Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 19

$$\eqalign{ & m = \frac{9}{2}\rho \cr & {M_x} = \frac{{99}}{5}\rho \cr & {M_y} = \frac{{27}}{4}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{2},\frac{{22}}{5}} \right) \cr} $$

$$\eqalign{ & y = - {x^2} + 4x + 2,{\text{ }}y = x + 2 \cr & - {x^2} + 4x + 2 = x + 2 \cr & - {x^2} + 3x = 0 \cr & x = 0,{\text{ }}x = 3 \cr & - {x^2} + 4x + 2 \geqslant x + 2{\text{ on the interval }}\left[ {0,3} \right],{\text{ then}} \cr & {\text{Let }}f\left( x \right) = - {x^2} + 4x + 2{\text{ and }}g\left( x \right) = x + 2,{\text{ on }}\left[ {0,3} \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_0^3 {\left[ {\left( { - {x^2} + 4x + 2} \right) - \left( {x + 2} \right)} \right]} dx \cr & m = \rho \int_0^3 {\left( { - {x^2} + 4x + 2 - x - 2} \right)} dx \cr & m = \rho \int_0^3 {\left( { - {x^2} + 3x} \right)} dx \cr & m = \rho \left[ { - \frac{1}{3}{x^3} + \frac{3}{2}{x^2}} \right]_0^3 = \rho \left[ { - \frac{1}{3}{{\left( 3 \right)}^3} + \frac{3}{2}{{\left( 3 \right)}^2}} \right] - 0 \cr & m = \frac{9}{2}\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_0^3 {\left[ {\frac{{\left( { - {x^2} + 4x + 2} \right) + \left( {x + 2} \right)}}{2}} \right]} \left( { - {x^2} + 3x} \right)dx \cr & {M_x} = \frac{1}{2}\rho \int_0^3 {\left( { - {x^2} + 5x + 4} \right)\left( { - {x^2} + 3x} \right)} dx \cr & {M_x} = \frac{1}{2}\rho \int_0^3 {\left( {{x^4} - 8{x^3} + 11{x^2} + 12x} \right)} dx \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{5}{x^5} - 2{x^4} + \frac{{11}}{3}{x^3} + 6{x^2}} \right]_0^3 \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{5}{{\left( 3 \right)}^5} - 2{{\left( 3 \right)}^4} + \frac{{11}}{3}{{\left( 3 \right)}^3} + 6{{\left( 3 \right)}^2}} \right] - \frac{1}{2}\rho \left[ 0 \right] \cr & {M_x} = \frac{{99}}{5}\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_0^3 x \left[ {\left( { - {x^2} + 4x + 2} \right) + \left( {x + 2} \right)} \right]dx \cr & {M_y} = \rho \int_0^3 x \left( { - {x^2} + 3x} \right)dx \cr & {M_y} = \rho \int_0^3 {\left( { - {x^3} + 3{x^2}} \right)} dx \cr & {M_y} = \rho \left[ { - \frac{1}{4}{x^4} + {x^3}} \right]_0^3 \cr & {M_y} = \rho \left[ { - \frac{1}{4}{{\left( 3 \right)}^4} + {{\left( 3 \right)}^3}} \right] - \rho \left[ { - \frac{1}{4}{{\left( 0 \right)}^4} + {{\left( 0 \right)}^3}} \right] \cr & {M_y} = \frac{{27}}{4}\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{{27}}{4}\rho }}{{\frac{9}{2}\rho }} = \frac{3}{2} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{99}}{5}\rho }}{{\frac{9}{2}\rho }} = \frac{{22}}{5} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{2},\frac{{22}}{5}} \right) \cr & \cr & {\text{Summary}} \cr & m = \frac{9}{2}\rho \cr & {M_x} = \frac{{99}}{5}\rho \cr & {M_y} = \frac{{27}}{4}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{2},\frac{{22}}{5}} \right) \cr} $$