Answer
$\left( {\overline x ,\overline y } \right) = \left( {\frac{1}{2}\pi ,\frac{{\left( {\pi + 2} \right)\sqrt 2 }}{{16}}} \right)$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \sin x{\text{ and }}g\left( x \right) = 0,{\text{ }}\frac{\pi }{4} \leqslant x \leqslant \frac{{3\pi }}{4} \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_{\pi /4}^{3\pi /4} {\left( {\sin x - 0} \right)} dx \cr
& m = \rho \int_{\pi /4}^{3\pi /4} {\sin x} dx \cr
& m = \rho \left[ { - \cos x} \right]_{\pi /4}^{3\pi /4} \cr
& m = - \rho \left[ {\cos \left( {\frac{{3\pi }}{4}} \right) - \cos \left( {\frac{\pi }{4}} \right)} \right] \cr
& m = - \rho \left( { - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}} \right) \cr
& m = \sqrt 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_{\pi /4}^{3\pi /4} {\left[ {\frac{{\sin x}}{2}} \right]} \left[ {\sin x} \right]dx \cr
& {M_x} = \frac{1}{2}\rho \int_{\pi /4}^{3\pi /4} {{{\sin }^2}x} dx \cr
& {M_x} = \frac{1}{2}\rho \int_{\pi /4}^{3\pi /4} {\frac{{1 - \cos 2x}}{2}} dx \cr
& {M_x} = \frac{1}{4}\rho \int_{\pi /4}^{3\pi /4} {\left( {1 - \cos 2x} \right)} dx \cr
& {M_x} = \frac{1}{4}\rho \left[ {x - \frac{1}{2}\sin 2x} \right]_{\pi /4}^{3\pi /4} \cr
& {M_x} = \frac{1}{4}\rho \left[ {\frac{\pi }{4} - \frac{1}{2}\sin 2\left( {\frac{{3\pi }}{4}} \right)} \right] - \frac{1}{4}\rho \left[ {\frac{\pi }{4} - \frac{1}{2}\sin 2\left( {\frac{\pi }{4}} \right)} \right] \cr
& {M_x} = \frac{1}{4}\rho \left( {\frac{{3\pi }}{4} + \frac{1}{2}} \right) - \frac{1}{4}\rho \left( {\frac{\pi }{4} - \frac{1}{2}} \right) \cr
& {M_x} = \frac{{3\pi }}{{16}}\rho + \frac{1}{8}\rho - \frac{\pi }{{16}}\rho + \frac{1}{8}\rho \cr
& {M_x} = \frac{\pi }{8}\rho + \frac{1}{4}\rho \cr
& {M_x} = \left( {\frac{\pi }{8} + \frac{1}{4}} \right)\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_{\pi /4}^{3\pi /4} x \left[ {\sin x - 0} \right]dx \cr
& {M_y} = \rho \int_{\pi /4}^{3\pi /4} {x\sin x} dx \cr
& {M_y} = \rho \left[ {\sin x - x\cos x} \right]_{\pi /4}^{3\pi /4} \cr
& {M_y} = \rho \left[ {\sin \left( {\frac{{3\pi }}{4}} \right) - \left( {\frac{{3\pi }}{4}} \right)\cos \left( {\frac{{3\pi }}{4}} \right)} \right] \cr
& - \rho \left[ {\sin \left( {\frac{\pi }{4}} \right) - \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{4}} \right)} \right] \cr
& {M_y} = \rho \left( {\frac{{\sqrt 2 }}{2} + \frac{{3\sqrt 2 }}{8}\pi } \right) - \rho \left( {\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{8}\pi } \right) \cr
& {M_y} = \frac{{\sqrt 2 }}{2}\rho + \frac{{3\sqrt 2 }}{8}\pi \rho - \frac{{\sqrt 2 }}{2}\rho + \frac{{\sqrt 2 }}{8}\pi \rho \cr
& {M_y} = \frac{{\sqrt 2 \pi }}{2}\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{{\sqrt 2 \pi }}{2}\rho }}{{\sqrt 2 \rho }} = \frac{1}{2}\pi \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\left( {\frac{\pi }{8} + \frac{1}{4}} \right)\rho }}{{\sqrt 2 \rho }} = \frac{{\left( {\frac{{\pi + 2}}{8}} \right)\rho }}{{\sqrt 2 \rho }} = \frac{{\left( {\pi + 2} \right)\sqrt 2 }}{{16}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{1}{2}\pi ,\frac{{\left( {\pi + 2} \right)\sqrt 2 }}{{16}}} \right) \cr} $$
