Question

The lamina in the figure that follows is bounded above by the lines with equations $y=x+2a$and $y=-x+2a$and below by thex-axis on the interval $-a\le x\le a.$The density of the lamina is constant.

Short answer

The Center of mass of lumina is at$\left(\overline{x},\overline{y}\right)=\left(0,\frac{7a}{9}\right).$

Step-by-step solution

Step 1. Given information.   

The given lamina is the following.

The density of the given lamina is constant.

Step 2. x coordinate Center of mass of the left lamina  

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

role="math" localid="1650352640964" $$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ {\overline{x}}_1=\frac{{\int }_{-a}^0{\int }_0^{x+2a}xkdydx}{{\int }_{-a}^0{\int }_0^{x+2a}kdydx} \\ {\overline{x}}_1=\frac{k{\int }_{-a}^0[y{]}_0^{x+2a}xdx}{k{\int }_{-a}^0[y{]}_0^{x+2a}dx} \\ \overline{x_1}=\frac{{\displaystyle k}{\int }_{-a}^0[x^2+2ax]dx}{{\displaystyle k}{\int }_{-a}^0[x+2a]dx} \\ {\overline{x}}_1=\frac{{\displaystyle {[\frac{{\displaystyle x^3}}{{\displaystyle 3}}+a{x}^2]}_{-a}^0}}{{\displaystyle {[\frac{{\displaystyle x^2}}{{\displaystyle 2}}+2ax]}_{-a}^0}} \\ \overline{x_1}=\frac{{\displaystyle [-\frac{{\displaystyle 2a^3}}{{\displaystyle 3}}]}}{{\displaystyle \left[\frac{{\displaystyle 3a^2}}{{\displaystyle 2}}\right]}}=-\frac{{\displaystyle 4}}{{\displaystyle 9}}a \end{gathered}$$

Step 3. y coordinate the Center of mass of the left lamina  

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

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

Step 4. x coordinate Center of mass of the right lamina 

substituting $\rho (x,y)=k$in the formula of the center of mass $\overline{x}$for the right lamina.
role="math" localid="1650352880244" $$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ \overline{x_2}=\frac{{\int }_0^a{\int }_0^{-x+2a}xkdydx}{{\int }_0^a{\int }_0^{-x+2a}kdydx} \\ \overline{x_2}=\frac{k{\int }_0^a[y{]}_0^{-x+2a}xdx}{k{\int }_0^a[y{]}_0^{-x+2a}dx} \\ \overline{x_2}=\frac{{\int }_0^a[-x^2+2ax]dx}{{\int }_0^a[-x+2a]dx} \\ \overline{x_1}=\frac{{[-\frac{x^3}{3}+a{x}^2]}_0^a}{{[-\frac{x^2}{2}+2ax]}_0^a} \\ \overline{x_2}=\frac{[-\frac{a^3}{3}+a^3]}{[-\frac{a^2}{2}+2a^2]} \\ \overline{x_2}=\frac{4}{9}a \end{gathered}$$

Step 5. y coordinate the Center of mass of the right lamina 

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

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

Step 6. Center of mass of composition of the lamina.    

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

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

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