Solución 
1.- determinar el tipo de centroide
 \(y=\frac{\int_.^.y\ dl}{\int_.^.dl}\)         \(x=\frac{\int_.^.x\ dl}{dl}\)
\(x=R\cos\theta\)      \(y=R\sin\theta\)
2.-elemento diferencial
\(dl=Rd\theta\)
3.- resolver integrales y obtener resultado
\(x=\frac{\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}R^2COS\theta d\theta}{\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}Rd\theta}\)  = \(\frac{R\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\cos\theta d\theta}{\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}d\theta}\)
\(y=\frac{\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}R^2\sin\theta d\theta}{\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}Rd\theta}=\frac{R\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\sin\theta d\theta}{d\theta}\)
\(x=\frac{R\left[\sin\theta\right]_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}}{\left[\theta\right]_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}}=\frac{R\sqrt{3}}{\left[\frac{2\pi}{3}+\frac{2\pi}{3}\right]}=\frac{3\sqrt{3}R}{4\pi}=0.124m\)
\(y=\frac{R\left[-\cos\theta\right]_{\frac{2\pi}{3}}^{\frac{2\pi}{3}}}{\left[\theta\right]_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}}=\frac{R\left[0.5+\left(-0.5\right)\right]}{\frac{4\pi}{3}}=0\)