Example 2.72. Calculate the moment of inertia of a plane figure bounded by the lines $y^{2}=4 a^{2} x$ and the line $x=a^{2}$ with respect to the $O Y$ axis. The surface density of the material of the figure is $\rho=\rho_{0}+k x$.
Solution. Divide the interval $\left[0, a^{2}\right]$ into elementary intervals of length $d x:[x, x+d x]$. Highlight the elementary strips cut from the figure by lines passing through the ends of each interval $[x, x+d x]$ and parallel to the $O Y$ axis (Fig. 2.12). The elementary strip can be approximately considered as a rectangle with area $d S=2|y| d x=2 \sqrt{4 a^{2} x} d x=4 a \sqrt{x} d x$. The density of the elementary strip can be approximately considered constant and equal to the value it takes at the left end of the interval $[x, x+d x], \rho=\rho_{0}+k x$. Then the mass $d m$ of the elementary strip is:

$$
d m=\left(\rho_{0}+k x\right) d S=\left(\rho_{0}+k x\right) 4 a \sqrt{x} d x
$$

and the differential of the moment of inertia relative to the $O Y$ axis is

$$
d I=x^{2} d m=x^{2}\left(\rho_{0}+k x\right) 4 a \sqrt{x} d x
$$

## Hence

$$
\begin{aligned}
& I_{y}=\int_{0}^{a} 4 a\left(\rho_{0} x^{5 / 2}+k x^{7 / 2}\right) d x=\left.\left(8 a \rho_{0} \frac{x^{7 / 2}}{7}+8 a k \frac{x^{9 / 2}}{9}\right)\right|_{0} ^{a}= \\
& =8 a^{9 / 2}\left(\frac{\rho_{0}}{7}+\frac{k a}{9}\right)
\end{aligned}
$$