Find the centroid of a homogeneous circular segment

Find the centroid of a homogeneous circular segment corresponding to a central angle $\alpha$.

Step 1 · Setup and symmetry

Place the circle of radius $R$ at the origin, the segment symmetric about the $x$-axis, chord at $x=R\cos\dfrac\alpha2$. By symmetry, $\bar y=0$.

Step 2 · Sector minus triangle

The sector (central angle $\alpha$) has $A_s=\dfrac{R^2\alpha}2$, $\bar x_s=\dfrac{4R\sin(\alpha/2)}{3\alpha}$ (doubleint-264), so $M_{y,s}=\dfrac{2R^3\sin(\alpha/2)}3$. The isosceles triangle with vertices $(0,0)$, $\left(R\cos\frac\alpha2,\pm R\sin\frac\alpha2\right)$ has $A_t=\dfrac{R^2}2\sin\alpha$, $\bar x_t=\dfrac{2R\cos(\alpha/2)}3$, so:

$$ M_{y,t} = \frac{R^3\sin\alpha\cos\dfrac\alpha2}3 = \frac{2R^3\sin\dfrac\alpha2\cos^2\dfrac\alpha2}3. $$

Step 3 · Area and static moment of the segment

$$ A = A_s-A_t = \frac{R^2}2(\alpha-\sin\alpha). $$

$$ M_y = M_{y,s}-M_{y,t} = \frac{2R^3\sin\dfrac\alpha2}3\left(1-\cos^2\frac\alpha2\right) = \frac{2R^3\sin^3\dfrac\alpha2}3. $$

$$ \bar x = \frac{M_y}A = \frac{4R\sin^3\dfrac\alpha2}{3(\alpha-\sin\alpha)}. $$

Step 4 · Limiting cases

For $\alpha=\pi$ (semicircle): $\bar x=\dfrac{4R}{3\pi}$, matching doubleint-255. As $\alpha\to0$ (thin segment): using $\sin\frac\alpha2\approx\frac\alpha2$ and $\alpha-\sin\alpha\approx\frac{\alpha^3}6$, $\bar x\to R$, i.e. the centroid approaches the arc.

Additional Notes

• This derivation matches the source exactly.
• Verified numerically (e.g. $R=1.9,\,\alpha=1.4$, via direct polar integration between the chord and the arc): $A\approx0.748263$, $\bar x\approx1.633853$, matching $\dfrac{R^2}2(\alpha-\sin\alpha)$ and $\dfrac{4R\sin^3(\alpha/2)}{3(\alpha-\sin\alpha)}$ exactly.