91Ó°ÊÓ

The given lamina is the following.

The density of the given lamina is constant.

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

role="math" localid="1650354896531" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ {\overline{x}}_1=\frac{{∫}_{−a}^0{∫}_0^{x+2a}xkydydx}{{∫}_{−a}^0{∫}_0^{x+2a}kydydx} \\ {\overline{x}}_1=\frac{{\displaystyle k{∫}_{−a}^0{\left[\frac{{\displaystyle y^2}}{{\displaystyle 2}}\right]}_0^{x+2a}}}{{\displaystyle k{∫}_{−a}^0{\left[\frac{{\displaystyle y^2}}{{\displaystyle 2}}\right]}_{−a}^{x+2a}}}dx \\ {\overline{x}}_1=\frac{{\displaystyle {∫}_{−a}^0(x^3+4a{x}^2+4a^{2}x)dx}}{{\displaystyle {∫}_{−a}^0(x^2+4ax+4a^2)dx}} \\ {\overline{x}}_1=\frac{{\displaystyle {[\frac{{\displaystyle x^4}}{{\displaystyle 4}}+\frac{{\displaystyle 4a{x}^3}}{{\displaystyle 3}}+\frac{{\displaystyle 4a^2{x}^2}}{{\displaystyle 2}}]}_{−a}^0}}{{\displaystyle {[\frac{{\displaystyle x^3}}{{\displaystyle 3}}+\frac{{\displaystyle 4a{x}^2}}{{\displaystyle 2}}+4a^{2}x]}_{−a}^0}} \\ {\overline{x}}_1=\frac{{\displaystyle −\left(\frac{{\displaystyle a^4}}{{\displaystyle 4}}−\frac{{\displaystyle 4a^4}}{{\displaystyle 3}}+\frac{{\displaystyle 4a^4}}{{\displaystyle 2}}\right)}}{{\displaystyle −\left(−\frac{{\displaystyle a^3}}{{\displaystyle 3}}+\frac{{\displaystyle 4a^3}}{{\displaystyle 2}}−4a^3\right)}} \\ {\overline{x}}_1=\frac{{\displaystyle -11a}}{{\displaystyle 28}} \end{gathered}$$

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

role="math" localid="1650354512152" $$\begin{gathered} \begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\end{array} \\ \overline{y_1}=\frac{{∫}_{−a}^0{∫}_0^{x+2a}ykydydx}{{∫}_{−a}^0{∫}_0^{x+2a}kydydx} \\ \overline{y_1}=\frac{k{∫}_{−a}^0{\left[\frac{y^3}{3}\right]}_0^{x+2a}dx}{k{∫}_{−a}^0{\left[\frac{y^2}{2}\right]}_0^{x+2a}dx} \\ \overline{y_1}=\frac{{∫}_{−a}^0\left[\frac{x^3+6a{x}^2+12a^{2}x+8a^3}{3}\right]dx}{{∫}_{−a}^0\left[\frac{x^2+4ax+4a^2}{2}\right]dx} \\ \overline{y_1}=\frac{{\left[\frac{\frac{x^4}{4}+2a{x}^3+6a^2{x}^2+8a^{3}x}{3}\right]}_{−a}^0}{{\left[\frac{\frac{x^3}{3}+2a{x}^2+4a^{2}x}{2}\right]}_{−a}^0} \\ \overline{y_1}=\frac{-\left(\frac{\frac{a^4}{4}−2a^4+6a^4−8a^4}{3}\right)}{-{\displaystyle \left(\frac{−\frac{a^3}{3}+2a^3−4a^3}{2}\right)}} \\ \overline{y_1}=\frac{15a}{14} \end{gathered}$$
So the center of mass of left lamina is at$\left(\overline{x_1},\overline{y_1}\right)=\left(\frac{-11a}{28},\frac{15a}{14}\right).$

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

role="math" localid="1650354858036" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ {\overline{x}}_2=\frac{{∫}_0^a{∫}_0^{−x+2a}xkydydx}{{∫}_0^a{∫}_0^{−x+2a}kydydx} \\ {\overline{x}}_2=\frac{{∫}_0^a{\left[\frac{y^2}{2}\right]}_0^{−x+2a}xdx}{k{∫}_0^a{\left[\frac{y^2}{2}\right]}_0^{−x+2a}dx} \\ \overline{{\overline{x}}_2}=\frac{{∫}_0^a(x^3−4a{x}^2+4a^{2}x)dx}{{∫}_0^a(x^2−4ax+4a^2)dx} \\ {\overline{x}}_2=\frac{{\left[\frac{x^4}{4}−\frac{4a{x}^3}{3}+\frac{4a^2{x}^2}{2}\right]}_0^a}{{\left[\frac{x^3}{3}−\frac{4a{x}^2}{2}+4a^{2}x\right]}_0^a} \\ {\overline{x}}_2=\frac{\frac{a^4}{4}−\frac{4a^4}{3}+\frac{4a^4}{2}}{\frac{a^3}{3}−\frac{4a^3}{2}+4a^3} \\ {\overline{x}}_2=\frac{11a}{28} \end{gathered}$$

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

role="math" localid="1650355216220" $$\begin{gathered} \begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\end{array} \\ {\overline{y}}_2=\frac{{∫}_0^a{∫}_0^{−x+2a}{y}^{2}kdydx}{{∫}_0^a{∫}_0^{−x+2a}kydydx} \\ {\overline{y}}_2=\frac{k{∫}_0^a{\left[\frac{y^3}{3}\right]}_0^{−x+2a}dx}{k{∫}_0^a{\left[\frac{y^2}{2}\right]}_0^{−x+2a}dx} \\ {\overline{y}}_2=\frac{{∫}_{−a}^0\left[\frac{−x^3+6a{x}^2−12a^{2}x+8a^3}{3}\right]dx}{{∫}_{−a}^0\left[\frac{x^2−4ax+4a^2}{2}\right]dx} \\ {\overline{y}}_2=\frac{{\left[\frac{−\frac{x^4}{4}+2a{x}^3−6a^2{x}^2+8a^{3}x}{3}\right]}_0^a}{{\left[\frac{\frac{x^3}{3}−2a{x}^2+4a^{2}x}{2}\right]}_0^a} \\ {\overline{y}}_2=\frac{\left(\frac{−\frac{a^4}{4}+2a^4−6a^4+8a^4}{3}\right)}{\left(\frac{\frac{a^3}{3}−2a^3+4a^3}{2}\right)} \\ {\overline{y}}_2=\frac{15a}{14} \end{gathered}$$

So the center of mass of right lamina is at$\left(\overline{x_2},\overline{y_2}\right)=\left(\frac{11a}{28},\frac{15a}{14}\right).$

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

$\begin{array}{rl}\stackrel{¯}{x}& =\frac{m_1{\stackrel{¯}{x}}_1+m_2{\stackrel{¯}{x}}_2}{m_1+m_2}\\ \stackrel{¯}{x}& =\frac{m(−\frac{11}{28}a)+m\left(\frac{11}{28}a\right)}{m+m}\\ \stackrel{¯}{x}& =0\\ \stackrel{¯}{y}& =\frac{m_1{\stackrel{¯}{y}}_1+m_2{\stackrel{¯}{y}}_2}{m_1+m_2}\\ \stackrel{¯}{y}& =\frac{m\left(\frac{15}{14}a\right)+m\left(\frac{15}{14}a\right)}{m+m}\\ \stackrel{¯}{y}& =\frac{15}{14}a\end{array}$

So the center of mass of the lamina is at$\left(\overline{x},\overline{y}\right)=\left(0,\frac{15a}{14}\right).$