Example 2. Find the volume of the body $\Omega$, bounded by the surfaces

$$
z=\frac{9}{2} \sqrt{x^{2}+y^{2}}, \quad z=\frac{11}{2}-x^{2}-y^{2}
$$
Solution.

1. Since $\Omega$ is a body of revolution around the $O Z$ axis, it is convenient to use cylindrical coordinates

$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \\
y=\varrho \sin \varphi \\
z=z
\end{array}\right.
$$

In this case, $(\varrho, \varphi, z) \in \Omega^{\prime}$, and the desired volume is determined by the formula

$$
V=\iiint_{\Omega^{\prime}} \varrho d \varrho d \varphi d z
$$

where the region $\Omega^{\prime}$ is bounded by the surfaces

$$
z=\frac{9}{2} \varrho, \quad z=\frac{11}{2}-\varrho^{2}
$$

2. Let's define the region $\Omega$ using inequalities. There are two possible cases: either $9 \varrho / 2 \leq z \leq 11 / 2-\varrho^{2}$, or $11 / 2-\varrho^{2} \leq z \leq 9 \varrho / 2$. In the first case, $\varrho \leq 1$, and in the second case, $\varrho \geq 1$, which means the region is unbounded, which is unacceptable.

Therefore,

$$
\Omega^{\prime}=\left\{\begin{array}{ll}
0 \leq \varrho \leq 1 \\
(\varrho, \varphi, z): & 9 \varrho / 2 \leq z \leq 11 / 2-\varrho^{2} \\
& 0 \leq \varphi \leq 2 \pi
\end{array}\right\}
$$

3. We calculate the volume using formula (2), reducing the triple integral to a repeated integral:

$$
V=\int_{0}^{2 \pi} d \varphi \int_{0}^{1} \varrho d \varrho \int_{9 \varrho / 2}^{11 / 2-\varrho^{2}} d z=2 \pi
$$

Answer. $\quad V=2 \pi$ cubic units.