Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 13

$$\eqalign{ & m = \rho \cr & {M_x} = \frac{1}{3}\rho \cr & {M_y} = \frac{4}{3}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{4}{3},\frac{1}{3}} \right) \cr} $$

$$\eqalign{ & y = \frac{1}{2}x,{\text{ }}y = 0,{\text{ }}x = 2 \cr & {\text{Let }}f\left( x \right) = \frac{1}{2}x{\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_0^2 {\left( {\frac{1}{2}x - 0} \right)} dx \cr & m = \frac{1}{4}\rho \left[ {{x^2}} \right]_0^2 = \frac{1}{4}\rho \left[ {4 - 0} \right] \cr & m = \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 + 0}}{2}} \right]} \left[ {\frac{1}{2}x - 0} \right]dx \cr & {M_x} = \frac{1}{8}\rho \int_0^2 {{x^2}} dx \cr & {M_x} = \frac{1}{8}\rho \left[ {\frac{1}{3}{x^3}} \right]_0^2 = \frac{1}{8}\rho \left[ {\frac{1}{3}{{\left( 2 \right)}^3} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr & {M_x} = \frac{1}{3}\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 - 0} \right]dx \cr & {M_y} = \frac{1}{2}\rho \int_0^2 {{x^2}} dx \cr & {M_y} = \frac{1}{2}\rho \left[ {\frac{1}{3}{x^3}} \right]_0^2 = \frac{1}{2}\rho \left[ {\frac{1}{3}{{\left( 2 \right)}^3} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr & {M_y} = \frac{4}{3}\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{{\frac{4}{3}\rho }}{\rho } = \frac{4}{3} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{1}{3}\rho }}{\rho } = \frac{1}{3} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{4}{3},\frac{1}{3}} \right) \cr & \cr & {\text{Summary}} \cr & m = \rho \cr & {M_x} = \frac{1}{3}\rho \cr & {M_y} = \frac{4}{3}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{4}{3},\frac{1}{3}} \right) \cr} $$