Find the centroid of a homogeneous circular sector
Find the centroid of a homogeneous circular sector of radius $R$ with central angle $\alpha$.
Explanation of Method
Placing the sector symmetrically about the $x$-axis makes $\bar y=0$ immediate; the static moment $M_y$ is then a short polar-coordinate integral.
- Use symmetry to get $\bar y=0$.
- Compute the area $A$.
- Compute the static moment $M_y$ in polar coordinates.
- Check limiting cases.
Step 1 · Setup and symmetry
Place the center at the origin, spanning $\theta\in\left[-\dfrac\alpha2,\dfrac\alpha2\right]$. By symmetry about the $x$-axis, $\bar y=0$. The area is $A=\dfrac12R^2\alpha$.
Step 2 · Static moment $M_y$
$$ M_y = \int_{-\alpha/2}^{\alpha/2}\int_0^Rr\cos\theta\cdot r\,dr\,d\theta = \int_{-\alpha/2}^{\alpha/2}\cos\theta\,d\theta\cdot\int_0^Rr^2\,dr = 2\sin\frac\alpha2\cdot\frac{R^3}3 = \frac{2R^3}3\sin\frac\alpha2. $$
Step 3 · Centroid coordinate $\bar x$
$$ \bar x = \frac{M_y}A = \frac{4R\sin\dfrac\alpha2}{3\alpha}. $$
Step 4 · Limiting cases
For $\alpha=\pi$ (semicircle): $\bar x=\dfrac{4R}{3\pi}$, matching doubleint-255. For $\alpha=2\pi$ (full disk): $\bar x=0$, as expected by symmetry. As $\alpha\to0$ (thin sector): $\bar x\to\dfrac{2R}3$, the centroid of a thin isosceles triangle of height $R$.
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $R=2.3,\,\alpha=1.1$): $\bar x\approx1.45719$, matching $\dfrac{4R\sin(\alpha/2)}{3\alpha}$ exactly.
Result
The centroid of the circular sector is: