Example 6. Find the flux of the vector field $\mathbf{a} = \mathbf{i} - \mathbf{j} + x y z \mathbf{k}$ through the circle $S$, obtained by the intersection of the sphere $x^{2} + y^{2} + z^{2} \leqslant R^{2}$ with the plane $y = x$. Take the side of the circle facing the positive part of the $O x$ axis.
Solution. Since the plane $y=x$ is perpendicular to the coordinate plane $x O y$, the circle $S$ lying on this plane projects onto the plane $x O y$ as a segment $A_{1} A_{1}$, and thus the uniqueness of the projection is violated. On the other coordinate planes, the circle $S$ projects uniquely. Projecting the circle, for example, onto the plane $x O z$, we obtain the region $D_{x z}$ bounded by an ellipse (Fig. 20). The equation of the ellipse is found by eliminating $y$ from the system of equations

$$
\left\{\begin{aligned}
x^{2}+y^{2}+z^{2} & =R^{2} \\
y & =x
\end{aligned}\right.
$$

From this,

$$
2 x^{2}+z^{2}=R^{2}, \text { or } \frac{x^{2}}{R^{2} / 2}+\frac{z^{2}}{R^{2}}=1
$$

By the condition, the normal to the circle $S$ forms an obtuse angle $\beta$ with the axis $O y$, so we have

$$
\mathbf{n}=-\operatorname{grad}(y-x)=\mathbf{i}-\mathbf{j}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-24.jpg?height=294&width=287&top_left_y=160&top_left_x=1393)

Fig. 20 from this

$$
\left.\mathbf{n}^{0}=\frac{n}{|n|}=\frac{1}{\sqrt{2}} \right\rvert\,-\frac{1}{\sqrt{2}} \boldsymbol{j}
$$

From the last equality, we have $\cos \beta=-\frac{1}{\sqrt{2}}<0$. The area element $d S$ of the circle is

$$
d S=\frac{d x d z}{|\cos \beta|}=\sqrt{2} d x d z
$$

We find the scalar product: $\left(a, n^{0}\right)=\sqrt{2}$.

The sought flux by formula (5) is

$$
\Pi=\iint_{D_{x z}} 2 d x d z=\iint_{D_{x t}} d x d z=2 \cdot \frac{\pi R^{2}}{\sqrt{2}}=\sqrt{2} R^{2} \pi
$$

since the area $Q$ of the region $D_{x z}$, bounded by the ellipse with semi-axes $a=\frac{\boldsymbol{R}}{\sqrt{2}}$
and $b=R$, is

$$
Q=\iint_{D_{x z}} d x d z=\pi a b=\frac{\pi R^{2}}{\sqrt{2}}
$$