91Ó°ÊÓ

Vertices of the triangular region are $(0,0),(1,1),and(1,-1)$.

The objective of this problem is to find the center of mass of the triangular region.

The density at each point is proportional to the point's distance from the $y$- axis. Density $\mathrm{ÒÏ}(x,y)=kx$

Use formula for center of mass

localid="1650641181816" $\stackrel{¯}{x}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}$ and localid="1650641192893" $\stackrel{¯}{y}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}$

localid="1650641380814" $$\begin{gathered} \stackrel{¯}{x}=\frac{{∫}_0^1{∫}_{-x}^{x}xkxdydx}{{∫}_0^1{∫}_{-x}^{x}kxdydx} \\ \stackrel{¯}{x}=\frac{{∫}_0^1{∫}_{-x}^{x}k{x}^{2}dydx}{{∫}_0^1{∫}_{-x}^{x}kxdydx} \\ \stackrel{¯}{x}=\frac{{∫}_0^{1}k{x}^2[y{]}_{-x}^{x}dx}{{∫}_0^{1}kx[y{]}_{-x}^{x}dx} \\ \stackrel{¯}{x}=\frac{{∫}_0^{1}k{x}^2\left[2x\right]dx}{{∫}_0^{1}kx\left[2x\right]dx} \\ \stackrel{¯}{x}=\frac{{∫}_0^{1}k{x}^{3}dx}{{∫}_0^{1}k{x}^{2}dx} \\ \stackrel{¯}{x}=\frac{{\left[\frac{x^4}{4}\right]}^1}{k{\left[\frac{x^3}{3}\right]}_0^1} \\ \stackrel{¯}{x}=\frac{3}{4} \end{gathered}$$

Now

$$\begin{gathered} \stackrel{¯}{y}=\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{∫}_0\mathrm{ÒÏ}(x,y)dA} \\ \stackrel{¯}{y}=\frac{{∫}_0^1{∫}_{-x}^{x}ykxdydx}{{∫}_0^1{∫}_{-x}^{x}kxdydx} \\ \stackrel{¯}{y}=\frac{{∫}_0^{1}kx{\left[\frac{y^2}{2}\right]}_{-x}^{x}dx}{{∫}_0^{1}kx[y{]}_{-x}^{x}dx} \\ \stackrel{¯}{y}=\frac{{∫}_0^{1}kx\left[0\right]dx}{{∫}_0^{1}2k{x}^2} \\ \stackrel{¯}{y}=\frac{{∫}_0^{1}kx\left[0\right]dx}{k{\left[\frac{2}{3}{x}^3\right]}_0^1} \\ \stackrel{¯}{y}=\frac{0}{\frac{2}{3}k}=0 \\ \stackrel{¯}{y}=0 \end{gathered}$$

Thus, the centroid of the triangular region is

$\stackrel{¯}{x}=\frac{3}{4},\stackrel{¯}{y}=0$