Question

Show that when the density of the region is proportional to the distance from the $y$-axis, the first moment about the $x$-axis is

$M_x={\int }_1^2{\int }_{-x+2}^{2x-1}kxydydx=\frac{27}{8}k$

Short answer

Thus, the first moment of the mass in $\Omega$ about the$x$ axis is

$M_x=\frac{17}{8}k$

Step-by-step solution

Given information

The expression is

$M_x={\int }_1^2{\int }_{-x+2}^{2x-1}kxydydx=\frac{27}{8}k$

Calculation

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 $BC$

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

And equation of$CA$

$y=2x-1$

First moment of the mass in $\Omega$about the $x$axis is

$M_x=\iint y\rho (x,y)dA$

Where $\rho (x,y)$is the density of the region $\Omega$.

Here $\rho (x,y)$is proportional to the distance from $y$- axis.

Assume$\rho (x,y)=kx$. Then

$M_x=\iint kxydydx$

Impose the limits on integrals.

$M_x={\int }_1^2{\int }_{-x+2}^{2x-1}kxydydx$

Integrate the inner integral first

$M_x=k{\int }_1^2\left[{\int }_{-x+2}^{2x-1}xydy\right]dx$

Integrate with respect to $y$

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

Substitute the limits

$M_x=k{\int }_1^2\left[\frac{x(2x-1{)}^2}{2}-\frac{x(-x+2{)}^2}{2}\right.$

$M_x=k{\int }_1^2\left[\frac{4x^3-4x^2+x}{2}-\frac{x^3-4x^2}{2}\right.$

$M_x=\frac{k}{2}{\int }_1^2\left(3x^3-3x^2\right)dx\text{[Simplify]}$

Integrate with respect to $x$

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

Substitute the limits

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

Thus, the first moment of the mass in$\Omega$about the $x$ axis is

$M_x=\frac{17}{8}k$