Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 16

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

$$\eqalign{ & y = \frac{1}{2}{x^2},{\text{ }}y = 0,{\text{ }}x = 2 \cr & {\text{Let }}f\left( x \right) = \frac{1}{2}{x^2}{\text{ and }}g\left( x \right) = 0,{\text{ on the interval }}\left[ {0,2} \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_0^2 {\left( {\frac{1}{2}{x^2} - 0} \right)} dx \cr & m = \frac{1}{2}\rho \int_0^2 {{x^2}} dx \cr & m = \frac{1}{6}\rho \left[ {{x^3}} \right]_0^2 = \frac{1}{6}\rho \left[ {{{\left( 2 \right)}^3} - {{\left( 0 \right)}^3}} \right] \cr & m = \frac{4}{3}\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_0^2 {\left[ {\frac{{\frac{1}{2}{x^2} - 0}}{2}} \right]} \left[ {\frac{1}{2}{x^2} - 0} \right]dx \cr & {M_x} = \frac{1}{8}\rho \int_0^2 {{x^4}} dx \cr & {M_x} = \frac{1}{{40}}\rho \left[ {{x^5}} \right]_0^2 \cr & {M_x} = \frac{1}{{40}}\rho \left[ {{{\left( 2 \right)}^5} - {{\left( 0 \right)}^3}} \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} - 0} \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[ {{{\left( 2 \right)}^4} - {{\left( 0 \right)}^4}} \right] \cr & {M_y} = 2\rho \cr & \cr & *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{{2\rho }}{{\frac{4}{3}\rho }} = \frac{3}{2} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{4}{5}\rho }}{{\frac{4}{3}\rho }} = \frac{3}{5} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{2},\frac{3}{5}} \right) \cr & \cr } $$