Example 1. Find the volume of the body $\Omega$, bounded by the surfaces

$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=\frac{1}{2}-y, \quad z=0
$$
Solution.

1. Define the region $\Omega$ by inequalities. Since $17 \sqrt{2 y} \geq 2 \sqrt{2 y}$, for $x$ we have the inequalities $2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y}$. Since $y$ appears under the square root, $y \geq 0$. For $z$, the possible inequalities are $0 \leq z \leq 1 / 2-y$ or $1 / 2-y \leq z \leq 0$. In the first case, $0 \leq y \leq 1 / 2$. In the second case, $y \geq 1 / 2$, i.e., the region is unbounded, which is unacceptable.

Thus,

$$
\Omega=\left\{\begin{array}{cc} 
& 2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y} \\
(x, y, z): & 0 \leq y \leq 1 / 2 \\
& 0 \leq z \leq 1 / 2-y
\end{array}\right\}
$$

2. We compute the volume using formula (1), reducing the triple integral to an iterated integral:

$$
V=\iiint_{\Omega} 1 \cdot d x d y d z=\int_{0}^{1 / 2} d y \int_{2 \sqrt{2 y}}^{17 \sqrt{2 y}} d x \int_{0}^{1 / 2-y} d z=1
$$

Answer. $V=1$ unit of volume.