Next, we need to compute the two moments. We didn’t include the density in the computations below because it will only cancel out in the final step.

\[\begin{align*}{M_x} & = \int_{0}^{2}{{\frac{1}{2}{{\left( {4 - {x^2}} \right)}^2}\,dx}} = \int_{0}^{2}{{\frac{1}{2}\left( {16 - 8{x^2} + {x^4}} \right)\,dx}} = \left. {\frac{1}{2}\left( {16x - \frac{8}{3}{x^3} + \frac{1}{5}{x^5}} \right)} \right|_0^2 = \frac{{128}}{{15}}\\ {M_y} & = \int_{0}^{2}{{x\left( {4 - {x^2}} \right)\,dx}} = \int_{0}^{2}{{4x - {x^3}\,dx}} = \left. {\left( {2{x^2} - \frac{1}{4}{x^4}} \right)} \right|_0^2 = 4\end{align*}\] Show Step 3

Finally, the coordinates of the center of mass is,

\[\overline{x} = \frac{{{M_y}}}{M} = \frac{{\rho \left( 4 \right)}}{{\rho \left( {\frac{{16}}{3}} \right)}} = \frac{3}{4}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\overline{y} = \frac{{{M_x}}}{M} = \frac{{\rho \left( {\frac{{128}}{{15}}} \right)}}{{\rho \left( {\frac{{16}}{3}} \right)}} = \frac{8}{5}\]

The center of mass is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\frac{3}{4},\frac{8}{5}} \right)}}\).