Question

Show that when the density of the region is constant, the first moment about the $y$-axis is

$M_y={\int }_1^2{\int }_{-x+2}^{2x-1}xdydx=\frac{5}{2}$

Short answer

the first moment of the mass in $\Omega$about the$y$ axis is

$M_y=\frac{5}{2}$

Step-by-step solution

Given information

the density of the region is constant, the first moment about the $y$-axis is

$M_y={\int }_1^2{\int }_{-x+2}^{2x-1}xdydx=\frac{5}{2}$

simplification

The objective of this problem is to show that when the density of region is constant the first moment about the $y$axis is

localid="1650648276014" $M_y={\int }_1^2{\int }_{-x+2}^{2x-1}xdydx=\frac{5}{2}.$

Plot the vertices localid="1650648279131" $(1,1),(2,0)$, and localid="1650648283039" $(2,3)$and join them.

Plot of triangle

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

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

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

Here $\rho (x,y)$is constant.

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

$M_y=\iint xkdA$

Impose the limits on integrals.

$M_y={\int }_1^2{\int }_{-x+2}^{2x-1}kxdydx$

Integrate the inner integral first

$\left.M_y=k{\int }_1^2\left[{\int }_{-x+2}^{2x-1}xdy\right]dx\text{[Take}k=1\right]$

Integrate with respect to y

$M_y={\int }_1^{2}x[y{]}_{-x+2}^{2x-1}dx$

Substitute the limits

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

Step 2:Simplification

Substitute the limits

$$\begin{gathered} M_y=\left[(2{)}^3-\frac{3}{2}(2{)}^2-(1{)}^3+\frac{3}{2}(1{)}^2\right] \\ {M}_y=\left[8-6-1+\frac{3}{2}\right] \\ {M}_y=\frac{5}{2}[\text{Simplify]} \end{gathered}$$

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

$M_y=\frac{5}{2}$

Integrate with respect to $

$M_y={\int }_1^{2}x[y{]}_{-x+2}^{2x-1}dx$

Substitute the limits

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

Integrate with respect to x

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

Substitute the limits

$$\begin{gathered} M_y=\left[(2{)}^3-\frac{3}{2}(2{)}^2-(1{)}^3+\frac{3}{2}(1{)}^2\right] \\ {M}_y=\left[8-6-1+\frac{3}{2}\right] \end{gathered}$$

$M_y=\frac{5}{2}[\text{Simplify]}$

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

$M_y=\frac{5}{2}$

Given information

the region is constant, the first moment about the $y$-axis is

$M_y={\int }_1^2{\int }_{-x+2}^{2x-1}xdydx=\frac{5}{2}$