Q. 64 - Calculus

Question

In the following lamina, all angles are right angles and the density is constant:

Step-by-Step Solution

Answer

The Center of mass of lamina is $\left(0,\frac{15}{2}\right).$

Step 1. Given information.

Given lamina is a composition of rectangles.

density is constant. 

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

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

$$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ \begin{array}{rl}\overline{x_1}& =\frac{k{\int }_{-3}^3{\int }_4^{6}xdydx}{k{\int }_{-3}^3{\int }_4^{6}dydx}\\ \overline{x_1}& =\frac{{\int }_{-3}^3[y{]}_4^{3}xdx}{{\int }_{-3}^3[y{]}_4^{6}dx}\\ \overline{x_1}& =\frac{2{\int }_{-3}^{3}xdx}{2{\int }_{-3}^{3}dx}\\ \overline{x_1}& =\frac{2\left({\left[\frac{x^2}{2}\right]}_{-3}^3\right)}{2\left([x{]}_{-3}^3\right)}\\ \overline{x_1}& =0\end{array} \end{gathered}$$

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

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

$\begin{array}{rl}\overline{y}& =\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\\ \overline{y_1}& =\frac{{\int }_{-3}^3{\int }_4^{6}kydydx}{{\int }_{-3}^3{\int }_4^{6}kdydx}\\ \overline{y_1}& =\frac{{\int }_{-3}^3{\left[\frac{y^2}{2}\right]}_{-4}^{6}dx}{{\int }_{-3}^3[y{]}_4^{6}dx}\\ \overline{y_1}& =\frac{10{\int }_{-3}^{3}dx}{2{\int }_{-3}^{3}dx}\\ \overline{y_1}=& \frac{10\left([x{]}_{-3}^3\right)}{2\left([x{]}_{-3}^3\right)}\\ \overline{y_1}=& 5\end{array}$

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

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

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

$$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ \begin{array}{rl}\overline{x_2}& =\frac{{\int }_{-1}^1{\int }_1^{4}kx dydx}{{\int }_{-1}^1{\int }_1^{4}k dydx}\\ \overline{x_2}& =\frac{{\int }_{-1}^1[y{]}_1^{4}xdx}{{\int }_{-1}^1[y{]}_1^{4}dx}\\ \overline{x_2}& =\frac{3{\int }_{-1}^{1}xdx}{3{\int }_{-1}^{1}dx}\\ \overline{x_2}& =\frac{3\left({\left[\frac{x^2}{2}\right]}_{-1}^1\right)}{3\left([x{]}_{-1}^1\right)}\\ \overline{x_2}& =0\end{array} \end{gathered}$$

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

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

$\begin{array}{rl}\overline{y}& =\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\\ {\overline{y}}_2& =\frac{{\int }_{-1}^1{\int }_1^{4}yk dydx}{{\int }_{-1}^1{\int }_1^{4}k dydx}\\ {\overline{y}}_2& =\frac{{\int }_{-1}^1{\left[\frac{y^2}{2}\right]}_1^4}{{\int }_{-1}^1[y{]}_1^{4}dx}\\ {\overline{y}}_2& =\frac{\frac{15}{2}{\int }_{-1}^{1}dx}{3{\int }_{-1}^{1}dx}\\ {\overline{y}}_2& =\frac{\frac{15}{2}[x{]}_{-1}^1}{3[x{]}_{-1}^1}\\ {\overline{y}}_2& =\frac{5}{2}\end{array}$

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

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

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

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

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