Example 2. Find the mass of the body $\Omega$ with density $\mu=z$, bounded by the surfaces

$$
x^{2}+y^{2}=4, \quad z=0, \quad z=\frac{x^{2}+y^{2}}{2}
$$
## Solution.

1. The mass of the body $\Omega$ with density $\mu=z$ is determined by the formula

$$
m=\iiint_{\Omega} z d x d y d z
$$

Since $\Omega$ is a body of revolution around the $O Z$ axis, it is convenient to switch to 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 mass is determined by the formula

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

We replace $x$ with $\varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the surface equations. We get

$$
\varrho=2, \quad z=0, \quad z=\frac{\varrho^{2}}{2}
$$

2. Define the region $\Omega^{\prime}$ by a system of inequalities

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

3. We calculate $m$, reducing the triple integral to a repeated one:

$$
m=\int_{0}^{2 \pi} d \varphi \int_{0}^{2} \varrho d \varrho \int_{0}^{\varrho^{2} / 2} z d z=\frac{8 \pi}{3}
$$

Answer. $V=8 \pi / 3$ units of mass.