Question

In Exercises $24-30$, let role="math" localid="1650649060037" $\mathcal{T}$be the triangular region with vertices $(0,0),(1,1)$,and$(1,-1)$.

Find the centroid of role="math" localid="1650649066645" $\mathcal{T}$.

Short answer

Thus, the centroid of the triangular region is$\overline{x}=\frac{3}{4},\overline{y}=0$

Step-by-step solution

Given information

Centroid of a plane figure is defined as the point of intersection of medians.

Calculation

For example, take a triangle with vertices $(0,0),(1,1),$and $(1,-1)$.

Use the formula for centroid

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

$\rho (x,y)$is the uniform density of the lamina. If $\rho (x,y)$is proportional to the point's distance from the y - axis.

$\rho (x,y)=kx$

$\overline{x}=\frac{{\int }_0^1{\int }_{-x}^{x}xkxdydx}{{\int }_0^1{\int }_{-x}^{x}kxdydx}$

localid="1650649551169" $\overline{x}=\frac{{\int }_0^1{\int }_{-x}^{x}k{x}^{2}dydx}{{\int }_0^1{\int }_{-x}^{x}kxdydx}$

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

Now

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

Thus, the centroid of the triangular region is $\overline{x}=\frac{3}{4},\overline{y}=0$