91Ó°ÊÓ

Determine x coordinate Center of mass of the horizontal lamina where x varies from $-3\mathrm{to}3$and y varies from $4\mathrm{to}6.$

role="math" localid="1650346135185" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ \begin{array}{rl}\stackrel{¯}{x_1}& =\frac{k{∫}_{−3}^3{∫}_4^{6}xdydx}{k{∫}_{−3}^3{∫}_4^{6}dydx}\\ \stackrel{¯}{x_1}& =\frac{{∫}_{−3}^3[y{]}_4^{3}xdx}{{∫}_{−3}^3[y{]}_4^{6}dx}\\ \stackrel{¯}{x_1}& =\frac{2{∫}_{−3}^{3}xdx}{2{∫}_{−3}^{3}dx}\\ \stackrel{¯}{x_1}& =\frac{2\left({\left[\frac{x^2}{2}\right]}_{−3}^3\right)}{2\left([x{]}_{−3}^3\right)}\\ \stackrel{¯}{x_1}& =0\end{array} \end{gathered}$$

Determine y coordinate Center of mass of the horizontal lamina where x varies from $-3\mathrm{to}3$and y varies from $4\mathrm{to}6.$

role="math" localid="1650346174670" $\begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\\ \stackrel{¯}{y_1}& =\frac{{∫}_{−3}^3{∫}_4^{6}kydydx}{{∫}_{−3}^3{∫}_4^{6}kdydx}\\ \stackrel{¯}{y_1}& =\frac{{∫}_{−3}^3{\left[\frac{y^2}{2}\right]}_{−4}^{6}dx}{{∫}_{−3}^3[y{]}_4^{6}dx}\\ \stackrel{¯}{y_1}& =\frac{10{∫}_{−3}^{3}dx}{2{∫}_{−3}^{3}dx}\\ \stackrel{¯}{y_1}=& \frac{10\left([x{]}_{-3}^3\right)}{2\left([x{]}_{−3}^3\right)}\\ \stackrel{¯}{y_1}=& 5\end{array}$

So the center of mass of horizontal lamina is$\left(\overline{x_1},\overline{y_1}\right)=\left(0,5\right).$

Determinex coordinate Center of mass of the horizontal lamina where x varies from $-1\mathrm{to}1$and y varies from $1\mathrm{to}4.$

role="math" localid="1650346297780" $$\begin{gathered} \stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA} \\ \begin{array}{rl}\stackrel{¯}{x_2}& =\frac{{∫}_{−1}^1{∫}_1^{4}kxdydx}{{∫}_{−1}^1{∫}_1^{4}kdydx}\\ \stackrel{¯}{x_2}& =\frac{{∫}_{−1}^1[y{]}_1^{4}xdx}{{∫}_{−1}^1[y{]}_1^{4}dx}\\ \stackrel{¯}{x_2}& =\frac{3{∫}_{−1}^{1}xdx}{3{∫}_{−1}^{1}dx}\\ \stackrel{¯}{x_2}& =\frac{3\left({\left[\frac{x^2}{2}\right]}_{−1}^1\right)}{3\left([x{]}_{−1}^1\right)}\\ \stackrel{¯}{x_2}& =0\end{array} \end{gathered}$$

Determiney coordinate Center of mass of the horizontal lamina where x varies from $-1\mathrm{to}1$and y varies from $1\mathrm{to}4$

$\begin{array}{rl}\stackrel{¯}{y}& =\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\\ {\stackrel{¯}{y}}_2& =\frac{{∫}_{−1}^1{∫}_1^{4}ykdydx}{{∫}_{−1}^1{∫}_1^{4}kdydx}\\ {\stackrel{¯}{y}}_2& =\frac{{∫}_{−1}^1{\left[\frac{y^2}{2}\right]}_1^4}{{∫}_{−1}^1[y{]}_1^{4}dx}\\ {\stackrel{¯}{y}}_2& =\frac{\frac{15}{2}{∫}_{−1}^{1}dx}{3{∫}_{−1}^{1}dx}\\ {\stackrel{¯}{y}}_2& =\frac{\frac{15}{2}[x{]}_{−1}^1}{3[x{]}_{−1}^1}\\ {\stackrel{¯}{y}}_2& =\frac{5}{2}\end{array}$

So the center of mass of vertical lamina is$\left(\overline{x_2},\overline{y_2}\right)=\left(0,\frac{5}{2}\right).$

add coordinates of all center of mass of all laminas to determine the Center of mass of composition of the lamina.

$\begin{array}{rl}(\stackrel{¯}{X},\stackrel{¯}{Y})& =({\stackrel{¯}{x}}_1,{\stackrel{¯}{y}}_1)+({\stackrel{¯}{x}}_2,{\stackrel{¯}{y}}_2)\\ (\stackrel{¯}{X},\stackrel{¯}{Y})& =(0,5)+(0,\frac{5}{2})\\ (\stackrel{¯}{X},\stackrel{¯}{Y})& =\left((0+0),\left(5+\frac{5}{2}\right)\right)\\ (\stackrel{¯}{X},\stackrel{¯}{Y})& =(0,\frac{15}{2})\end{array}$

So the center of mass of lamina is $\left(0,\frac{15}{2}\right).$