Example 3. Calculate the circulation of the vector $\mathbf{a}=\sqrt{1+x^{2}+y^{2}} \mathbf{i}+y[x y+$ $\left.\ln \left(x+\sqrt{1+x^{2}+y^{2}}\right)\right]$ along the circle $x^{2}+y^{2}=R^{2}$.
Solution. The circulation of the given vector is equal to

$$
\mu=\oint_{L}(\mathbf{2} . d \mathbf{r})=\oint_{L} \sqrt{1+x^{2}+y^{2}} d x+y\left[x y+\ln \left(x+\sqrt{1+x^{2}+y^{2}}\right)\right] d y .
$$

Here

$$
P=\sqrt{1+x^{2}+y^{2}}, \quad Q=x y^{2}+y \ln \left(x+\sqrt{1+x^{2}+y^{2}}\right)
$$

We find the partial derivatives

$$
\frac{\partial P}{\partial y}=\frac{y}{\sqrt{1+x^{2}+y^{2}}}, \quad \frac{\partial Q}{\partial x}=y^{2}+\frac{y}{\sqrt{1+x^{2}+y^{2}}}
$$

Applying Green's formula, we get

$$
\Lambda=\iint_{D}\left(y^{2}+\frac{y}{\sqrt{1+x^{2}+y^{2}}}-\frac{y}{\sqrt{1+x^{2}+y^{2}}}\right) d x d y=\iint_{D} y^{2} d x d y
$$

Switching to polar coordinates

$$
x=\rho \cos \varphi, \quad y=\rho \sin \varphi
$$

we will have

$$
\mu=\iint_{D} \rho^{2} \sin ^{2} \varphi \rho d \rho d \varphi=\iint_{D} \rho^{3} \sin ^{2} \varphi d \rho d \varphi
$$

Since $0 \leqslant \varphi<2 \pi, 0 \leqslant \rho \leqslant R$, we get

$$
\mu=\int_{0}^{2 \pi} \sin ^{2} \varphi d \varphi \int_{0}^{R} \rho^{3} d \rho=\frac{\pi R^{4}}{4}
$$

## Problems for Independent Solution

Calculate the circulation of the following vector fields along the given contours using Green's formula:

209. $\mathrm{a}=(y+x) \mathrm{i}+(y-x)\} ; \quad L: x+y=1, x=0 . y=0$.
210. $\left.=\left(x-y^{2}\right) \mid+2 x y\right)$

$L: y=x, y=x^{2}$.

211. $=x \ln \left(1+y^{2}\right)!+\frac{x^{2} y}{1+y^{2}} \mathbf{j}$;

L: $x^{2}+y^{2}=2 x$

212. $=y^{2} \mid-x^{2}$ I;

L: $x+y=-1, x=0, y=0$.

213. $a=\frac{\left.\left(3 x-y^{3} \sqrt{1+x^{2}+4 y^{3}}\right) I+\left(18 y^{2}+x^{3} \sqrt{1+x^{2}+4 y^{3}}\right)\right]}{3 \sqrt{1+x^{2}+4 y^{3}}} ; \quad: x^{2}+y^{2}=1$.
214. Using Green's formula, calculate the difference between the integrals

$$
I_{1}=\int_{A m B}(x+y)^{2} d x-(x-y)^{2} d y \quad \text { and } \quad I_{2}=\int_{A n B}(x+y)^{2} d x-(x-y)^{2} d y
$$

where $A \pi B$ is the line segment connecting points $A(0,0)$ and $B(1,1)$, and $A n B$ is the arc of the parabola $y=x^{2}$

215. Prove that the value of the integral $\oint(2 x+y) d x+2 x d y$, where $I$ is a closed contour, gives the area of the region bounded by this contour.
216. Using Green's formula, calculate the line integral $\int_{2}(a, d r)$ of the vector field $\left.R=\left(e^{2} \sin y-y\right)\right)+\left(e^{2} \cos y-1\right)$, where the line $\mathbf{L}$ is the upper semicircle $x^{2}+y^{2}=2 x$, traversed from point $A(2,0)$ to point $O(0,0)$.
CHAPTER

4

## Potential Field

## § 18. Criteria for Potentiality of a Field

Definition. A vector field $\mathbf{a}(M)=P(x, y, z) \mid+Q(x, y, z)]+$ $\boldsymbol{R}(\boldsymbol{x}, y, z) \mathbf{k}$, defined in a spatial domain $V$, is called potential if there exists a scalar function $\varphi(M)$ such that the equality

$$
a(M)=\operatorname{grad} \varphi(M)
$$

holds at all points in the domain $V$.

The function $\varphi(M)=\varphi(x, y, z)$, satisfying the equality (1) in the domain $V$, is called the potential (or potential function) of the vector field $\mathbf{a}$.

The relation (1) is equivalent to the following three scalar equations:

$$
P(x, y, z)=\frac{\partial \varphi}{\partial x}, \quad Q(x, y, z)=\frac{\partial \varphi}{\partial y}, \quad R(x, y, z)=\frac{\partial \varphi}{\partial z}
$$

The potential of a field is defined up to an additive constant.

Remark. For force fields, the function $\varphi(M)$ is usually called the force function, and the potential is called the function $-\varphi(M)$.