Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis.
It'sIts moment of inertia is $r d\theta dr (r \sin (\theta ))^2$.
The moment of inertia about the $x$ axis of the complete sector:
$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$
Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis.
It's moment of inertia is $r d\theta dr (r \sin (\theta ))^2$.
The moment of inertia about the $x$ axis of the complete sector:
$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$
Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis.
Its moment of inertia is $r d\theta dr (r \sin (\theta ))^2$.
The moment of inertia about the $x$ axis of the complete sector:
$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$
Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis.
It's moment of inertia is $r d\theta dr (r \sin (\theta ))^2$.
The moment of inertia about the $x$ axis of the complete sector:
$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$
