2.20. Calculate the rotor of the following vector fields:
a) $\vec{A}=x \dot{i}-z^{2} \vec{j}+y^{2} \dot{k}$;

b) $\dot{A}=y z \dot{i}+x z \dot{j}+x y \dot{k}$.
Solution.

a) By definition

$$
\begin{aligned}
& \operatorname{rot} \dot{A}=\nabla \times \dot{A}=\left|\begin{array}{ccc}
i & j & \dot{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x & -z^{2} & y^{2}
\end{array}\right|=\dot{i}(2 y+2 z) \\
& \text { b) } \operatorname{rot} \dot{A}=\nabla \times \dot{A}=\left|\begin{array}{ccc}
i & j & \dot{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y z & x z & x y
\end{array}\right|=\dot{0}
\end{aligned}
$$

The field $\ddot{A}$, defined in $R_{3}$, turned out to be irrotational, hence this field is potential and can be represented as the gradient of some scalar field $\varphi$, i.e.

$$
\dot{A}=\operatorname{grad} \varphi
$$

Let's find this scalar field $\varphi$ using a line integral of the second kind (this field is determined up to an arbitrary constant):

$$
\varphi=\int_{0}^{x} 0 d t+\int_{0}^{y} 0 d t+\int_{0}^{z} x y d t+C=x y z+C
$$

In these calculations, the initial point $(x_{0}, y_{0}, z_{0})$ is taken as the point $(0,0,0)$.