91Ó°ÊÓ

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

$\mathrm{ÒÏ}(x,y)=k{x}^2$

The required formula is

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

Use $\mathrm{ÒÏ}(x,y)=k{x}^2$

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

$\stackrel{¯}{x}=\frac{{∫}_1^2{∫}_{-x+1}^{x-1}k{x}^{3}dydx}{{∫}_1^2{∫}_{-x+1}^{x-1}k{x}^{2}dydx}$

Solving inner integral first

$\stackrel{¯}{x}=\frac{{∫}_1^{2}k{x}^3[y{]}_{-x+1}^{x-1}dx}{{∫}_1^{2}k{x}^2[y{]}_{-x+1}^{x-1}dx}$

$\stackrel{¯}{x}=\frac{{∫}_1^{2}k{x}^3\left[\right(x-1)-(-x+1\left)\right]dx}{{∫}_1^{2}k{x}^2\left[\right(x-1)-(-x+1\left)\right]dx}=\frac{{∫}_1^{2}k{x}^3[2x-2]dx}{{∫}_1^{2}k{x}^2[2x-2]dx}=\frac{{∫}_1^{2}k\left(x^4-x^3\right)dx}{{∫}_1^{2}k\left(x^3-x^2\right)dx}$

Simplifying further

$\stackrel{¯}{x}=\frac{k{\left[\frac{x^5}{5}-\frac{x^4}{4}\right]}_1^2}{k{\left[\frac{x^4}{4}-\frac{x^3}{3}\right]}_1^2}=\frac{k\left[\frac{49}{20}\right]}{k\left[\frac{17}{12}\right]}=\frac{147}{85}$

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

$\stackrel{¯}{y}=\frac{{∫}_1^2{∫}_{-x+1}^{x-1}yk{x}^{2}dydx}{{∫}_1^2{∫}_{-x+1}^{x-1}k{x}^{2}dydx}$

$\stackrel{¯}{y}=\frac{{∫}_1^{2}k{x}^2{\left[\frac{y^2}{2}\right]}_{-x+1}^{x-1}dx}{{∫}_1^{2}k{x}^2[y{]}_{-x+1}^{x-1}dx}=\frac{{∫}_1^{2}k{x}^2\left[\frac{(x-1{)}^2-(-x+1{)}^2}{2}\right]dx}{{∫}_1^{2}2k\left(x^3-x^2\right)dx}=\frac{{∫}_1^{2}k{x}^2\left[0\right]dx}{{∫}_1^{2}2k\left(x^3-x^2\right)dx}$

$\stackrel{¯}{x}=\frac{147}{85},\stackrel{¯}{y}=0$

Hence, Center of Mass is$\stackrel{¯}{x}=\frac{147}{85},\stackrel{¯}{y}=0$