Example 13. Calculate the circulation of the vector field given in spherical coordinates: $2=r \mathbf{e}_{r}+(R+r) \sin \theta \mathbf{e}_{\varphi}$, along the circle $L:\{r=$ $\left.R, \theta=\frac{\pi}{2}\right\}$ in the direction of increasing angle $\varphi$ directly and using Stokes' theorem.
Solution. In the given example

$$
a_{r}=r_{1} \quad a_{9}=0, a_{\varphi}=(R+r) \sin \theta
$$

1) Direct calculation of circulation. According to formula (14), the desired circulation is

$$
\square=\oint_{L} r d r+(R+r) \sin \theta \cdot r \sin \theta d \varphi=\oint_{L} r d r+r(R+r) \sin ^{2} \theta d \varphi
$$

On the given circle $\boldsymbol{L}$, the center of which is at the origin of coordinates, we have:

$$
r=R, \quad d r=0 ; \quad \theta=\frac{\pi}{2} ; \quad 0 \leqslant \varphi<2 \pi
$$

and, therefore,

$$
\zeta \zeta=2 R^{2} \oint_{L} d \varphi=2 R^{2} \int_{0}^{2 \pi} d \varphi=4 \pi R^{2}
$$

2) Calculation of circulation using Stokes' theorem. The desired circulation according to Stokes' theorem is

$$
I=\oint_{\Sigma}(\mathrm{z}, d r)=\iint_{S}\left(\mathrm{rot}, \pi^{0}\right) d S
$$

where $d S$ is the surface stretched over the circle $L$.

We find the rotor of the given vector

$$
\operatorname{rot} \mathbb{R}=\frac{1}{r^{2} \sin \theta}\left|\begin{array}{ccc}
\mathrm{e}_{r} & r e_{\theta} & r \sin \theta e_{\varphi} \\
\frac{\theta}{\partial r} & \frac{\theta}{\partial \theta} & \frac{\theta}{\theta_{\varphi}} \\
r & 0 & \left(R r+r^{2}\right) \sin ^{2} \theta
\end{array}\right|=
$$

As the surface $S$ stretched over the circle $L$, we can take, for example, the upper hemisphere of radius $R: r=R, 0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant \varphi<2 \pi$. The unit vector $\mathbf{n}^{\text {" }}$ normal to the outer side of the hemisphere $S$ is directed along the vector $\mathbf{e}_r$, so we take $\mathbf{n}^{\text {" }}=\mathbf{e}_r$. We find the scalar product

$$
\left(r o t a, r^{\circ}\right)=\left(\frac{2(R+r)}{r} \cos \theta e_{r}-\frac{R+2 r}{r} \sin \theta e_{g}, e_{r}\right)=\frac{2(R+r)}{r} \cos \theta
$$

since $\left(e_{r}, e_{r}\right)=1,\left(e_{r}, r_{0}\right)=0$.

Considering that $r=R$ on the surface $S$, the expression for the desired flux is

$$
\Pi=\iint_{s} \frac{2(R+r)}{r} \cos \theta d S=4 \iint_{S} \cos \theta d S
$$

In spherical coordinates, the element of area $d S$ of the coordinate surface $r=\boldsymbol{R}$, i.e., the hemisphere $S$, will be

$$
d S=R^{2} \sin \theta d \theta d \varphi
$$

and, consequently,

$$
\Pi=4 \iint_{s} \cos \theta \cdot R^{2} \sin \theta d \theta d \varphi=4 R^{2} \int_{0}^{\pi / 2} \cos \theta \sin \theta d \theta \int_{0}^{2 \pi} d \varphi=4 \pi R^{2}
$$

By Stokes' theorem, we obtain $\boldsymbol{L}=4 \pi \boldsymbol{R}^{2}$

As the surface $S$ stretched over the circle $L$, we could have taken the lower hemisphere, the unit normal vector of which is $\mathbf{n}^{0}=-\mathbf{e}_{r}$; the result would be the same: $\boldsymbol{L}=4 \pi \boldsymbol{R}^{2}$

Note that the surface $S$ stretched over the circle $L$ should not be taken as the circle bounded by this circle, as this circle contains the point $r=0$ (the center of the circle), where the rotor of the given vector has a discontinuity.

## Problems for Independent Solution

Calculate the line integral over the given lines $L$ from the vector fields given in cylindrical coordinates.
298. $a=z e_{\rho}+\rho \varphi e_{s}+\cos \varphi e_{x}, \quad L-$ arc of a straight line: $\{\rho=a, \varphi=0,0 \leqslant z \leqslant 1\}$. 299. $=\rho \mathrm{e}_{\rho}+2 \rho \varphi \mathrm{e}_{\psi}+z \mathrm{c}_{2}, \quad L-$ semicircle: $\{\rho=1, z=0,0 \leqslant \varphi \leqslant \pi\}$. 300. $a=e^{p} \cos \varphi e_{p}+\rho \sin \varphi \mathrm{e}_{p}+\rho e_{z}, \quad L-$ helical turn $0 \leqslant \varphi \leqslant 2 \pi\}$

Calculate the line integral over the given line $L$ from the vector fields given in spherical coordinates
301. $\mathbf{n}=e^{r} \cos \theta e,+2 \theta \cos \varphi e_{\text {e }}+\varphi e_{\varphi}, \quad L-$ semicircle: $\{\rho=1, \varphi=0$, $0 \leqslant \theta \leqslant \pi\}$

302. $a=4 r^{3} \lg \frac{\varphi}{2} e_{r}+\theta \varphi \varphi_{r}+\cos ^{2} \varphi e_{s}, \quad L-$ segment of a straight line $\left\{\varphi=\frac{\pi}{2}, \theta=\frac{\pi}{4}\right.$, $0 \leqslant r \leqslant 1\}$
303. $a=\sin ^{2} \theta e+\sin \theta e_{g}+r \varphi \theta e_{w}, \quad L-$ segment of a straight line: $\left\{\varphi=\frac{\pi}{2}, r=\frac{1}{\sin \theta}\right.$, $\left.\frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}\right\}$

Calculate the circulation of the vector fields given in cylindrical coordinates over the given contours directly and using Stokes