Example 1. Find the volume of the body bounded by the surfaces

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

1. By formula (1) with $f_{2}=1 / 2-y$ and $f_{1}=0$, the desired volume is

$$
V=\iint_{D}\left(\frac{1}{2}-y\right) d x d y
$$

where $D$ is the projection of the body onto the $X O Y$ plane.

2. To find $D$, we define the body using inequalities and eliminate $z$ from them. In this case, the body is defined by the system of inequalities

$$
\left\{\begin{array}{l}
x \leq 17 \sqrt{2 y} \\
x \geq 2 \sqrt{2 y} \\
0 \leq z \leq 1 / 2-y
\end{array}\right.
$$

Therefore,

$$
D=\left\{\begin{array}{ll}
(x, y): \begin{array}{l}
2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y} \\
0 \leq 1 / 2-y, \quad y \geq 0
\end{array}
\end{array}\right\}
$$

The inequality $y \geq 0$ is necessary because $y$ is under the square root.

3. We compute the double integral:

$$
\begin{aligned}
& V=\iint_{D}\left(\frac{1}{2}-y\right) d x d y=\int_{0}^{1 / 2} d y \int_{2 \sqrt{2 y}}^{17 \sqrt{2 y}}\left(\frac{1}{2}-y\right) d x= \\
&= 15 \sqrt{2} \int_{0}^{1 / 2}\left(\frac{1}{2}-y\right) \sqrt{y} d y=1
\end{aligned}
$$

Answer. $V=1$ unit of volume.