Question

Let $R$be rectangle with coordinates $(0,0),(b,0),(0,h),\text{and}(b,h)$

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

Short answer

The center of mass is$\overline{x}=\frac{3}{4}b,\overline{y}=\frac{h}{2}$.

Step-by-step solution

Given Information

The vertices of rectangular region is $(0,0),(b,0),(0,h)\text{and}(b,h)$.

$\rho (x,y)=k{x}^2$

Calculation of x-

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

Also $\rho (x,y)=k{x}^2$

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

$\overline{x}=\frac{{\int }_0^b{\int }_0^{h}k{x}^{3}dydx}{{\int }_0^b{\int }_0^{h}k{x}^{2}dydx}$

$\overline{x}=\frac{{\int }_0^{b}k{x}^3[y{]}_0^{h}dx}{{\int }_0^{b}k{x}^2[y{]}_0^{h}dx}=\frac{{\int }_0^{b}kh{x}^{3}dx}{{\int }_0^{b}kh{x}^{2}dx}=\frac{kh{\int }_0^b{x}^{3}dx}{kh{\int }_0^b{x}^{2}dx}$

$\overline{x}=\frac{{\left[\frac{x^4}{4}\right]}_0^b}{{\left[\frac{x^3}{3}\right]}_0^b}=\frac{\left[\frac{b^4}{4}\right]}{\left[\frac{b^3}{3}\right]}$

$\overline{x}=\frac{3}{4}b$

Calculation of y-

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

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

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

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

$\overline{y}=\frac{h}{2}$

Hence,$\overline{x}=\frac{3}{4}b,\overline{y}=\frac{h}{2}$