Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 22

$$\eqalign{ & m = \frac{{128}}{5}\rho \cr & {M_x} = \frac{{512}}{7}\rho \cr & {M_y} = 0 \cr & \left( {\overline x ,\overline y } \right) = \left( {0,\frac{{20}}{7}} \right) \cr} $$

$$\eqalign{ & y = {x^{2/3}},{\text{ }}y = 4 \cr & y = y \cr & {x^{2/3}} = 4 \cr & x = \pm 8 \cr & 4 \geqslant {x^{2/3}}{\text{ on the interval }}\left[ { - 8,8} \right],{\text{ then}} \cr & {\text{Let }}f\left( x \right) = 4{\text{ and }}g\left( x \right) = {x^{2/3}},{\text{ on }}\left[ { - 8,8} \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_{ - 8}^8 {\left( {4 - {x^{2/3}}} \right)} dx \cr & {\text{By symmetry}} \cr & m = 2\rho \int_0^8 {\left( {4 - {x^{2/3}}} \right)} dx \cr & m = 2\rho \left[ {4x - \frac{3}{5}{x^{5/3}}} \right]_0^8 = 2\rho \left[ {4\left( 8 \right) - \frac{3}{5}{{\left( 8 \right)}^{5/3}}} \right] - 0 \cr & m = \frac{{128}}{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_{ - 8}^8 {\left( {\frac{{4 + {x^{2/3}}}}{2}} \right)} \left( {4 - {x^{2/3}}} \right)dx \cr & {M_x} = \frac{1}{2}\rho \int_{ - 8}^8 {\left( {16 - {x^{4/3}}} \right)} dx \cr & {M_x} = \rho \int_0^8 {\left( {16 - {x^{4/3}}} \right)} dx \cr & {M_x} = \rho \left[ {16x - \frac{3}{7}{x^{7/3}}} \right]_0^8 \cr & {M_x} = \rho \left[ {16\left( 8 \right) - \frac{3}{7}{{\left( 8 \right)}^{7/3}}} \right] \cr & {M_x} = \frac{{512}}{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_{ - 8}^8 x \left[ {4 - {x^{2/3}}} \right]dx \cr & {M_y} = \rho \int_{ - 8}^8 {\left( {4x - {x^{5/3}}} \right)} dx \cr & {M_y} = \rho \left[ {2{x^2} - \frac{3}{8}{x^{8/3}}} \right]_{ - 8}^8 \cr & {M_y} = \rho \left[ {2{{\left( 8 \right)}^2} - \frac{3}{8}{{\left( 8 \right)}^{8/3}}} \right] - \rho \left[ {2{{\left( { - 8} \right)}^2} - \frac{3}{8}{{\left( { - 8} \right)}^{8/3}}} \right] \cr & {M_y} = \rho \left( {32} \right) - \rho \left( {32} \right) \cr & {M_y} = 0 \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{0}{{\frac{{128}}{5}\rho }} = 0 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{512}}{7}\rho }}{{\frac{{128}}{5}\rho }} = \frac{{20}}{7} \cr & \left( {\overline x ,\overline y } \right) = \left( {0,\frac{{20}}{7}} \right) \cr & \cr & {\text{Summary}} \cr & m = \frac{{128}}{5}\rho \cr & {M_x} = \frac{{512}}{7}\rho \cr & {M_y} = 0 \cr & \left( {\overline x ,\overline y } \right) = \left( {0,\frac{{20}}{7}} \right) \cr} $$