If the density at each point in $\mathcal{T}$ is proportional to the point's distance from the x-axis, find the mass of $\mathcal{T}$.

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 $\rho (x,y)=kx$

Use formula for center of mass

localid="1650641181816" $\overline{x}=\frac{{\iint }_{\Omega }x\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}$ and localid="1650641192893" $\overline{y}=\frac{{\iint }_{\Omega }y\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}$

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

Now

$$\begin{gathered} \overline{y}=\frac{{\iint }_{\Omega }y\rho (x,y)dA}{{\int }_0\rho (x,y)dA} \\ \overline{y}=\frac{{\int }_0^1{\int }_{-x}^{x}ykxdydx}{{\int }_0^1{\int }_{-x}^{x}kxdydx} \\ \overline{y}=\frac{{\int }_0^{1}kx{\left[\frac{y^2}{2}\right]}_{-x}^{x}dx}{{\int }_0^{1}kx[y{]}_{-x}^{x}dx} \\ \overline{y}=\frac{{\int }_0^{1}kx\left[0\right]dx}{{\int }_0^{1}2k{x}^2} \\ \overline{y}=\frac{{\int }_0^{1}kx\left[0\right]dx}{k{\left[\frac{2}{3}{x}^3\right]}_0^1} \\ \overline{y}=\frac{0}{\frac{2}{3}k}=0 \\ \overline{y}=0 \end{gathered}$$

Thus, the centroid of the triangular region is

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