Solid Square Moment

Let the lower left corner be the origin $O$ and upper left corner and centre of the square be $P$ and $Q$ respectively. Finding the moment of inertia $I$ of the square is equal to $4 \times$ the moment of inertia of $\triangle OPQ$ . Let the mass per area be $\lambda$ . Then we have:

$\begin{aligned} I & = 4 \int_0^\frac L{\sqrt 2} \int_0^x {\color{#3D99F6}\lambda} y^2 \ dy \ dx & \small \color{#3D99F6} \text{Note that }\lambda = \frac M{L^2} \\ & = 4 \cdot {\color{#3D99F6}\frac M{L^2}} \int_0^\frac L{\sqrt 2} \int_0^x y^2 \ dy \ dx \\ & = \frac {4M}{L^2} \int_0^\frac L{\sqrt 2} \frac {x^3}3 \ dx \\ & = \frac {4M}{3L^2} \frac {x^4}4 \bigg|_0^\frac L{\sqrt 2} \\ & = \frac M{3L^2} \left(\frac L{\sqrt 2}\right)^4 \\ & = \frac {ML^2}{12} \end{aligned}$

$\implies \alpha = \boxed{12}$

Consider an axis of rotation along one of its lengths. Consider an strip of width $dx$ at a distance x from the axis, it's moment of inertia would be $dI=dm\cdot x^2=\frac{Mdx}L\cdot x^2$

Thus $I={ML^2}/{3}$ , by integrating the above expression between $0$ and $L$ .

Using the perpendicular axis theorem, the moment of inertia about the corner $=2I$ . Applying parallel axis theorem we get, $2I=I_{COM}+\frac{ML^2}{2}\\I_{COM}=2\frac{Ml^2}3-\frac{ML^2}2=\frac{ML^2}6$

Applying perpendicular axis theorem about the center of mass wraps the process, we get $2I_{\text{diagonal}} = I_{COM} \\I_{\text{diagonal}} =\frac{ML^2}{12}$

Edit: Since the body is a square I have used the fact that $I_x=I_y=I$ and $I_x+I_y=2I$ .

Kindly change a line in the question,uniform mass density as a function of area,it's ambiguous and may lead to reporting