Answer
$\overline {x} =\pi -\dfrac{1}{2}; \overline {y}=\dfrac{9}{8}$
Work Step by Step
We have $\overline {x}=\dfrac{1}{4 \pi } \int_{0}^{2 \pi } x \cdot (2+\sin x) dx$
or, $\overline {x}=(\dfrac{1}{4 \pi }) \times [x^2+\sin x-x \cos x]_0^{2 \pi}$
or, $\overline {x}=\pi -\dfrac{1}{2}$
$\overline {y}=\dfrac{1}{4 \pi } \int_{0}^{2 \pi } (2+\sin x)
\times [\dfrac{2+\sin x}{2}] dx $
or, $\overline {y}=\dfrac{1}{8\pi } \int_{0}^{2 \pi } 4+4 \sin x+\dfrac{1-\cos 2x}{2} dx$
or, $\overline {y}=\dfrac{9}{8}$
