$ y = x^2 – 3x$ and y = 2x if the density of the lamina at any point is given by $\frac{24}{25} xy$

Solution:

The curve $y=x^{2}-3 x$ i.e. $y+\frac{9}{4}=\left(x-\frac{3}{2}\right)^{2}$ is parabola intersecting the x -axis in $x=0$ and $x=3 .$ The line $y=2 x$ intersects this parabola at $x^{2}$ $-3 x=2 x$ i.e. $x^{2}-5 x=0$ i.e. $x=0, x=5 .$ Therefore, points of intersection are $(0,0)$ and $(5,10) .$ The surface density is $\rho=(24 / 25) \mathrm{xy}$ . Taking the elementary strip parallel to the y-axis, on the strip y varies from $y=x^{2}-3 x$ to $y=2 x$ and then x varies from $x=0$ to $x=5$ .

$\therefore$ Mass of lamina $=\int_{0}^{5} \int_{x^{2}-3 x}^{2 x} \frac{24}{25} x y d x d y$

$=\frac{24}{25} \int_{0}^{5} x\left[\frac{y^{2}}{2}\right]_{x^{2}-3 x}^{2 x} d x$

$=\frac{24}{50} \int_{0}^{5} 4 x^{3}-x\left(x^{4}-6 x^{3}+9 x^{2}\right) ] d x$

$=\frac{24}{50} \int_{0}^{5}-5 x^{3}+6 x^{4}-x^{5} ) ] d x$

$=\frac{24}{50}\left[\frac{-x^{6}}{b}+\frac{6 x^{5}}{5}-\frac{5 x^{4}}{4}\right]_{0}^{5}$

$=\frac{24}{50} \cdot 5^{4}\left[-\frac{25}{6}+6-\frac{5}{4}\right]$

$=\frac{24}{50} \cdot 5^{4} \cdot \frac{7}{12}$

$\therefore$ Mass of lamina $=175 .$