Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 15

$$\left( {\overline x ,\overline y } \right) = \left( {\frac{{12}}{5},\frac{3}{4}} \right)$$

$$\eqalign{ & y = \sqrt x ,{\text{ }}y = 0,{\text{ }}x = 4 \cr & {\text{Let }}f\left( x \right) = \sqrt x {\text{ and }}g\left( x \right) = 0,{\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 - 0} \right)} dx \cr & m = \rho \int_0^4 {{x^{1/2}}} dx \cr & m = \frac{2}{3}\rho \left[ {{x^{3/2}}} \right]_0^4 = \frac{2}{3}\rho \left[ {{4^{3/2}} - {0^{3/2}}} \right] \cr & m = \frac{{16}}{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 - 0}}{2}} \right]} \left[ {\sqrt x - 0} \right]dx \cr & {M_x} = \frac{1}{2}\rho \int_0^4 x dx \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{2}{x^2}} \right]_0^4 \cr & {M_x} = \frac{1}{4}\rho \left[ {{{\left( 4 \right)}^2} - {{\left( 0 \right)}^2}} \right] \cr & {M_x} = 4\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 - 0} \right]dx \cr & {M_y} = \rho \int_0^4 {{x^{3/2}}} dx \cr & {M_y} = \frac{2}{5}\rho \left[ {{x^{5/2}}} \right]_0^4 \cr & {M_y} = \frac{2}{5}\rho \left[ {{{\left( 4 \right)}^{5/2}} - {{\left( 0 \right)}^{5/2}}} \right] \cr & {M_y} = \frac{{64}}{5}\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{{64}}{5}\rho }}{{\frac{{16}}{3}\rho }} = \frac{{12}}{5} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{4\rho }}{{\frac{{16}}{3}\rho }} = \frac{3}{4} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{{12}}{5},\frac{3}{4}} \right) \cr & \cr & {\text{Summary}} \cr & m = \frac{{16}}{3}\rho \cr & {M_x} = 4\rho \cr & {M_y} = \frac{{64}}{5}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{{12}}{5},\frac{3}{4}} \right) \cr} $$