6.2.2. Find the volume of the body bounded by the surfaces: $2 z=x^{2}+y^{2} ; z=2 ; x=0 ; y=2 x$.
Solution. The body is bounded from below by the surface $z=\frac{1}{2}\left(x^{2}+y^{2}\right)$, and from above by $-z=2$. On the plane $x O y$, we construct the region $D$ (Fig. 63).

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-267.jpg?height=476&width=610&top_left_y=414&top_left_x=399)

Fig. 63

We will transition to cylindrical coordinates:

$$
\begin{aligned}
& x=\rho \cos \phi \\
& y=\rho \sin \phi \\
& z=z
\end{aligned}
$$

The equation of the paraboloid will have the form:

$$
2 z=\rho^{2} \cos ^{2} \phi+\rho^{2} \cdot \sin ^{2} \phi, \text { or } 2 z=\rho^{2}, \text { or } z=\frac{\rho^{2}}{2} \text {. }
$$

The radius $\rho$ varies from 0 to $\rho=\sqrt{2 z}=\sqrt{2 \cdot 2}=2$.

We will find the limits of variation of the angle $\phi$ (Fig. 64).

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-267.jpg?height=370&width=502&top_left_y=1436&top_left_x=457)

Fig. 64
$\operatorname{tg} \theta=\frac{A B}{O B}=\frac{y}{x}=\frac{2 x}{x}=2$, then $\theta=63.4349^{\circ}$ and $\phi=90^{\circ}+\theta=$ $=153.4349^{\circ}=0.852 \pi$, i.e., $\phi$ varies from 0 to $153.4349^{\circ}=0.852 \pi$. We find the volume:

$$
\begin{gathered}
V=\iiint_{V} d x d y d z=\iiint_{V} \rho d \rho d \phi d z=\int_{0}^{0.852 \pi} d \phi \int_{0}^{2} \rho d \rho \int_{\frac{\rho^{2}}{2}}^{2} d z= \\
=\left.\int_{0}^{0.852 \pi} d \phi \int_{0}^{2} \rho d \rho z\right|_{\frac{\rho^{2}}{2}} ^{2}=\int_{0}^{0.852 \pi} d \phi \int_{0}^{2} \rho 2-\frac{\rho^{2}}{2} d \rho=\int_{0}^{0.852 \pi} d \phi \int_{0}^{2} 2 \rho-\frac{1}{2} \rho^{3} d \rho= \\
=\int_{0}^{0.852 \pi} d \phi 2 \cdot \frac{\rho^{2}}{2}-\left.\frac{1}{2} \cdot \frac{1}{4} \rho^{4}\right|_{0} ^{2}=2 \int_{0}^{0.852 \pi} d \phi=\left.2 \phi\right|_{0} ^{0.852 \pi}=1.704 \pi \text { (cubic units). }
\end{gathered}
$$

## 6.3. Curvilinear Integrals

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-268.jpg?height=909&width=486&top_left_y=884&top_left_x=187)

Fig. 65