Q.14 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The first moment of the mass in $\Omega$ about the $x$ axis is$M_x=2$

Step 1: Given information

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

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

Step 2:Simplification

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

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

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

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 constant.

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

$M_x=\iint ykdA$

Impose the limits on integrals.

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

Integrate the inner integral first

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

Integrate with respect to $y$

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

Substitute the limits

$$\begin{gathered} M_x={\int }_1^2\left[\frac{(2x-1{)}^2}{2}-\frac{(-x+1{)}^2}{2}\right]dx \\ {M}_x={\int }_1^2\left[\frac{4x^2-4x+1}{2}-\frac{x^2-2x+1}{2}\right]dx \\ {M}_x=\frac{1}{2}{\int }_1^2\left(3x^2-2x\right)dx[\text{ Simplify]} \end{gathered}$$

Integrate with respect to $x$

$M_x=\frac{1}{2}{\left[\frac{3}{3}{x}^3-\frac{2}{2}{x}^2\right]}_1^2$

Substitute the limits

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

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

$M_x=2$