Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 21

$$\eqalign{ & m = \frac{{96}}{5}\rho \cr & {M_x} = \frac{{192}}{7}\rho \cr & {M_y} = 96\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {5,\frac{{10}}{7}} \right) \cr} $$

$$\eqalign{ & y = {x^{2/3}},{\text{ }}y = 0,{\text{ }}x = 8 \cr & {\text{Let }}f\left( x \right) = {x^{2/3}}{\text{ and }}g\left( x \right) = 0,{\text{ on }}\left[ {0,8} \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_0^8 {\left( {{x^{2/3}} - 0} \right)} dx \cr & m = \rho \left[ {\frac{3}{5}{x^{5/3}}} \right]_0^8 = \rho \left[ {\frac{3}{5}{{\left( 8 \right)}^{5/3}}} \right] - \rho \left[ 0 \right] \cr & m = \frac{{96}}{5}\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_0^8 {\left( {\frac{{{x^{2/3}} + 0}}{2}} \right)} \left( {{x^{2/3}} - 0} \right)dx \cr & {M_x} = \frac{1}{2}\rho \int_0^8 {{{\left( {{x^{2/3}}} \right)}^2}} dx \cr & {M_x} = \frac{1}{2}\rho \int_0^8 {{x^{4/3}}} dx \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{{3{x^{7/3}}}}{7}} \right]_0^8 \cr & {M_x} = \frac{3}{{14}}\rho \left[ {{{\left( 8 \right)}^{7/3}} - {{\left( 0 \right)}^{7/3}}} \right] \cr & {M_x} = \frac{{192}}{7}\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_0^8 x \left[ {{x^{2/3}} - 0} \right]dx \cr & {M_y} = \rho \int_0^8 {{x^{5/3}}} dx \cr & {M_y} = \rho \left[ {\frac{3}{8}{x^{5/3}}} \right]_0^8 \cr & {M_y} = \frac{3}{8}\rho \left[ {{{\left( 8 \right)}^{8/3}} - {{\left( 0 \right)}^{5/3}}} \right] \cr & {M_y} = 96\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{{96\rho }}{{\frac{{96}}{5}\rho }} = 5 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{192}}{7}\rho }}{{\frac{{96}}{5}\rho }} = \frac{{10}}{7} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{{54}}{5},\frac{5}{2}} \right) \cr & \cr & {\text{Summary}} \cr & m = \frac{{96}}{5}\rho \cr & {M_x} = \frac{{192}}{7}\rho \cr & {M_y} = 96\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {5,\frac{{10}}{7}} \right) \cr} $$