Example 8. Find the potential of the vector field given in cylindrical coordinates: $\mathbf{a}=\left(\frac{\operatorname{arctg} z}{\rho}+\cos \varphi\right) \mathbf{e}_{\rho}-\sin \varphi \mathbf{e}_{\varphi}+\frac{\ln \rho}{1+z^{2}} \mathbf{e}_{z}$.
Solution. Using formula (4), we find

$$
\operatorname{rot} a=\left|\begin{array}{ccc}
\frac{1}{\rho} \mathrm{e}_{\rho} & \mathrm{e}_{\varphi} & \frac{1}{\rho} \mathbf{c}_{z} \\
\frac{\partial}{\partial \rho} & \frac{\partial}{\partial \varphi} & \frac{\partial}{\partial z} \\
\frac{\operatorname{arctg} z}{\rho}+\cos \varphi & -\rho \sin \varphi & \frac{\ln \rho}{1+z^{2}}
\end{array}\right|=0(\rho>0) \text {, }
$$

i.e., the given field is potential. The sought potential \( u = u(\rho, \varphi, z) \) is the solution to the following system of differential equations:

$$
\left\{\begin{array}{l}
\frac{\partial u}{\partial \rho}=\frac{\operatorname{arctg} z}{\rho}+\cos \varphi \\
\frac{\partial u}{\partial \varphi}=-\sin \varphi \cdot \rho \\
\frac{\partial u}{\partial z}=\frac{\ln \rho}{1+z^{2}}
\end{array}\right.
$$

From the first equation, integrating with respect to \(\rho\), we find that

$$
u=\ln \rho-\operatorname{arctg} z+\rho \cos \varphi+C(\varphi, z)
$$

Differentiating (8) with respect to \(\varphi\), we get

$$
\frac{\partial u}{\partial \varphi}=-\rho \sin \varphi+\frac{\partial C}{\partial \varphi}
$$

and since \(\frac{\partial u}{\partial \varphi}=-\rho \sin \varphi\), it follows that \(\frac{\partial C}{\partial \varphi} \equiv 0\), i.e., \(C=C_{1}(z)\). Thus,

$$
u=\ln \rho \cdot \operatorname{arctg} z+\rho \cos \varphi+C_{1}(\varphi, z)
$$

From this,

$$
\frac{\partial u}{\partial z}=\frac{\ln \rho}{1+z^{2}}+C_{1}^{\prime}(z)
$$

By the third equation of the system, we have

$$
\frac{\ln \rho}{1+z^{2}}=\frac{\ln \rho}{1+z^{2}}+C_{1}^{\prime}(z)
$$

i.e., \(C_{1}^{\prime}(z) \equiv 0\), hence \(C_{1}(z) \equiv C=\) const.

Thus, the potential of the given field is

$$
u(\rho, \varphi, z)=\ln \rho \cdot \operatorname{arctg} z+\rho \cos \varphi+C
$$

## Problems for Independent Solution

In the following problems, verify that the vector fields given in cylindrical coordinates are potential and find their potentials:

$$
\begin{aligned}
& \text { 28B. } a=e_{\rho}+\frac{1}{\rho} e_{\varphi}-e_{z} . \quad 289 . a=\rho e_{\rho}+\frac{\varphi}{\rho} e_{\varphi}+z e_{z} . \\
& 290 .=\varphi z e_{\rho}+z e_{\varphi}+\rho \varphi e_{z} . \quad 291 . \approx=e^{\rho} \sin \varphi e_{\rho}+\frac{1}{\rho} e^{\rho} \cos \varphi \mathrm{e}_{\varphi}+2 z e_{z} . \\
& 292 . a=\varphi \cos z e_{\rho}+\cos z e_{\varphi}-\rho \varphi \sin z e_{z} .
\end{aligned}
$$

Example 9. Find the potential of the vector field given in spherical coordinates: \( a=\frac{1}{r} e^{\theta \varphi} \mathrm{e}_{r}+\frac{\ln r}{r} \varphi e^{\theta \varphi} \mathrm{e}_{\theta}+\frac{\theta \ln r}{r \sin \theta} e^{\theta \varphi} \mathbf{e}_{\varphi} \).

Solution. Using formula (4'), we get

$$
\operatorname{rot}=\left|\begin{array}{ccc}
\mathrm{e}_{r} & r \mathrm{e}_{\theta} & r \sin \theta \mathrm{e}_{\varphi} \\
\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi} \\
\frac{1}{r} e^{\theta \varphi} & \varphi \ln r \cdot e^{\theta \varphi} & \theta \ln r \cdot e^{\theta \varphi}
\end{array}\right|=0
$$

The given field is potential in the region where \( r>0, \theta \neq n \pi (\pi=0, \pm 1, \ldots) \).

The system of differential equations (7) for finding the potential \( u = u(r, \theta, \varphi) \) is

$$
\left\{\begin{array}{l}
\frac{\partial u}{\partial r}=\frac{1}{r} e^{\theta \varphi}, \\
\frac{\partial u}{\partial \theta}=\varphi e^{\theta \varphi} \ln r, \\
\frac{\partial u}{\partial \varphi}=\theta e^{\theta \varphi} \ln r .
\end{array}\right.
$$

Integrating the first equation of the system (9), we get

$$
u=e^{\theta \varphi} \ln r+C(\varphi, \theta)
$$

Differentiating (10) with respect to \(\theta\) and using the second equation of the system, we have

$$
\varphi e^{\theta \varphi} \ln r=\varphi e^{\theta \varphi} \ln r+\frac{\partial C}{\partial \theta}
$$

i.e., \(\frac{\partial C}{\partial \theta} \equiv 0\), hence \(C(\varphi, \theta) \equiv C_{1}(\varphi)\), and thus,

$$
u=e^{\theta \varphi} \ln r+C_{1}(\varphi)
$$

Differentiating (11) with respect to \(\varphi\) and using the third equation of the system (9), we find

$$
\theta e^{\theta \varphi} \ln r=\theta e^{\theta \varphi} \ln r+C_{1}^{\prime}(\varphi)
$$

or \(C_{1}^{\prime}(\varphi) \equiv 0\), hence \(C_{1}(\varphi) \equiv C=\) const. The sought potential is

$$
u(r, \theta, \varphi)=e^{\theta \varphi} \ln r+C
$$

## Problems for Independent Solution

Verify the potentiality of the following vector fields given in spherical coordinates and find their potentials.
293. \( a=\theta e_{r}+e_{r} \).
294. \( a=2 r e_{r}+\frac{1}{r \sin \theta} e_{r}+\frac{1}{r} e_{r} \).

295. \( a=\frac{1}{2} \varphi^{2} e_{r}+\frac{\varphi}{\sin \theta} e_{\varphi}+\frac{\theta}{r} e_{\theta} \).
296. \( a=\cos \varphi \sin \theta \mathrm{e}_{r}+\cos \varphi \cos \theta \mathrm{e}_{\theta}-\sin \varphi \mathrm{e}_{\varphi} \). 297. \( a=e^{r} \sin \theta e_{r}+\frac{1}{r} e^{r} \cos \theta e_{\theta}+\frac{2 \varphi}{\left(1+\varphi^{2}\right) r \sin \theta} e_{\varphi} \).

## \(7^{\circ}\). Calculation of the line integral and circulation of a vector field in curvilinear coordinates. Let the vector field
$$
a(M)=a_{1}\left(q_{1}, q_{2}, q_{3}\right) e_{1}+a_{2}\left(q_{1}, q_{2}, q_{3}\right) e_{2}+a_{3}\left(q_{1