Q. 19

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

$I_{x} = \int_{1}^{2} \int_{- x + 2}^{2 x - 1} k x y^{2} d y d x = \frac{28}{5} k$

Step-by-Step Solution

Verified

Answer

The moment of inertia about x-axis is

$I_{x} = \frac{28}{5} k$

1Step 1: Given information

The expression is

$I_{x} = \int_{1}^{2} \int_{- x + 2}^{2 x - 1} k x y^{2} d y d x = \frac{28}{5} k$

2Step 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

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) y - 1 = \frac{0 - 1}{2 - 1} (x - 1) y = - x + 2$

Equation of $B C$

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

And equation of $C A$ is

$y = 2 x - 1$

The moment of inertia of the mass in $Ω$ about the $x$ axis is

$I_{x} = \iint_{Ω} y^{2} ρ (x, y) d A$

Where $ρ (x, y)$ is the density of the region $Ω .$

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

Assume $ρ (x, y) = k x$ . Then

$I_{x} = \iint_{Ω} k x y^{2} d y d x$

Impose the limits on integrals.

$I_{x} = \int_{1}^{2} \int_{- x + 2}^{2 x - 1} k x y^{2} d y d x$

Integrate the inner integral with respect to $y$ first.

$I_{x} = k \int_{1}^{2} x {[\frac{y^{3}}{3}]}_{- x + 2}^{2 x - 1} d x I_{x} = \frac{k}{3} \int_{1}^{2} x [(2 x - 1)^{3} - (- x + 2)^{3}] d x I_{x} = \frac{9 k}{3} \int_{1}^{2} [x^{4} - 2 x^{3} + 2 x^{2} - x] d x [Simplify]$

Integrate with respect to $x$ .

$I_{x} = 3 k {[\frac{1}{5} x^{5} - \frac{2}{4} x^{4} + \frac{2}{3} x^{3} - \frac{1}{2} x^{2}]}_{1}^{2}$

Substitute the limits

$I_{x} = 3 k [(\frac{1}{5} (2)^{5} - \frac{2}{4} (2)^{4} + \frac{2}{3} (2)^{3} - \frac{1}{2} (2)^{2}) - (\frac{1}{5} - \frac{2}{4} + \frac{2}{3} - \frac{1}{2})]$