Example 12. Calculate the circulation of the vector field given in cylindrical coordinates: $2=\rho \sin \varphi \mathrm{e}_{\rho}+\rho z \mathrm{e}_{\varphi}+\rho^{3} \mathrm{e}_{z}$, along the curve L: $\{\rho=\sin \varphi, z=0,0 \leqslant \varphi \leqslant \pi\}$ directly and using Stokes' theorem.
Solution. Coordinates of the given vector

$$
a_{p}=\rho \sin \varphi, \quad a_{v}=\rho z, \quad a_{n}=\rho^{3}
$$

The contour $L$ represents a closed curve located in the plane $z=0$ (Fig. 41).

1) Direct calculation of circulation.

Substituting the coordinates of the vector into formula (13), we get

$$
\boldsymbol{\mu}=\oint_{\mathbf{L}} \rho \sin \varphi d \rho+\rho^{2} z d \varphi+\rho^{3} d z
$$

On the curve $L$ we have:

$$
z=0, d z=0 ; \quad \rho=\sin \varphi, d \rho=\cos \varphi d \varphi ; \quad 0 \leqslant \varphi \leqslant \pi
$$

Therefore, the sought circulation will be equal to

$$
\boldsymbol{\mu}=\oint_{L} \rho \sin \varphi d \rho=\int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi d \varphi=0
$$

2) Calculation of circulation using Stokes' theorem. According to Stokes' theorem, the sought circulation is equal to

$$
\mu=\oint_{\mathbf{L}}(\mathrm{a}, d r)=\iint_{s}\left(\mathrm{rot} \mathrm{a}, \mathrm{n}^{0}\right) d S
$$

where $S$ is the surface stretched over the contour $L$.

We find the rotor of the given field

$$
\operatorname{rot} \alpha=\frac{1}{\rho}\left|\begin{array}{ccc}
e_{\phi} & \rho e_{\phi} & e_{z} \\
\frac{\theta}{\partial \rho} & \frac{\theta}{\partial \varphi} & \frac{\theta}{\partial z} \\
\rho \sin ^{2} \varphi & \rho^{2} z & \rho^{3}
\end{array}\right|=-\rho e_{\rho}-3 \rho^{2} e_{\varphi}+(2 z-\cos \varphi) e_{z} .
$$

At points where $\rho=0$, the value of the rotor is determined by continuity, assuming

$$
\text { rot } a(0, \varphi, z)=(2 z-\cos \varphi) e_{z} \text {. }
$$

Thus, the rotor is defined throughout the entire three-dimensional space. Since the curve $L$ lies in the plane $z=0$, we can take as the surface $S$ stretched over this curve, the part of the plane $z=0$ bounded by the curve $L$. Then, for the normal vector $\mathbf{n}^{\prime \prime}$ to the surface $S$, we can take the unit vector $\mathbf{e}_{z}$, i.e., $\mathbf{n}^{\prime \prime}=\mathbf{e}_{z}$. We find the scalar product:

$\left(\operatorname{rot} \mathrm{A}, \mathrm{n}^{0}\right)=\left(-\rho e_{\rho}-3 \rho^{2} \mathrm{e}_{\varphi}+(2 z-\cos \varphi) e_{z}, \mathrm{e}_{z}\right)=2 z-\cos \varphi_{.}$.

Since, due to the orthonormality of the basis vectors $e_{\rho}, \mathbf{e}_{\varphi}, \mathbf{e}_{z}$, we have

$$
\left(e_{\rho}, e_{z}\right)=\left(e_{\varphi}, e_{z}\right)=0, \quad\left(e_{z}, e_{z}\right)=1
$$

The sought circulation is equal to

$$
\zeta=\iint_{S}(2 z-\cos \varphi) d S
$$

Considering that $z=0$ on $S$ and the element of area $d S$ of the coordinate surface $z=0$ is equal to

$$
d S=\rho d \rho d \varphi
$$

we finally get

$$
\begin{aligned}
& \zeta=-\iint_{S} \cos \varphi d S=-\iint_{S} \cos \varphi \cdot \rho d \rho d \varphi= \\
& =-\int_{0}^{\pi} \cos \varphi d \varphi \int_{0}^{\sin \varphi} \rho d \rho=-\frac{1}{2} \int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi d \varphi=0
\end{aligned}
$$