Question

Let $T_2$be triangular region with vertices $(1,0),(2,1),\text{and}(2,-1)$

If the density at each point in $T_2$is proportional to the point’s distance from the $y$-axis, find the center of mass of $T_2$.

Short answer

The center of mass is$(\overline{x},\overline{y})=\left(\frac{17}{10},0\right)$

Step-by-step solution

Given Information

The vertices of triangular region is $(1,0),(2,1),\text{and}(2,-1)$

Density is$\rho (x,y)=kx$

Find x-

The formula is:

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

Limits of $y$varies from $-x+1$to $x+1$

$x$varies from $1-2$

Formula becomes $\overline{x}=\frac{{\int }_1^2{\int }_{-x+1}^{x-1}xkxdydx}{{\int }_1^2{\int }_{-x+1}^{x-1}kxdydx}$

$=\frac{{\int }_1^2{\int }_{-x+1}^{x-1}k{x}^{2}dydx}{{\int }_1^2{\int }_{-x+1}^{x-1}kxdydx}$

$\overline{x}=\frac{{\int }_1^2{x}^2[y{]}_{-x+1}^{x-1}dx}{{\int }_1^{2}x[y{]}_{-x+1}^{x-1}dx}$

$=\frac{{\int }_1^2{x}^2\left[\right(x-1)-(-x+1\left)\right]dx}{{\int }_1^{2}x\left[\right(x-1)-(-x+1\left)\right]dx}$

$=\frac{{\int }_1^2{x}^2[2x-2]dx}{{\int }_1^{2}x[2x-2]dx}$

$\overline{x}=\frac{{\int }_1^2\left(x^3-x^2\right)dx}{{\int }_1^2\left(x^2-x\right)dx}$

$=\frac{{\left[\frac{x^4}{4}-\frac{x^3}{3}\right]}_1^2}{{\left[\frac{x^3}{3}-\frac{x^2}{2}\right]}_1^2}$

$=\frac{\left[\left(\frac{2^4}{4}-\frac{2^3}{3}\right)-\left(\frac{1}{4}-\frac{1}{3}\right)\right]}{\left[\left(\frac{2^3}{3}-\frac{2^2}{2}\right)-\left(\frac{1}{3}-\frac{1}{2}\right)\right]}=\frac{7}{10}$

Find y-

The formula is

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

Using values of $\rho (x,y)$and values of $x,y$

$\overline{y}=\frac{{\int }_1^2{\int }_{-x+1}^{x-1}ykxdydx}{{\int }_1^2{\int }_{-x+1}^{x-1}kxdydx}$

$\overline{y}=\frac{{\int }_1^{2}kx{\left[\frac{y^2}{2}\right]}_{-x+1}^{x-1}dx}{{\int }_1^{2}kx[y{]}_{-x+1}^{x-1}dx}$

$=\frac{{\int }_1^{2}kx\left[\frac{(x-1{)}^2-(-x+1{)}^2}{2}\right]dx}{{\int }_1^{2}2k\left(x^2-x\right)dx}$

$=\frac{{\int }_1^{2}kx\left[0\right]dx}{{\int }_1^{2}2k\left(x^2-x\right)dx}=0$

Centroid is$(\overline{x},\overline{y})=\left(\frac{17}{10},0\right)$