Answer
$\left( {\overline x ,\overline y } \right) = \left( { - \frac{1}{2},\frac{8}{9}} \right)$
Work Step by Step
$$\eqalign{
& {\text{From the graph shown below:}} \cr
& {\text{Let }}f\left( x \right) = 2 - {x^2}{\text{ and }}g\left( x \right) = x{\text{ on the interval }}\left[ { - 2,1} \right] \cr
& \cr
& {\text{*The mass of the lamina is }} \cr
& m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& m = \rho \int_{ - 2}^1 {\left( {2 - {x^2} - x} \right)} dx \cr
& m = \rho \left[ {2x - \frac{1}{3}{x^3} - \frac{1}{2}{x^2}} \right]_{ - 2}^1 \cr
& m = \rho \left[ {2\left( 1 \right) - \frac{1}{3}{{\left( 1 \right)}^3} - \frac{1}{2}{{\left( 1 \right)}^2}} \right] - \rho \left[ {2\left( { - 2} \right) - \frac{1}{3}{{\left( { - 2} \right)}^3} - \frac{1}{2}{{\left( { - 2} \right)}^2}} \right] \cr
& m = \rho \left( {\frac{7}{6}} \right) - \rho \left( { - \frac{{10}}{3}} \right) \cr
& m = \frac{9}{2}\rho \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_x} = \rho \int_{ - 2}^1 {\left[ {\frac{{2 - {x^2} + x}}{2}} \right]} \left[ {2 - {x^2} - x} \right]dx \cr
& {M_x} = \frac{\rho }{2}\int_{ - 2}^1 {\left( {{x^4} - 5{x^2} + 4} \right)} dx \cr
& {M_x} = \frac{\rho }{2}\left[ {\frac{1}{5}{x^5} - \frac{5}{3}{x^3} + 4x} \right]_{ - 2}^1 \cr
& {M_x} = \frac{\rho }{2}\left[ {\frac{1}{5}{{\left( 1 \right)}^5} - \frac{5}{3}{{\left( 1 \right)}^3} + 4\left( 1 \right)} \right] - \frac{\rho }{2}\left[ {\frac{1}{5}{{\left( { - 2} \right)}^5} - \frac{5}{3}{{\left( { - 2} \right)}^3} + 4\left( { - 2} \right)} \right] \cr
& {M_x} = \frac{\rho }{2}\left( {\frac{8}{3}} \right) - \frac{\rho }{2}\left( { - \frac{{16}}{3}} \right) \cr
& {M_x} = 4\rho \cr
& \cr
& *{\text{The moment about the }}y{\text{ - axis is}} \cr
& {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_y} = \rho \int_{ - 2}^1 x \left[ {2 - {x^2} - x} \right]dx \cr
& {M_y} = \rho \int_{ - 2}^1 {\left( {2x - {x^3} - {x^2}} \right)dx} \cr
& {M_y} = \rho \left[ {{x^2} - \frac{1}{4}{x^4} - \frac{1}{3}{x^3}} \right]_{ - 2}^1 \cr
& {M_y} = \rho \left[ {{{\left( 1 \right)}^2} - \frac{1}{4}{{\left( 1 \right)}^4} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - \rho \left[ {{{\left( { - 2} \right)}^2} - \frac{1}{4}{{\left( { - 2} \right)}^4} - \frac{1}{3}{{\left( { - 2} \right)}^3}} \right] \cr
& {M_y} = \rho \left( {\frac{5}{{12}}} \right) - \rho \left( {\frac{8}{3}} \right) \cr
& {M_y} = - \frac{9}{4}\rho \cr
& \cr
& *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{ - \frac{9}{4}\rho }}{{\frac{9}{2}\rho }} = - \frac{1}{2} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{4\rho }}{{\frac{9}{2}\rho }} = \frac{8}{9} \cr
& \left( {\overline x ,\overline y } \right) = \left( { - \frac{1}{2},\frac{8}{9}} \right) \cr} $$
