Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 18

$$\eqalign{ & m = \frac{4}{3}\rho \cr & {M_x} = \frac{4}{3}\rho \cr & {M_y} = \frac{{32}}{{15}} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{8}{5},1} \right) \cr} $$

$$\eqalign{ & y = \sqrt x ,{\text{ }}y = \frac{1}{2}x \cr & \sqrt x = \frac{1}{2}x \cr & \sqrt x - \frac{1}{2}x = 0 \cr & x = 0,{\text{ }}x = 4 \cr & \sqrt x \geqslant \frac{1}{2}x{\text{ on the interval }}\left[ {0,4} \right],{\text{ then}} \cr & {\text{Let }}f\left( x \right) = \sqrt x {\text{ and }}g\left( x \right) = \frac{1}{2}x,{\text{ on the interval }}\left[ {0,4} \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_0^4 {\left( {\sqrt x - \frac{1}{2}x} \right)} dx \cr & m = \rho \left[ {\frac{2}{3}{x^{3/2}} - \frac{1}{4}{x^2}} \right]_0^4 = \rho \left[ {\frac{2}{3}{{\left( 4 \right)}^{3/2}} - \frac{1}{4}{{\left( 4 \right)}^2}} \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_0^4 {\left[ {\frac{{\sqrt x + \frac{1}{2}x}}{2}} \right]} \left[ {\sqrt x - \frac{1}{2}x} \right]dx \cr & {M_x} = \frac{1}{2}\rho \int_0^4 {\left( {x - \frac{1}{4}{x^2}} \right)} dx \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{2}{x^2} - \frac{1}{{12}}{x^3}} \right]_0^4 \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{2}{{\left( 4 \right)}^2} - \frac{1}{{12}}{{\left( 4 \right)}^3}} \right] - \frac{1}{2}\rho \left[ {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{{12}}{{\left( 0 \right)}^3}} \right] \cr & {M_x} = \frac{4}{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_0^4 x \left[ {\sqrt x - \frac{1}{2}x} \right]dx \cr & {M_y} = \rho \int_0^4 {\left( {{x^{3/2}} - \frac{1}{2}{x^2}} \right)} dx \cr & {M_y} = \rho \left[ {\frac{2}{5}{x^{5/2}} - \frac{1}{6}{x^3}} \right]_0^4 \cr & {M_y} = \rho \left[ {\frac{2}{5}{{\left( 4 \right)}^{5/2}} - \frac{1}{6}{{\left( 4 \right)}^3}} \right] - \rho \left[ {\frac{2}{5}{{\left( 0 \right)}^{3/2}} - \frac{1}{6}{{\left( 0 \right)}^3}} \right] \cr & {M_y} = \frac{{32}}{{15}} \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{{32}}{{15}}\rho }}{{\frac{4}{3}\rho }} = \frac{8}{5} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{4}{3}\rho }}{{\frac{4}{3}\rho }} = 1 \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{8}{5},1} \right) \cr & \cr } $$