91Ó°ÊÓ

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

Density is$\mathrm{ÒÏ}(x,y)=kx$

The formula is:

$\stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\text{and}\stackrel{¯}{y}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}$

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

$x$varies from $1-2$

Formula becomes $\stackrel{¯}{x}=\frac{{∫}_1^2{∫}_{-x+1}^{x-1}xkxdydx}{{∫}_1^2{∫}_{-x+1}^{x-1}kxdydx}$

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

$\stackrel{¯}{x}=\frac{{∫}_1^2{x}^2[y{]}_{-x+1}^{x-1}dx}{{∫}_1^{2}x[y{]}_{-x+1}^{x-1}dx}$

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

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

$\stackrel{¯}{x}=\frac{{∫}_1^2\left(x^3-x^2\right)dx}{{∫}_1^2\left(x^2-x\right)dx}$

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

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

The formula is

$\stackrel{¯}{y}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}$

Using values of $\mathrm{ÒÏ}(x,y)$and values of $x,y$

$\stackrel{¯}{y}=\frac{{∫}_1^2{∫}_{-x+1}^{x-1}ykxdydx}{{∫}_1^2{∫}_{-x+1}^{x-1}kxdydx}$

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

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

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

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