Q. 18 - Calculus

Question

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

$I_y={\int }_1^2{\int }_{-x+2}^{2x-1}k{x}^{3}dydx=\frac{147}{20}k$

Step-by-Step Solution

Answer

The moment of inertia about y- axis is

$I_y=\frac{147}{20}k$

Step 1: Given information

The expression is 

$I_y={\int }_1^2{\int }_{-x+2}^{2x-1}k{x}^{3}dydx=\frac{147}{20}k$

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 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 $CA$ is

$y=2 x-1$

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

$I_y={\iint }_{\Omega }{x}^2\rho (x,y)dA$

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

$I_y={\iint }_{\Omega }{x}^{3}dydx$

Impose the limits on integrals.

$I_y={\int }_1^2{\int }_{-x+2}^{2x-1}k{x}^{3}dydx$

Integrate the inner integral first

$I_y=k{\int }_1^2\left[{\int }_{-x+2}^{2x-1}{x}^{3}dy\right]dx$

Integrate with respect to y

$I_y=k{\int }_1^2{\left[x^{3}y\right]}_{-x+2}^{2x-1}dx$

Substitute the limits

$$\begin{gathered} I_y=k{\int }_1^2{x}^3\left[\right(2x-1)-(-x+2\left)\right]dx{I}_y \\ =k{\int }_1^2\left(3x^4-3x^3\right)dx\text{ [Simplify] } \end{gathered}$$

Integrate with respect to $x$

$I_y=k{\left[\frac{3}{5}{x}^5-\frac{3}{4}{x}^4\right]}_1^2$

Substitute the limits

$$\begin{gathered} I_y=k\left[\left(\frac{3}{5}(2{)}^5-\frac{3}{4}(2{)}^4\right)-\left(\frac{3}{5}-\frac{3}{4}\right)\right] \\ {I}_y=\frac{147}{20}k[\text{ Simplify]} \end{gathered}$$

Thus, the moment of inertia about the $y$ axis is

$I_y=\frac{147}{20}k$