91Ó°ÊÓ

The given information is,

$$\begin{gathered} x^2+y^2+2x=0 \\ F=yi-xj \end{gathered}$$

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

$$\begin{gathered} ForF=yi-xj \\ ∫F.dx=∫ydx-xdy \end{gathered}$$

Use Green’s Theorem.

${âˆ\neg }_A\left(\frac{∂Q}{∂x}-\frac{∂P}{∂y}\right)dxdy={âˆ\circledR }_{∂A}Pdx+Qdy$

Where $∂A$is the boundary of the area A.

It can be seen that

$$\begin{gathered} P=y→\frac{∂P}{∂y} \\ =1 \\ Q=-x→\frac{∂Q}{∂x} \\ =-1 \end{gathered}$$

Thus, the integral over some contour C that encloses an area A

$$\begin{gathered} {âˆ\circledR }_{C}ydx+xdy=∫{∫}_A\left(-1-1\right)dxdy \\ =-2A \end{gathered}$$

Use completing the square method to determine the radius of the given contour (circle) in order to find the area enclosed by the contour.

$$\begin{gathered} x^2+y^2+2x={\left(x+1\right)}^2+y^2-1 \\ →{\left(x+2\right)}^2+y^2 \\ =1 \end{gathered}$$

The contour is a circle centered at (-1,0) and of radius 1

$$\begin{gathered} ∫F.dr=-2\mathrm{π}{\left(1\right)}^2 \\ =-2\mathrm{π} \end{gathered}$$

Hence, The Solution to the problem is

$∫F.dr=-2\mathrm{π}$