91Ó°ÊÓ

The given lamina is the following.

The density of the given lamina is constant.

substituting $\mathrm{ÒÏ}(x,y)=k$in the formula of the center of mass $\overline{x}$for left lamina.

role="math" localid="1650352640964" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ {\stackrel{¯}{x}}_1=\frac{{∫}_{−a}^0{∫}_0^{x+2a}xkdydx}{{∫}_{−a}^0{∫}_0^{x+2a}kdydx} \\ {\stackrel{¯}{x}}_1=\frac{k{∫}_{−a}^0[y{]}_0^{x+2a}xdx}{k{∫}_{−a}^0[y{]}_0^{x+2a}dx} \\ \stackrel{¯}{x_1}=\frac{{\displaystyle k}{∫}_{−a}^0[x^2+2ax]dx}{{\displaystyle k}{∫}_{−a}^0[x+2a]dx} \\ {\stackrel{¯}{x}}_1=\frac{{\displaystyle {[\frac{{\displaystyle x^3}}{{\displaystyle 3}}+a{x}^2]}_{−a}^0}}{{\displaystyle {[\frac{{\displaystyle x^2}}{{\displaystyle 2}}+2ax]}_{−a}^0}} \\ \stackrel{¯}{x_1}=\frac{{\displaystyle [−\frac{{\displaystyle 2a^3}}{{\displaystyle 3}}]}}{{\displaystyle \left[\frac{{\displaystyle 3a^2}}{{\displaystyle 2}}\right]}}=−\frac{{\displaystyle 4}}{{\displaystyle 9}}a \end{gathered}$$

substituting $\mathrm{ÒÏ}(x,y)=k$in the formula of the center of mass $\overline{y}$for left lamina.

role="math" localid="1650352548682" $$\begin{gathered} \begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\end{array} \\ \overline{y_1}=\frac{{∫}_{−a}^0{∫}_0^{x+2a}ykdydx}{{∫}_{−a}^0{∫}_0^{x+2a}kdydx} \\ \overline{y_1}=\frac{k{∫}_{−a}^0{\left[\frac{y^2}{2}\right]}_0^{x+2a}dx}{k{∫}_{−a}^0[y{]}_0^{x+2a}dx} \\ \overline{y_1}=\frac{{∫}_{−a}^0\left[\frac{(x^2+4ax+4a^2)]}{2}\right]dx}{{∫}_{−a}^0[x+2a]dx} \\ \overline{y_1}=\frac{\frac{1}{2}{\left[\frac{x^3}{3}+2a{x}^2+4a^{2}x\right]}_{−a}^0}{{\left[\frac{x^2}{2}+2ax\right]}_{-a}^a} \\ \overline{y_1}=\frac{\frac{1}{2}(\frac{a^3}{3}−2a^3+4a^3)}{(−\frac{a^2}{2}−2a^2)} \\ \overline{y_1}=\frac{7}{9}a \end{gathered}$$
So the center of mass of the left lamina is at$\left(\overline{x_1},\overline{y_1}\right)=\left(\frac{-4a}{9},\frac{7a}{9}\right).$

substituting $\mathrm{ÒÏ}(x,y)=k$in the formula of the center of mass $\overline{x}$for the right lamina.
role="math" localid="1650352880244" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ \overline{x_2}=\frac{{∫}_0^a{∫}_0^{−x+2a}xkdydx}{{∫}_0^a{∫}_0^{−x+2a}kdydx} \\ \overline{x_2}=\frac{k{∫}_0^a[y{]}_0^{−x+2a}xdx}{k{∫}_0^a[y{]}_0^{−x+2a}dx} \\ \overline{x_2}=\frac{{∫}_0^a[−x^2+2ax]dx}{{∫}_0^a[−x+2a]dx} \\ \overline{x_1}=\frac{{[−\frac{x^3}{3}+a{x}^2]}_0^a}{{[−\frac{x^2}{2}+2ax]}_0^a} \\ \overline{x_2}=\frac{[−\frac{a^3}{3}+a^3]}{[−\frac{a^2}{2}+2a^2]} \\ \overline{x_2}=\frac{4}{9}a \end{gathered}$$

substituting $\mathrm{ÒÏ}(x,y)=k$in the formula of the center of mass $\overline{y}$for the right lamina.

role="math" localid="1650353155945" $$\begin{gathered} \begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\end{array} \\ \overline{y_2}=\frac{{∫}_0^a{∫}_0^{−x+2a}ykdydx}{{∫}_0^a{∫}_0^{−x+2a}kdydx} \\ \overline{y_2}=\frac{{\displaystyle {∫}_0^a{\left[\frac{{\displaystyle y^2}}{{\displaystyle 2}}\right]}_0^{−x+2a}}}{{\displaystyle k{∫}_0^a[y{]}_0^{−x+2a}dx}} \\ \overline{y_2}=\frac{{\displaystyle {∫}_0^a\left[\frac{{\displaystyle (x^2−4ax+4a^2)}}{{\displaystyle 2}}\right]dx}}{{\displaystyle {∫}_0^a[−x+2a]dx}} \\ \overline{y_2}=\frac{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle 2}}{[\frac{{\displaystyle x^3}}{{\displaystyle 3}}−2a{x}^2+4a^{2}x]}_0^a}}{{\displaystyle {[−\frac{{\displaystyle x^2}}{{\displaystyle 2}}+2ax]}_0^a}} \\ \overline{y_2}=\frac{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle 2}}(\frac{{\displaystyle a^3}}{{\displaystyle 3}}−2a^3+4a^3)}}{{\displaystyle (−\frac{{\displaystyle a^2}}{{\displaystyle 2}}+2a^2)}} \\ \overline{y_2}=\frac{{\displaystyle 7}}{{\displaystyle 9}}a \end{gathered}$$
So the center of mass of the right lamina is at$\left(\overline{x_2},\overline{y_2}\right)=\left(\frac{4a}{9},\frac{7a}{9}\right).$

The Center of mass of composition of the left and right lamina is following.

$\begin{array}{rl}\stackrel{¯}{x}& =\frac{m_1{\stackrel{¯}{x}}_1+m_2{\stackrel{¯}{x}}_2}{m_1+m_2}\\ \stackrel{¯}{x}& =\frac{m(−\frac{4}{9}a)+m\left(\frac{4}{9}a\right)}{m+m}\\ \stackrel{¯}{x}& =0\\ \stackrel{¯}{y}& =\frac{m_1{\stackrel{¯}{y}}_1+m_2{\stackrel{¯}{y}}_2}{m_1+m_2}\\ \stackrel{¯}{y}& =\frac{m\left(\frac{7}{9}a\right)+m\left(\frac{7}{9}a\right)}{m+m})\\ \stackrel{¯}{y}& =\frac{7}{9}a\end{array}$

So the center of mass of the lamina is at$\left(\overline{x},\overline{y}\right)=\left(0,\frac{7a}{9}\right).$