Chapter 8 - Section 8.3 - Applications to Physics and Engineering. - 8.3 Exercises - Page 585: 27

$\left( {\overline x ,\overline y } \right) = \left( {\frac{3}{2},\frac{3}{5}} \right)$

$$\eqalign{ & {\text{From the given graph}}{\text{, we can estimate the centroid at}} \cr & \left( {1.5,0.7} \right) \cr & \cr & y = \frac{1}{2}{x^2},{\text{ }}0 \leqslant x \leqslant 2 \cr & {\text{Let }}f\left( x \right) = \frac{1}{2}{x^2}{\text{ and }}g\left( x \right) = 0 \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_{ - 1}^1 {\left( {\frac{1}{2}{x^2} - 0} \right)} dx \cr & m = \rho \int_0^2 {\frac{1}{2}{x^2}} dx \cr & m = \rho \left[ {\frac{{{x^3}}}{6}} \right]_0^2 \cr & m = \frac{\rho }{6}\left( {8 - 0} \right) \cr & m = \frac{{4\rho }}{3} \cr & \cr & *{\text{The moment about the }}x{\text{ - axis is}} \cr & {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_x} = \rho \int_0^2 {\left[ {\frac{{\frac{1}{2}{x^2}}}{2}} \right]} \left[ {\frac{1}{2}{x^2}} \right]dx \cr & {M_x} = \frac{\rho }{8}\int_0^2 {{x^4}} dx \cr & {M_x} = \frac{1}{8}\rho \left[ {\frac{{{x^5}}}{5}} \right]_0^2 \cr & {M_x} = \frac{1}{{40}}\rho \left( {32} \right) \cr & {M_x} = \frac{4}{5}\rho \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^2 x \left[ {\frac{1}{2}{x^2}} \right]dx \cr & {M_y} = \frac{1}{2}\rho \int_0^2 {{x^3}} dx \cr & {M_y} = \frac{1}{8}\rho \left[ {{x^4}} \right]_0^2 \cr & {M_y} = \frac{1}{8}\rho \left( {16} \right) \cr & {M_y} = 2\rho \cr & \cr & *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{{2\rho }}{{\frac{{4\rho }}{3}}} = \frac{3}{2} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{4}{5}\rho }}{{\frac{{4\rho }}{3}}} = \frac{3}{5} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{2},\frac{3}{5}} \right) \cr} $$