Q. 15 - Calculus

Question

Show that when the density of the region is proportional to the distance from the $y$-axis, the mass of $\Omega$ is given by

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

Step-by-Step Solution

Answer

the mass of $\Omega$ is given by

$\frac{5}{2}k$

Step 1: Given information

The expression is ${\int }_1^2{\int }_{-x+2}^{2x-1}kxdydx=\frac{5}{2}k$

Step 2: Calculation

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

Obtain the equation of $A B$ 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 $B C$

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

And equation of $CA$

$y=2 x-1$

Mass of $\Omega$ can be computed by the integral

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

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

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

$$\begin{gathered} \rho (x,y)=kx.\text{ Then } \\ m=\iint kxdxdy \end{gathered}$$

Impose the limits on integrals.

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

Integrate the inner integral first

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

Integrate with respect to $y$

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

Substitute the limits

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

Integrate with respect to $x$

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

Substitute the limits

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

Thus, the mass of $\Omega$ is given by

$\frac{5}{2}k$