Q 31. - Calculus

Question

Let $T_2$ be triangular region with vertices $(1,0),(2,1),\text{ and }(2,-1)$

Find centroid of $T_2$

Step-by-Step Solution

Answer

The centroid is $\overline{x}=\frac{17}{10},\overline{y}=0$

Step 1: Given Information

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

Step 2: Finding x -

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

Density is uniform

$\rho (x,y)$ is proportional to point's distance from $y$ axis.

$\rho (x,y)=kx$

$\overline{x}=\frac{{\int }_1^2{\int }_{-x+1}^{x-1}xkxdydx}{{\int }_1^2{\int }_{-x+1}^{x-1}kxdydx}$

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

$\overline{x}=\frac{{\int }_1^{2}k{x}^2[y{]}_{-x+1}^{x-1}dx}{{\int }_1^{2}kx[y{]}_{-x+1}^{x-1}dx}$

$\overline{x}=\frac{{\int }_1^{2}k{x}^2[2x-2]dx}{{\int }_1^{2}kx[2x-2]dx}$

$\overline{x}=\frac{2k{\int }_1^2\left(x^3-x^2\right)dx}{2k{\int }_1^2\left(x^2-x\right)dx}$

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

$\overline{x}=\frac{\left[\left(\frac{16}{4}-\frac{8}{3}\right)-\left(\frac{1}{4}-\frac{1}{3}\right)\right]}{\left[\left(\frac{8}{3}-\frac{4}{2}\right)-\left(\frac{1}{3}-\frac{1}{2}\right)\right]}$

$\overline{x}=\frac{17}{10}$

Step 3: Finding y -

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

$\overline{y}=\frac{{\int }_1^2{\int }_{-x+1}^{x-1}ykxdydx}{{\int }_1^2{\int }_{-x+1}^{x-1}kxdydx}$

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

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

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

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

Hence

$\overline{x}=\frac{17}{10},\overline{y}=0$