Example 2. Compute the flux of the vector field a $=4 x i-y j+z \mathbf{k}$ through the surface of the torus.
Solution. Using the Ostrogradsky-Gauss theorem, we find that the sought flux $\Pi$ is equal to

$$
\begin{aligned}
\Pi & =\oiint_{s}\left(\mathrm{a}, \mathrm{n}^{0}\right) d \sigma= \\
& =\iiint_{V}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right) d v=4 V
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-30.jpg?height=203&width=300&top_left_y=526&top_left_x=1362)

Fig. 27 where $\boldsymbol{V}$ is the volume of the torus. To calculate the volume $\boldsymbol{V}$, we use Pappus's theorem for the volume of a solid of revolution, according to which this volume is equal to the product of the area of the rotating figure and the path described by the center of mass of this figure.

Let $R_{1}$ and $R_{2}$ be the inner and outer radii of the torus (Fig. 27). The area $S$ of the circle, which forms the torus upon rotation, is

$$
S=\pi\left(\frac{R_{2}-R_{1}}{2}\right)^{2}
$$

The length of the path described by the center of mass - the center of this circle - is the circumference $\frac{\left(\boldsymbol{R}_{1}+\boldsymbol{R}_{2}\right)}{2}$, i.e.,

$$
l=2 \pi \frac{R_{1}+R_{2}}{2}=\pi\left(R_{1}+R_{2}\right)
$$

Thus, the volume $V$ of the torus is

$$
V'=\pi\left(\frac{R_{2}-R_{1}}{2}\right)^{2} \pi\left(R_{1} + R_{2}\right)=\frac{\pi^{2}}{4}\left(R_{2}-R_{1}\right)^{2}\left(R_{2}+R_{1}\right)
$$

The sought flux is

$$
\Pi=\pi^{2}\left(R_{2}-R_{1}\right)^{2}\left(R_{2}+R_{1}\right)
$$