By integration
Calculate \(I_x\) and \(I_y\) and use the perpendicular axis theorem to get the \(I_o\). Finally you can use the parallel axis theorem to shift the origin to the centre of mass.
For a polygon \(A_0,A_1,A_2⋯A_{n−1}\) with \(A_i=(xi,yi)Ai=(x_i,y_i)\) and \(A_n=A_0\), those are equal to:
\[I_x =\rho \frac{1}{12}\sum_{i=0}^{n-1}\left(x_iy_{i+1}-x_{i+1}y_i\right)\left(y_i^2+y_iy_{i+1}+y_{i+1}^2\right)\]
\[I_y =\rho \frac{1}{12}\sum_{i=0}^{n-1}\left(x_iy_{i+1}-x_{i+1}y_i\right)\left(x_i^2+x_ix_{i+1}+x_{i+1}^2\right)\]
Now, denote by zz the line perpendicular to the \(xy\)-plane that passes through the origin. Then:
\[\begin{align}J_z &= I_x+I_y\\&= \rho \frac1{12}\sum_{i=0}^{n-1}(x_{i}y_{i+1}-x_{i+1}y_i)\left(x_{i+1}^2+x_{i+1}x_i+x_i^2 +y_{i+1}^2+y_{i+1}y_i+y_i^2\right)\end{align}\]
Then the Parallel Axis Theorem can be used to shift the reference axis to the centre of mass.
Useful References
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