Solution. The volume $V$ of a spatial body is equal to the triple integral over the spatial domain $W$ occupied by the body:

$$
V=\iint_{H} \int d x d y d z
$$

The first of the equations defining the boundary of the body in the problem statement is the equation of a plane parallel to $O y$ and intersecting the axes $O x$ and $O z$ at points $x=6$ and $z=6$. This plane intersects the horizontal plane $x O y$ along the line $x=6$ and bounds the body from above.

The second and third equations in the problem statement define parabolic cylinders with generators parallel to $O z$, defined for $x \geqslant 0, y \geqslant 0$ and serve as the side boundary surfaces of the body. Their directrices in the plane $x O y$ are the parabolas $y=\sqrt{x}$ and $y=2 \sqrt{x}$. The body is bounded from below by the horizontal coordinate plane $z=0$ (Fig. 3.17).

Each line passing through an internal point of the constructed spatial domain $W$ and parallel to $O z$ intersects the boundary of the domain only at two points: from below at $z=0$ and from above at $z=6-x$; these equalities define the lower and upper limits of integration with respect to $z$.

The limits of integration with respect to $y$ are determined by the region $D$ in the plane $x O y$ onto which the spatial domain $W$ is projected. The lower boundary of the region $D$ is the curve $y=\sqrt{x}$, and the upper boundary is the curve $y=2 \sqrt{x}$. Finally, integration with respect to $x$ is performed over the segment $[0 ; 6]$ of the $O x$ axis, onto which the region $D$ is projected.

$$
\begin{aligned}
V=\iiint_{W} d x d y d z= & \int_{0}^{6} d x \int_{\sqrt{x}}^{2 \sqrt{x}} d y \int_{0}^{6-x} d z=\int_{0}^{6} d x \int_{\sqrt{x}}^{2 \sqrt{x}}(6-x) d y= \\
& =\int_{0}^{6}(6-x) \sqrt{x} d x=\left.6\left(\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}\right)\right|_{0} ^{6}=\frac{48}{5} \sqrt{6}
\end{aligned}
$$

Answer, $V=\frac{48}{5} \sqrt{6}$.
Example 4. Calculate the volume of the body bounded by the surfaces: $x+z=6 ; y=\sqrt{x} ; y=2 \sqrt{x} ; z=0$.