# Calculate the centroid of region "R"?

Jul 13, 2018

$\overline{x} = \frac{{\int}_{0}^{2} x \cdot y \setminus \mathrm{dx}}{{\int}_{0}^{2} y \setminus \mathrm{dx}}$

$= \frac{{\int}_{0}^{2} 4 x - {x}^{3} \setminus \mathrm{dx}}{{\int}_{0}^{2} 4 - {x}^{2} \setminus \mathrm{dx}}$

$= \frac{{\left[2 {x}^{2} - {x}^{4} / 4\right]}_{0}^{2}}{{\left[4 x - {x}^{3} / 3\right]}_{0}^{2}} = \frac{4}{\frac{16}{3}} = \frac{3}{4} = 0.75$

$\overline{y} = \frac{{\int}_{0}^{2} \frac{y}{2} \cdot y \setminus \mathrm{dx}}{{\int}_{0}^{2} y \setminus \mathrm{dx}}$

$= \frac{\frac{1}{2} {\int}_{0}^{2} 16 - 8 {x}^{2} + {x}^{4} \setminus \mathrm{dx}}{\frac{16}{3}}$

$= \frac{1}{2} \frac{{\left[16 x - \frac{8}{3} {x}^{3} + {x}^{5} / 5\right]}_{0}^{2}}{\frac{16}{3}}$

$= \frac{1}{2} \cdot \frac{32 - \frac{64}{3} + \frac{32}{5}}{\frac{16}{3}} = \frac{8}{5} = 1.6$