1.27. The body $\Omega$ is bounded by the planes $x=2 ; y=0 ; y=1 ; z=0$ and the cylinder $z^{2}=6 x$. The volume density of the material at each point is proportional to the distance from the plane $X O Y$. Find the moment of inertia of the body $\Omega$ about the $z$-axis.
Solution. By definition, the moment of inertia of a non-uniform body

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-076.jpg?height=351&width=341&top_left_y=130&top_left_x=794)

Fig. 1.43 of $\Omega$ relative to the $z$-axis is calculated by the formula

$$
I_{O z}=\iint_{\varsigma} \int_{2}\left(x^{2}+y^{2}\right) p(M) d x d y d z,
$$

where $p(M)=k \cdot z, k=$ const (volumetric density of the material).

The body $\Omega$ is shown in Fig. 1.43. Let's perform the necessary calculations:

$$
\begin{gathered}
I_{O Z}=k \int_{0}^{2} d x \int_{0}^{1} d y \int_{0}^{\sqrt{6 x}}\left(x^{2}+y^{2}\right) z d z=\left.k \int_{0}^{2} d x \int_{0}^{1}\left(x^{2}+y^{2}\right) \frac{z^{2}}{2}\right|_{0} ^{\sqrt{6 x}} d y= \\
=3 k \int_{0}^{2}\left(x^{3}+\frac{x}{3}\right) d x=\left.3 k\left(\frac{x^{4}}{4}+\frac{x^{2}}{6}\right)\right|_{0} ^{2}=14 k
\end{gathered}
$$