91Ó°ÊÓ

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

role="math" localid="1650347825085" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ \begin{array}{rl}{\stackrel{¯}{x}}_1& =\frac{{∫}_{−3}^3{∫}_4^{6}xkydydx}{{∫}_{−3}^3{∫}_4^{6}kydydx}\\ {\stackrel{¯}{x}}_1& =\frac{k{∫}_{−3}^3{\left[\frac{y^2}{2}\right]}_4^{6}xdx}{k{∫}_{−3}^3{\left[\frac{y^2}{2}\right]}_{−3}^{6}dx}\end{array} \\ \begin{array}{rl}{\stackrel{¯}{x}}_1=& \frac{2k{∫}_{−3}^{3}xdx}{2k{∫}_{−3}^{3}dx}\\ {\stackrel{¯}{x}}_1=& \frac{2k\left({\left[\frac{x^2}{2}\right]}_{−3}^3\right)}{2k\left([x{]}_{−3}^3\right)}\\ {\stackrel{¯}{x}}_1& \begin{array}{rl}=0& \end{array}\end{array} \end{gathered}$$

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

role="math" localid="1650348806680" $$\begin{gathered} \begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\end{array} \\ \overline{y_1}=\frac{{∫}_{−3}^3{∫}_4^{6}ykydydx}{k{∫}_{−3}^3{∫}_4^{6}kydydx} \\ \overline{y_1}=\frac{k{∫}_{−3}^3{\left[\frac{y^3}{3}\right]}_{−4}^6}{k{∫}_{−3}^3{\left[y^2\right]}_{-4}^6}dx \\ \overline{y_1}=\frac{76k{∫}_{−3}^{3}dx}{15k{∫}_{−3}^{3}dx} \\ \overline{y_1}=\frac{76k[x{]}_{−3}^3}{15k[x{]}_{−3}^3} \\ \overline{y_1}=\frac{76}{15} \end{gathered}$$

So the center of mass of horizontal lamina is$\left(\overline{x_1},\overline{y_1}\right)=\left(0,\frac{76}{15}\right).$

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

role="math" localid="1650348765861" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ \overline{x_2}=\frac{{∫}_{−1}^1{∫}_0^{4}xkydydx}{k{∫}_{−1}^1{∫}_0^{4}kydydx} \\ \overline{x_2}=\frac{k{∫}_{−1}^1{\left[\frac{y^2}{2}\right]}_0^{4}xdx}{k{∫}_{−1}^1{\left[\frac{y^2}{2}\right]}_0^{4}dx} \\ \overline{x_2}=\frac{8k{∫}_{−1}^{1}xdx}{8k{∫}_{−1}^{1}dx} \\ \overline{x_2}=\frac{8k{\left[\frac{x^2}{2}\right]}_{−1}^1}{8k[x{]}_{−1}^1} \\ \overline{x_2}=0 \end{gathered}$$

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

$$\begin{gathered} \overline{y_2}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ \overline{y_2}=\frac{{∫}_{−1}^1{{∫}_0^4}_{\mathrm{Î\copyright }}ykydydx}{{∫}_{−1}^1{∫}_0^{4}kydydx} \\ \overline{y_2}=\frac{k{∫}_{−1}^1{\left[\frac{y^3}{3}\right]}_0^{4}dx}{k{∫}_{−1}^1{\left[\frac{y^2}{2}\right]}_0^4}dx \\ \overline{y_2}=\frac{{\displaystyle \frac{64}{3}}{∫}_{−1}^{1}dx}{8{∫}_{−1}^{1}dx} \\ \overline{y_2}=\frac{8[x{]}_{−1}^1}{3[x{]}_{−1}^1} \\ \overline{y_2}=\frac{8}{3} \end{gathered}$$

So the center of mass of vertical lamina is$\left(\overline{x_2},\overline{y_2}\right)=\left(0,\frac{8}{3}\right).$

Considering the mass of each lamina is m then the Center of mass of composition of the lamina is following.

$\begin{array}{rl}\stackrel{¯}{x}& =\frac{m_1{\stackrel{¯}{x}}_1+m_2{\stackrel{¯}{x}}_2}{m_1+m_2}\\ \stackrel{¯}{x}& =\frac{m\left(0\right)+m\left(0\right)}{m+m}\\ \stackrel{¯}{x}& =0\\ \stackrel{¯}{y}& =\frac{m_1{\stackrel{¯}{y}}_1+m_2{\stackrel{¯}{y}}_2}{m_1+m_2}\\ \stackrel{¯}{y}& =\frac{m\left(\frac{76}{15}\right)+m\left(\frac{8}{3}\right)}{m+m}\\ \stackrel{¯}{y}& =\frac{\left(\frac{116}{15}\right)}{2}\\ \stackrel{¯}{y}& =\frac{58}{15}\end{array}$

So the center of mass of the lamina is at$\left(\overline{X},\overline{Y}\right)=\left(0,\frac{58}{15}\right).$