Solución:
\(x=\frac{\int_1^{ }dL}{\int_1^{ }dL}\)
\(y=\frac{\int_1^{ }y\ dL}{\int_1^{ }dL}\)
\(x=R\cos\theta\)
\(y=R\sin\theta\)
\(dL=Rd\theta\)
\(x=\frac{\int_{-\frac{2\Pi}{3}}^{\frac{2\Pi}{3}}R^{x\ }\cos\theta\ d\theta}{\int_{-\frac{2\Pi}{3}}^{\frac{2\Pi}{3}}Rd\theta}=\frac{R\int_{-\frac{2\Pi}{3}}^{\frac{2\Pi}{3}}\cos\theta\ d\theta}{\int_{-\frac{2}{3}}^{\frac{2\Pi}{3}}d\theta}\)
\(y=\frac{\int_{-\frac{2\Pi}{3}}^{2\Pi3}R^2\sin\theta\ d\theta}{\int_{-\frac{2\Pi}{3}}^{\frac{2\Pi}{3}}R\ d\theta}=\frac{R\int_{-\frac{2\Pi}{3}}^{\frac{2\Pi}{3}}\sin\theta\ d\theta}{\int_{-\frac{2\Pi}{3}}^{\frac{2\Pi}{3}}d\theta}\)
\(x=\frac{R\left[\sin\theta\right]\frac{\frac{2\Pi}{3}}{-\frac{2\Pi}{3}}}{\left[\theta\right]\frac{\frac{2\Pi}{3}}{-\frac{2\Pi}{3}}}=\frac{R\sqrt{3}}{\left[\frac{2\Pi}{3}+\frac{2\Pi}{3}\right]}=\frac{3\sqrt{3\ }R}{4\Pi}=0.124\ m\)
\(y=\frac{R\left[-\cos\theta\right]\frac{\frac{2\Pi}{3}}{-\frac{2\Pi}{3}}}{\left[\theta\right]\frac{\frac{2\Pi}{3}}{-\frac{2\Pi}{3}}}=\frac{\left[0.5+\left(-0.5\right)\right]}{\frac{4\Pi}{3}}=0\)
El Resultado es 0.124 m
Ejercicio #2