91Ó°ÊÓ

Th expression is

Plot the vertices $(1,1),(2,0)$, and $(2,3)$and join them.

Obtain the equation of $AB$by using the formula of coordinate geometry

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ y-1=\frac{0-1}{2-1}(x-1) \\ y=-x+2 \end{gathered}$$

Equation of $BC$

$$\begin{gathered} y-0=\frac{3-0}{2-2}(x-2) \\ y-0=\frac{3}{0}(x-2) \\ x-2=0[\text{Cross multiply]} \end{gathered}$$

And equation of$CA$

$y=2x-1$

First moment of the mass in $\mathrm{Î\copyright }$about the $y$axis is

$M_y=âˆ\neg x\mathrm{ÒÏ}(x,y)dA$

Where $\mathrm{ÒÏ}(x,y)$is the density of the region $\mathrm{Î\copyright }$.

Here $\mathrm{ÒÏ}(x,y)$is proportional to the distance from the $y$-axis

$$\begin{gathered} \mathrm{ÒÏ}(x,y)=xk.\text{Then} \\ {M}_y=âˆ\neg k{x}^{2}dA \end{gathered}$$

Impose the limits on integrals.

$M_y={∫}_1^2{∫}_{-x+2}^{2x-1}k{x}^{2}dydx$

Integrate the inner integral first

$M_y=k{∫}_1^2\left[{∫}_{-x+2}^{2x-1}{x}^{2}dy\right]dx$

Integrate with respect to $$y$$

$M_y=k{∫}_1^2\left[x^2[y{]}_{-x+2}^{2x-1}\right]dx$

Substitute the limits

$$\begin{gathered} M_y=k{∫}_1^2\left[\right(2x-1)-(-x+2\left)\right]x^{2}dx \\ {M}_y=k{∫}_1^2\left[\right(2x-1)-(-x+2\left)\right]x^{2}dx \\ {M}_y=k{∫}_1^2\left(3x^3-3x^2\right)dx[\text{Simplify]} \end{gathered}$$

Integrate with respect to $x$

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

Substitute the limits

$$\begin{gathered} M_y=k\left[\left(\frac{3}{4}(2{)}^4-(2{)}^3\right)-\left(\frac{3}{4}-1\right)\right] \\ {M}_y=\left[(12-8)+\frac{1}{4}\right] \\ {M}_y=\frac{17}{4}k[\text{Simplify]} \end{gathered}$$

Thus, the first moment of the mass in $\mathrm{Î\copyright }$about the $y$axis is

$M_y=\frac{17}{4}k$