Question

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

If the density at each point in $T$is proportional to the point’s distance from the $x$-axis, find the center of mass of $T$.

Short answer

Mass is $\frac{k}{3}$.

Moment of Mass are $M_y=\frac{k}{4}$and $M_x=\frac{k}{6}$.

Center of mass are $\overline{x}=\frac{3}{4},\overline{y}=\frac{1}{2}$

Step-by-step solution

Given Information

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

Density is proportional to point's distance from $x$axis.

$\rho (x,y)=ky$

Mass of Triangular Region

Mass of triangle = Twice the mass of upper triangle.

$m=2{\int }_0^1{\int }_0^{x}kydydx\left[m={\iint }_{\Omega }\rho (x,y)dA\right]$

$m=2k{\int }_0^1{\left[\frac{y^2}{2}\right]}_0^{x}dx$

$m=k{\int }_0^1{x}^{2}dx$

$m=k{\left[\frac{x^3}{3}\right]}_0^1$

$m=\frac{k}{3}$

First Moment of Mass about y axis

$M_y={\iint }_{\Omega }x\rho (x,y)dA$

Putting limits

$M_y={\int }_0^1{\int }_{-x}^{x}xkydydx\left[\rho \right(x,y)=ky]$

Do half the limit and twice the integral as per the given figure.

$M_y=2{\int }_0^1{\int }_0^{x}xkydydx$

$M_y=2k{\int }_0^1{\left[\frac{y^2}{2}\right]}_0^{x}xdx$

$M_y=k{\int }_0^1{x}^{3}dx$

$M_y={\left[\frac{x^4}{4}\right]}_0^{1}k=\frac{k}{4}$

First Moment of Mass about x axis

$M_x={\iint }_{\Omega }y\rho (x,y)dA$

Putting limits

$M_x={\int }_0^1{\int }_{-x}^{x}k{y}^{2}dydx\left[\rho \right(x,y)=ky]$

Solving inner integral first

$M_x=k{\int }_0^1{\left[\frac{y^3}{3}\right]}_{-x}^{x}dx$

$M_x=k{\int }_0^1\left[\frac{(x{)}^3-(-x{)}^3}{3}\right]dx$

$M_x=\frac{2}{3}k{\int }_0^1{x}^{3}dx$

$M_x=\frac{2}{3}k{\left[\frac{x^4}{4}\right]}_0^1=\frac{k}{6}$

Center of Mass

The coordinates are $\overline{x}=\frac{M_y}{m},\overline{y}=\frac{M_x}{m}$

$\overline{x}=\frac{\frac{k}{k}}{\frac{k}{3}},\overline{y}=\frac{\frac{6}{k}}{\frac{k}{3}}$

$\overline{x}=\frac{3}{4},\overline{y}=\frac{1}{2}$