Find the moment of inertia of a circular sector of radius
Find the moment of inertia of a circular sector of radius $r$ with central angle $\alpha$ with respect to one of its bounding radii.
Explanation of Method
Place one bounding radius along the $x$-axis, evaluate $I_x=\iint y^2\,dA=\int_0^\alpha\int_0^r(\rho\sin\theta)^2\rho\,d\rho\,d\theta$ in polar coordinates, and separate the integrals.
Step 1: Set up
\[I_x=\int_0^\alpha\sin^2\theta\,d\theta\int_0^r\rho^3\,d\rho\]
Step 2: Radial integral
\[\int_0^r\rho^3\,d\rho=\frac{r^4}4\]
Step 3: Angular integral
\[\int_0^\alpha\sin^2\theta\,d\theta=\frac12\left[\theta-\frac{\sin2\theta}2\right]_0^\alpha=\frac12\big(\alpha-\sin\alpha\cos\alpha\big)\]
Step 4: Combine
\[I_x=\frac{r^4}4\cdot\frac12(\alpha-\sin\alpha\cos\alpha)=\frac{r^4}8(\alpha-\sin\alpha\cos\alpha)=\frac{r^4(2\alpha-\sin2\alpha)}{16}\]
\[\boxed{I_x=\frac{r^4}8(\alpha-\sin\alpha\cos\alpha)}\]
Step 5: Special cases
Semicircle ($\alpha=\pi$): $I_x=\dfrac{\pi r^4}8$, half the full-circle diameter result — consistent. Quarter circle ($\alpha=\pi/2$): $I_x=\dfrac{\pi r^4}{16}$.
Result
Final answer:
\[\boxed{I_x = \dfrac{r^4}{8}(\alpha-\sin\alpha\cos\alpha) = \dfrac{r^4(2\alpha-\sin2\alpha)}{16}}\]