Answer
$\left( {\overline x ,\overline y } \right) = \left( {\pi - \frac{{3\sqrt 3 }}{2},\frac{3}{8}\sqrt 3 } \right)$
Work Step by Step
$$\eqalign{
& y = \sin 2x,{\text{ }}y = \sin x,{\text{ 0}} \leqslant x \leqslant \frac{\pi }{3} \cr
& {\text{Let }}f\left( x \right) = \sin 2x{\text{ and }}g\left( x \right) = \sin 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_0^{\pi /3} {\left( {\sin 2x - \sin x} \right)} dx \cr
& m = \rho \left[ { - \frac{1}{2}\cos 2x + \cos x} \right]_0^{\pi /3} \cr
& m = \rho \left[ { - \frac{1}{2}\cos 2\left( {\frac{\pi }{3}} \right) + \cos \left( {\frac{\pi }{3}} \right)} \right] - \rho \left[ { - \frac{1}{2}\cos \left( 0 \right) + \cos \left( 0 \right)} \right] \cr
& m = \rho \left( {\frac{3}{4}} \right) - \rho \left( {\frac{1}{2}} \right) \cr
& m = \frac{1}{4}\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_0^{\pi /3} {\left( {\frac{{\sin 2x + \sin x}}{2}} \right)} \left( {\sin 2x - \sin x} \right)dx \cr
& {M_x} = \rho \int_0^{\pi /3} {\left( {{{\sin }^2}2x - {{\sin }^2}x} \right)} dx \cr
& {\text{Integrating}} \cr
& {M_x} = \rho \int_0^{\pi /3} {\left( {\frac{{1 - \cos 4x}}{2} - \frac{{1 - \cos 2x}}{2}} \right)} dx \cr
& {M_x} = \frac{1}{2}\rho \int_0^{\pi /3} {\left( {\frac{{\cos 2x}}{2} - \frac{{\cos 4x}}{2}} \right)} dx \cr
& {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{4}\sin 2x - \frac{1}{8}\sin 4x} \right]_0^{\pi /3} \cr
& {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{4}\sin \left( {\frac{{2\pi }}{3}} \right) - \frac{1}{8}\sin \left( {\frac{{4\pi }}{3}} \right)} \right] - \frac{1}{2}\rho \left[ {\frac{1}{4}\sin \left( 0 \right) - \frac{1}{8}\sin \left( 0 \right)} \right] \cr
& {M_x} = \frac{{3\sqrt 3 }}{{32}}\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_0^{\pi /3} x \left( {\sin 2x - \sin x} \right)dx \cr
& {M_y} = \rho \underbrace {\int_0^{\pi /3} {\left( {x\sin 2x - x\sin x} \right)} dx}_{{\text{Integrating by parts each term}}} \cr
& {M_y} = \rho \left[ {\frac{1}{4}\sin 2x - \frac{1}{2}x\cos 2x - \sin x + x\cos x} \right]_0^{\pi /3} \cr
& {\text{Evaluating}}{\text{, we obtain}} \cr
& {M_y} = \rho \left( {\frac{1}{4}\sin \left( {\frac{{2\pi }}{3}} \right) - \frac{1}{2}\left( {\frac{\pi }{3}} \right)\cos \left( {\frac{{2\pi }}{3}} \right) - \sin \left( {\frac{\pi }{3}} \right) + \frac{\pi }{3}\cos \left( {\frac{\pi }{3}} \right)} \right) \cr
& {M_y} = \rho \left( {\frac{{\sqrt 3 }}{8} + \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{2} + \frac{\pi }{6}} \right) \cr
& {M_y} = \rho \left( {\frac{1}{4}\pi - \frac{{3\sqrt 3 }}{8}} \right) \cr
& \cr
& *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{\rho \left( {\frac{1}{4}\pi - \frac{{3\sqrt 3 }}{8}} \right)}}{{\frac{1}{4}\rho }} = \frac{{\frac{\rho }{4}\left( {\pi - \frac{{3\sqrt 3 }}{2}} \right)}}{{\frac{1}{4}\rho }} = \pi - \frac{{3\sqrt 3 }}{2} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{3\sqrt 3 }}{{32}}\rho }}{{\frac{1}{4}\rho }} = \frac{3}{8}\sqrt 3 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\pi - \frac{{3\sqrt 3 }}{2},\frac{3}{8}\sqrt 3 } \right) \cr} $$
