$\mathbf{\oint }\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$around the circle over the curved part of the hemisphere in Problem 24, if ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{0}$, where $\mathit{F}\mathbf{=}\mathit{y}\mathit{i}\mathbf{-}\mathit{x}\mathit{j}$.

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.

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

Use Green’s Theorem.

${\iint }_A\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy={\oint }_{\partial A}Pdx+Qdy$

Where $\partial A$is the boundary of the area A.

It can be seen that

$$\begin{gathered} P=y\to \frac{\partial P}{\partial y} \\ =1 \\ Q=-x\to \frac{\partial Q}{\partial x} \\ =-1 \end{gathered}$$

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

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

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

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

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

$$\begin{gathered} \int F.dr=-2\pi {\left(1\right)}^2 \\ =-2\pi \end{gathered}$$

Hence, The Solution to the problem is

$\int F.dr=-2\pi$