left[0,\frac{\sqrt{a^2+b^2}}2\right]$ look at the circle centered at the axis of radius $r$ and take only those parts which intersect the rectangle; for a fixed $r$ you will get a circle or a union of circular arcs.
We can hence calculate $$I_{\text{rectangle}} =\int_{r=0}^{\sqrt{a^2+b^2}}dI_{\text{rectangle $\cap$ ring}}(r) = \int_{r=0}^{\sqrt{a^2+b^2}}m(r)r^2\,dr$$
where $m(r)$ is the mass of the ring-like object at radius $r$. It remains to simply calculate its mass for any fixed $r$. Denote $\sigma = \frac{M}{ab}$ the areal density of our rectangle. Clearly, the mass of our ring-like object is then simply
$$m(r) = \sigma\cdot \text{area of the object} = \sigma\,dr\cdot\text{length of the object}$$
since its thickness is infinitesimal. Hence it remains to calculate its length. There are three cases:

*

*If $r \in \left[0,\frac{b}2\right]$ then the ring of radius $r$ lies entirely inside the rectangle so its length is simply $2r\pi$.


*If $r \in \left[\frac{b}2,\frac{a}2\right]$ then the ring of radius $r$ intersects our rectangle in two circular arcs of equal length (left and right). Half of the central angle of one arc is given by the relation $\sin\phi = \frac{b}{2r}$ so by symmetry, the entire length is
$$4\arcsin\left(\frac{b}{2r}\right).$$


*If $r \in \left[\frac{a}2,\frac{\sqrt{a^2+b^2}}2\right]$ then the ring of radius $r$ intersects our rectangle in four circular arcs of equal length (one near each corner of the rectangle). Similary to the last example, one can see that the entire length is
$$4\arcsin\left(\frac{b}{2r}\right) - 4\arccos\left(\frac{a}{2r}\right).$$
Plugging this in, we get
$$I_{\text{rectangle}} =\int_{r=0}^{\frac{b}2}2r^3\pi\sigma\,dr +  \int_{r=\frac{b}2}^{\frac{a}2} 4\arcsin\left(\frac{b}{2r}\right)r^2\sigma\,dr + \int_{r=\frac{a}2}^{\frac{\sqrt{a^2+b^2}}2 } 4\left(\arcsin\left(\frac{b}{2r}\right) - \arccos\left(\frac{a}{2r}\right)\right)r^2\sigma\,dr.$$
These integrals are not easy but they can be calculated using e.g. Wolfram Mathematica, reproducing the standard result listed here.
