Q 36. - Calculus

Question

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

If the density at each point in $T_2$ is proportional to the square of the point’s distance from the $y$-axis, find the center of mass of $T_2$.

Step-by-Step Solution

Answer

The centroid is $\overline{x}=\frac{147}{85},\overline{y}=0$.

Step 1: Given Information

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

$\rho (x,y)=k{x}^2$

Step 2: Finding Center of Mass

The required formula is

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

Use $\rho (x,y)=k{x}^2$

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

$\overline{x}=\frac{{\int }_1^2{\int }_{-x+1}^{x-1}k{x}^{3}dydx}{{\int }_1^2{\int }_{-x+1}^{x-1}k{x}^{2}dydx}$

Solving inner integral first

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

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

Simplifying further

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

Step 3: Simplification

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

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

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

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

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