Example 8. Calculate the flux of the vector field $\mathbf{a}=x \mathbf{i}+y \mathbf{j}+\sqrt{x^{2}+y^{2}-1} \mathbf{k}$ through the outer side of the one-sheet hyperboloid $z=\sqrt{x^{2}+y^{2}-1}$, bounded by the planes $z=0, z=\sqrt{3}$
Solution. The given surface is projected unambiguously onto the $x O y$ plane in the region $D_{x y}$, bounded by a circle

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-25.jpg?height=77&width=333&top_left_y=388&top_left_x=1001)

We find the outer normal $\mathbf{n}$:

$z= \pm \operatorname{grad}\left(z-\sqrt{x^{2}+y^{2}-1}\right)= \pm\left(\frac{-x \mathbf{i}-y \mathbf{j}}{\sqrt{x^{2}+y^{2}-1}}+\mathbf{k}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-25.jpg?height=349&width=232&top_left_y=175&top_left_x=1431)

Fig. 22
Since $\mathbf{n}$ forms an obtuse angle $\gamma$ with the $O z$ axis (Fig. 22), we take the negative sign, and thus,

$$
\mathbf{n}=\frac{x \mathbf{i} + y \mathbf{j}}{\sqrt{x^{2}+y^{2}-1}} - \mathbf{k}
$$

We find the scalar product

$$
(\mathbf{a}, \mathbf{n})=\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}-1}}-\sqrt{x^{2}+y^{2}-1}=\frac{1}{\sqrt{x^{2}+y^{2}-1}}
$$

Applying formula (7), we get

$$
\mathrm{n}=\iint_{S}\left(\mathbf{a}, \mathbf{n}^{0}\right) d S=\iint_{D_{x y}} \frac{d x d y}{\sqrt{x^{2}+y^{2}-1}}
$$

Switching to polar coordinates $x=\rho \cos \varphi, y=\rho \sin \varphi$, we have

$$
\Pi=\iint_{D_{x y}} \frac{\rho d \rho d \varphi}{\sqrt{\rho^{2}-1}}=\int_{0}^{2 \pi} d \varphi \int_{1}^{2} \frac{\rho d \rho}{\sqrt{\rho^{2}-1}}=\left.2 \pi \sqrt{\rho^{2}-1}\right|_{1} ^{2}=2 \sqrt{3} \pi
$$