Example 2. Consider the line integral

$$
\int_{L} \frac{-y d x}{x^{2}+y^{2}}+\frac{x d y}{x^{2}+y^{2}}
$$
Solution. The given expression makes sense everywhere except at the point $O(0,0)$; therefore, we exclude this point. In the remaining part of the plane (which will already be a disconnected region), the coefficients of $d x$ and $d y$ are continuous and have continuous partial derivatives, and they are identically equal to

$$
\frac{\partial}{\partial x}\left(\frac{x}{x^{2}+y^{2}}\right) \equiv \frac{\partial}{\partial y}\left(-\frac{y}{x^{2}+y^{2}}\right)
$$

On the other hand, if we compute the integral along the circle $\boldsymbol{L}: \boldsymbol{x}^{2}+\boldsymbol{y}^{2}=\boldsymbol{R}^{2}$, parameterizing the equation of this circle, we will have

$$
\oint_{L} \frac{-y d x+x d y}{x^{2}+y^{2}}=\int_{0}^{2 \pi} \frac{R^{2} \sin ^{2} t+R^{2} \cos ^{2} t}{R^{2}} d t=\int_{0}^{2 \pi} d t=2 \pi
$$

We obtained that the circulation is different from zero, and therefore, the line integral depends on the path of integration.

## Problems for Independent Solution

Determine in which of the specified vector fields the integral does not depend on the shape of the integration path:

206. $\mathbf{a}=z^{2} \mathbf{i}+x^{2} \mathbf{j}+y^{\prime} \mathbf{k} . \quad 207 . \quad \mathrm{a}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\sqrt{1+x^{2}+y^{2}+z^{2}}} . \quad 208 . \mathrm{a}=\frac{y \mathbf{i}-x \mathbf{j}+z \mathbf{k}}{x^{2}+y^{2}+z^{2}}$

## Green's Formula

Let in the region $D$ with boundary $\boldsymbol{L}$ there be a plane vector field

$$
\mathbf{a}=P(x, y) \mathbf{i}+Q(x, y) \mathbf{j}
$$

where the coordinates $P(x, y), Q(x, y)$ are continuous and have continuous partial derivatives $\frac{\partial P}{\partial y}$

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-42.jpg?height=211&width=392&top_left_y=885&top_left_x=476)

Fig. 33 and $\frac{\partial Q}{\partial x}$

Then Green's formula is valid

$$
\int_{2} P d x+Q d y=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y
$$

In this case, the boundary $L$ is traversed so that the region $D$ remains on the left.
The region $D$ can be disconnected, so its boundary can consist of several components (see Fig. 33). In this case, under the integral

$$
\int_{L} P d x+Q d y
$$

we understand the sum of integrals over all components of the boundary of the region $D$.

Green's formula (2) is a special case of Stokes' theorem (see § 16).

Green's formula allows simplifying the computation of the circulation of a vector field in some cases.