Example 6. Calculate the circulation of the vector field $\mathbf{a} = y e^{x y} \mathbf{i} + x e^{x y} \mathbf{j} + x y z \mathbf{k}$ along the curve $L$, obtained by the intersection of the cone $x^{2} + y^{2} = (z-1)^{2}$ with the coordinate planes (Fig. 31).
Solution. Line $\boldsymbol{L}$ consists of two segments BC and CA, located on the coordinate planes $y O z$ and $z O z$ respectively, and the arc AB of the circle $x^{2}+y^{2}=\mathrm{L}_{1} z=0$. Therefore, the circulation of the given vector field will be

$\mu=\oint_{L}(\mathrm{a}, d r)=\int_{B C}(\mathrm{a}, d r)+\int_{C_{A}}(\mathrm{a}, d r)+\int_{\lambda B}(\mathrm{a}, d r)$

I) On the segment $B C$ we have

$=0, d x=0 ; \quad z=1-y, d x=-d y ; \quad 1 \geqslant y \geqslant 0$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-37.jpg?height=270&width=254&top_left_y=601&top_left_x=1409)

Fig. 31

$$
\int_{B C}(a, d r)=\int_{B C} y d x=0
$$

2) On the segment $C A$ we have

$$
y=0, d y=0 ; \quad z=1-x, d z=-d x ; \quad 0 \leqslant x \leqslant 1
$$

Therefore,

$$
\int_{C_{A}}(\mathbf{m}, d \mathbf{r})=\int_{C_{A}} x d y=0
$$

3) On the arc $\breve{A B}$ of the circle $x^{2}+y^{2}=1, z=0$ we have $z=0, d z=0$, and thus $\int_{\mid \delta}(a, d r)=\int_{\lambda \delta} e^{x y}(y d x+x d y)=\int_{\lambda x} e^{x y} d(y x)=\int_{\lambda x} d\left(e^{x y}\right)=\left.e^{x y}\right|_{\lambda(1,0)} ^{B(0,1)}=1-1=0$.

Example 7. Calculate the circulation of the vector field $\mathbf{a}=x y \mathbf{i}+y z \mathbf{j}+x z \mathbf{k}$, where $L:\left\{\begin{aligned} x^{2}+y^{2} & =1, \\ x+y+z & =1 .\end{aligned}\right.$

Solution. We have

$$
\mu=\oint_{L}(\mathrm{a}, \mathrm{dr})=\int_{L} x y d x+y z d y+x z d z
$$

The line $\boldsymbol{L}$ is an ellipse resulting from the intersection of the cylinder $\boldsymbol{x}^{2}+\boldsymbol{y}^{2}=1$ with the plane $x+y+z=1$. Let's find the parametric equations of this line. The projection of any point on this line onto the $x O y$ plane lies on the circle $x^{2}+y^{2}=1$. From this, we get $x=\cos t, y=\sin t$. But the ellipse lies on the plane $z+y+z=1$, so $z=1-x-y$ or $z=1-\cos t-\sin t$. Therefore, the parametric equations of the line $\mathbf{L}$ are:

$$
\left\{\begin{array}{l}
x=\cos t \\
y=\sin t \\
z=1-\cos t-\sin t \quad(0 \leqslant t<2 \pi)
\end{array}\right.
$$

From this, we find

$$
d x=-\sin t d t, \quad d y=\cos t d t, \quad d z=(\sin t-\cos t) d t,
$$

and thus the circulation will be equal to

$$
\begin{aligned}
\mu= & \int_{0}^{2 \pi}\left[-\cos t \sin ^{2} t+\sin t(1-\cos t-\sin t) \cos t+\right. \\
& \quad+\cos t(1-\cos t-\sin t)(\sin t-\cos t)] d t= \\
& =\int_{0}^{2 \pi}\left(-3 \sin ^{2} t \cos t+\sin 2 t-\cos ^{2} t \sin t-\cos ^{2} t-\cos ^{3} t\right) d t=-\int_{0}^{2 \pi} \cos ^{2} t d t=-\pi \quad
\end{aligned}
$$

## Problems for independent solving

Calculate the circulation $D$ of the vector field $\mathbf{a}$ along the given line $\boldsymbol{L}$:

175. $\mathrm{a}=(x z+y) \mathbf{i}+(y z-x) \mathbf{j}-\left(x^{2}+y^{2}\right) \mathbf{k}$

$$
\boldsymbol{L}:\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
z=3
\end{array}\right.
$$

176. $a=y^{2} \mathbf{i}+z^{2} \mathbf{j}+x^{2} \mathbf{k}$

$$
L:\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=R^{2} \\
x^{2}+y^{2}=R x \quad(z \geqslant 0)
\end{array}\right.
$$

177. $\mathrm{a}=(2 x+z) \mathbf{i}+(2 y-z) \mathbf{j}+x y z \mathbf{k} ; \boldsymbol{L}-$ the line of intersection of the paraboloid of revolution $x^{2}+y^{2}=1-z$ with the coordinate planes.
178. Show that if the circulation of a vector field along any closed contour is zero, then in such a field there cannot be any closed vector lines.

## § 15. Rotor (vortex) of a vector field

Let us have a vector field

$$
\mathbf{a}(M)=P(x, y, x) \mathbf{i}+Q(x, y, z) \mathbf{j}+R(x, y, z) \mathbf{k}
$$

We will assume that the coordinates $P, Q, R$ of the vector $\mathbf{a}(M)$ are continuous and have continuous first-order partial derivatives with respect to all their arguments.

Definition 1. The rotor of the vector $\mathbf{a}(M)$ is a vector denoted by the symbol rot $\mathbf{a}$ and defined by the equality

$$
\text { rot } \mathbf{a}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) \mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) \mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \mathbf{k}
$$

or in a symbolic, convenient for memorization, form

$$
\operatorname{rot} \mathbf{a}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R
\end{array}\right|
$$

This determinant is usually expanded along the elements of the first row, with the operations of multiplying the elements of the second row by the elements of the third row being interpreted as differentiation operations, for example, $\frac{\partial}{\partial x} \cdot Q=\frac{\partial Q}{\partial x}$

Definition 2. If in some region $G$ we have rot a $=0$, then the vector field $\mathbf{a}$ in the region $G$ is called irrotational.