Q 40. - Calculus

Question

Let $R$ be rectangular region with vertices $(0,0),(b,0),(0,h),\text{ and }(b,h)$

If the density at each point in $R$ is proportional to the point’s distance from the $y$-axis, find the center of mass of $R$ .

Step-by-Step Solution

Answer

The center of mass is $\overline{x}=\frac{2}{3}b,\overline{y}=\frac{h}{2}$

Step 1: Given Information

It is given that vertices of rectangular region are $\overline{x}=\frac{2}{3}b,\overline{y}=\frac{h}{2}$

Also $\text{ Density }\rho (x,y)=kx$

Step 2: Calculating x -

The formula is $\overline{x}=\frac{{\iint }_{\Omega }x\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}$

As $\rho (x,y)=kx$

$\overline{x}=\frac{{\int }_0^b{\int }_0^{h}xkxdydx}{{\int }_0^b{\int }_0^{h}kxdydx}=\frac{{\int }_0^b{\int }_0^{h}k{x}^{2}dydx}{{\int }_0^b{\int }_0^{h}kxdydx}$

Solving further

$\overline{x}=\frac{{\int }_0^{b}k{x}^2[y{]}_0^{h}dx}{{\int }_0^{b}kx[y{]}_0^{h}dx}=\frac{{\int }_0^{b}kh{x}^{2}dx}{{\int }_0^{b}khxdx}=\frac{kh{\int }_0^b{x}^{2}dx}{kh{\int }_0^{b}xdx}$

$\Rightarrow \overline{x}=\frac{{\left[\frac{x^3}{3}\right]}_0^b}{{\left[\frac{x^2}{2}\right]}_0^b}==\frac{\left[\frac{b^3}{3}\right]}{\left[\frac{b^2}{2}\right]}$

Hence, $\overline{x}=\frac{2}{3}b$

Step 3: Calculating y -

Similarly the formula is $\overline{y}=\frac{{\iint }_{\Omega }y\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}$

$\overline{y}=\frac{{\int }_0^b{\int }_0^{h}ykxdydx}{{\int }_0^b{\int }_0^{h}kxdydx}=\frac{{\int }_0^{b}kx{\left[\frac{y^2}{2}\right]}_0^{h}dx}{{\int }_0^{b}kx[y{]}_0^{h}dx}=\frac{{\int }_0^{b}kx\left[\frac{h^2}{2}\right]dx}{{\int }_0^{b}khxdx}$

$\overline{y}=\frac{k{h}^2{\int }_0^{b}xdx}{2kh{\int }_0^{b}xdx}==\frac{h{\left[\frac{x^2}{2}\right]}_0^b}{2{\left[\frac{x^2}{2}\right]}_0^b}=\frac{h\left[\frac{b^2}{2}\right]}{2\left[\frac{b^2}{2}\right]}$

Hence, $y=\frac{h}{2}$

Center of Mass is $\overline{x}=\frac{2}{3}b,\overline{y}=\frac{h}{2}$