Moment of inertia for a n sided regular polygon It is as simple as the title. We have an area $A_n$ of an homogeneous n-sided polygon with the density of $\rho$. With its center in $(0,0)$, and one of it's points in $(a,0)$, $a>0$.
Calculate $I_n$ aroudn its $z$-axis via the area $A_n$'s mass. $\iint_{A_n} dm$ ,and express it with regards to $n$ and $a$.
Obviously $I_n=\frac{\iint_{A_n} r^2dm}{\iint_{A_n}dm}$
So I can use the symmetry of the problem to get to
$A_n=4n\int_0^{\frac{\pi}{n}}d\phi\int \rho rdr$ 
But here is also where i get stuck. How do I set the integral boundries (not sure if that is correctly translated to English) for the second integral?
Probably a simple thing that a long walk and sleep would cure but currently am stuck.

Consider one of the $2n$ right triangles which form the $n$-regular polygon of circumradius $a$:
$$T_n:=\{(x,y):,0\leq x \leq a\cos(\pi/n), 0\leq y\leq \tan(\pi/n)x\}.$$
Hence, the moment of inertia with respect to $z$-axis of the $n$-regular polygon is
\begin{align*}
I_n&=2n\delta\int_{x=0}^{a\cos(\pi/n)}\int_{y=0}^{\tan(\pi/n)x}(x^2+y^2) dydx
\\&=
\frac{2M}{a^2 \sin(\pi/n)\cos(\pi/n)}\int_{\theta=0}^{\pi/n}\int_{\rho=0}^{a\cos(\pi/n)/\cos(\theta)}
\!\!\!\!\!\!\!\!\rho^2\cdot \rho d\rho d\theta\\
&=\cdots=\frac{Ma^2}{6}\left(1+2\cos^2(\pi/n)\right)
\end{align*}
where $M$ is the mass of the polygon.
P.S. Note that $\lim_{n\to\infty}I_n=\frac{Ma^2}{2}$ which is the moment of inertia of a homogeneous disk.
P.P.S. If $l$ is the side of the $n$-regular polygon then $l=2a\sin(\pi/n)$ and 
$$I_n=\frac{Ml^2}{24}\left(1+3\cot^2(\pi/n)\right).$$
