Example 1. Find the vector potential $\mathbf{b}=\mathbf{b}(x, y, z)$ for a solenoidal field given by the vector $\mathbf{a} = 2 y \mathbf{i} - z \mathbf{j} + 2 x \mathbf{k}$.
Solution. Generate it in the form

$$
\mathbf{b}=\mathbf{h}(x, y, z)=Q_{1}(x, y, z) \mathbf{j}+R_{1}(x, y, z) \mathbf{k}
$$

where $Q_{1}(x, y, z)$ and $R_{1}(x, y, z)$ are found using formulas (6) and (7). Since in this

| Find the vector potentials of solenoidal fields: |  |
| :---: | :---: |
| 256. $a=i+\mathbf{j}+\mathbf{k}$.  258. $a=\left(e^{x}-e^{y}\right) \mathbf{k}$ | 257. $a=2 y \mathbf{i}+2 z j$  259. $a=6 y^{2} \mathbf{i}+6 z j+6 x k$ |
| 260. $\mathbf{a}=3 y^{3} \mathbf{i}-x^{2} \mathbf{j}-\left(y^{2}+2 x\right) \mathbf{k}$ | 261. $a=y e^{x^{i}}+2 y z j-\left(2 x y z e^{x^{2}}+z^{2}\right) k$. |

case $P=2 y, Q=-z, R=2 x$, we will have

$Q_{1}(x, y, z)=\int 2 x d x=x^{2}, \quad R_{1}(x, y, z)=\int z d z+\int 2 y d y=x z-y^{2}$. Thus,

$$
\mathbf{b}(\cdot x, y, z)=x^{2} \mathbf{j}+\left(x z+y^{2}\right) \mathbf{k}
$$

By direct verification, we confirm that $\mathbf{b}=\mathbf{a}$, and therefore, this vector is a vector potential of the given field.

Remark. Due to the arbitrariness in choosing the vector $\mathbf{h}$, instead of the condition $P_{1}(x, y, z) \equiv 0$, we can require that $Q_{1}(x, y, z) \equiv 0$ or $R_{1}(x, y, z) \equiv 0$. The system of equations (2) and formulas (5), (6), (7) will accordingly change.

## Problems for Independent Solution

If the vector field $\mathbf{a}=\mathbf{a}(M)$ is solenoidal in the domain $G$, which is star-shaped (see § 19, Chapter IV) with the center at the origin of coordinates $O(0,0,0)$ (the field $\mathbf{a}(M)$ may be undefined at the point $O$), then one of the vector potentials $\mathbf{b}=\mathbf{b}(M)$ can be found using the formula

$$
\mathbf{b}(M)=\int_{0}^{1}\left[\mathbf{a}\left(M^{\prime}\right), \mathbf{r}(M)\right] t d t
$$

where $\mathbf{r}(M)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$ is the radius vector of the point $M(x, y, z)$, and the point $M^{\prime}(t x, t y, t z)$, as the parameter $t$ varies from 0 to 1, traverses the segment $OM$.

Example 2. Using formula (8), find the vector potential of the solenoidal field $\mathbf{a}=2 y \mathbf{i}-z \mathbf{j}+2 x \mathbf{k}$.

Solution. The given vector field is defined over the entire three-dimensional space, which is a star-shaped domain with the center at the origin of coordinates, so we can use formula (8) to find the vector potential. At the point $M^{\prime}(t x, t y, t z)$, we have

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

We find the vector product

$$
\left[\mathbf{a}\left(M^{\prime}\right), \mathbf{r}(M)\right]=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 t y & -t z & 2 t x \\
x & y & z
\end{array}\right|=\left[-\left(2 x y-z^{2}\right) \mathbf{i}+\left(2 x^{2}-2 y z\right) \mathbf{j}+\left(2 y^{2}+x z\right) \mathbf{k}\right] t
$$

Using formula (8), we find

$$
\begin{aligned}
& \mathbf{b}(M)=\int_{0}^{1}\left[-\left(2 x y-z^{2}\right) \mathbf{i}+\left(2 x^{2}-2 y z\right) \mathbf{j}+\left(2 y^{2}+x z\right) \mathbf{k}\right] t^{2} d t= \\
&=-\frac{1}{3}\left(2 x y-z^{2}\right) \mathbf{i}+\frac{2}{3}\left(x^{2}-y z\right) \mathbf{j}+\frac{1}{3}\left(2 y^{2}+x z\right) \mathbf{k}
\end{aligned}
$$

It is easy to verify that $\operatorname{rot} \mathbf{b}(M)=\mathbf{a}(M)$.

Remark. In Examples 1 and 2, for the same solenoidal field $\mathbf{a}=2 y \mathbf{i}-z \mathbf{j}+2 x \mathbf{k}$, different vector potentials were obtained;

$$
\begin{aligned}
& \mathbf{b}_{1}(M)=x^{2} \mathbf{j}+\left(x z-y^{2}\right) \mathbf{k} \\
& \mathbf{b}_{2}(M)=-\frac{1}{3}\left(2 x y-z^{2}\right) \mathbf{i}+\frac{2}{3}\left(x^{2}-y z\right) \mathbf{j}+\frac{1}{3}\left(2 y^{2}+x z\right) \mathbf{k}
\end{aligned}
$$

They differ from each other by a term equal to the gradient of some scalar field $f(M)$. This term plays the role of an arbitrary constant (in the sense of its rotor). It can be represented as the gradient of some scalar function $f(M)$. Let's find this function in our case. We have

$\operatorname{grad} f(M)=\mathbf{b}_{1}(M)-\mathbf{b}_{2}(M)=\frac{1}{3}\left(2 x y+z^{2}\right) \mathbf{i}+\frac{1}{3}\left(x^{2}+2 y z\right) \mathbf{j}+\frac{1}{3}\left(2 x z+y^{2}\right) \mathbf{k}$.

To find the scalar field $f(M)$, we apply formula (3) from § 19, taking the point $\left(x_{0}, y_{0}, z_{0}\right)$ as the origin of coordinates $O(0,0,0)$. Then we get

$f(M)=\int_{0}^{x} 0 d x+\int_{0}^{y} \frac{1}{3} x^{2} d y+\int_{0}^{z} \frac{1}{3}\left(2 x z+y^{2}\right) d z+C=\frac{1}{3}\left(x^{2} y+y^{2} z+z^{2} x\right)+C$, where $C$ is an arbitrary constant.