91Ó°ÊÓ

$F=\left(2x-3y\right)i-\left(3x-2y\right)j$

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.

$$\begin{gathered} F=\left(2x-3y\right)i-\left(3x-2y\right)j \\ ∫F·dr=∫\left(2x-3y\right)dx+\left(3x-2y\right)dy \end{gathered}$$

Use Green’s Theorem.

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

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

It can be seen that

$\begin{array}{rcl}P& =& 2x-3y→\frac{∂P}{∂y}\\ & =& -3\\ Q& =& 2y-3x→\frac{∂Q}{∂x}\\ & =& -3\end{array}$

Thus, the integral over some contour C the encloses area A is,

$\begin{array}{rcl}{âˆ\circledR }_C\left(2x-3y\right)dx+\left(2y-3x\right)dy& =& ∫{∫}_A-3-(-3)dxdy\\                                                                     & =& 0\end{array}$

So, considering the sector C of the circle $x^2+y^2=2$and the points (0,0), (1,1), (1,-1).

$\begin{array}{rcl}{âˆ\circledR }_C\left(2x-3y\right)dx+\left(2y-3x\right)dy& =& {∫}_{\left(0,0\right)}^{\left(1,1\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy\\ & & +{∫}_{\left(1,1\right)}^{\left(1,-1\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy\\ & & +{∫}_{\left(1,-1\right)}^{\left(0,0\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy\\ & =& 0\end{array}$

Over the line from (0,0) to (1,1), the relation between x and y is $x=y→dx=dy$.

$\begin{array}{rcl}{∫}_{\left(0,0\right)}^{\left(1,1\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy& =& {∫}_0^1\left(2x-3y\right)dx+\left(2y-3x\right)dy\\ & =& {∫}_0^1-2xdx\\ & =& {\left.-x^2\right|}_0^1\\ & =& -1\end{array}$

Over the line from (1,-1) to (0,0), the relation between x and y is .$y=-x→dy=-dx$

$\begin{array}{rcl}{∫}_{\left(1,-1\right)}^{\left(0,0\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy& =& {∫}_{-1}^0\left(2x-3y\right)dx+\left(2y-3x\right)dy\\ & =& {∫}_{-1}^{0}10xdx\\ & =& {\left.5x^2\right|}_{-1}^0\\ & =& -5\end{array}$

Thus,

${∫}_{\left(0,0\right)}^{\left(1,1\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy+{∫}_{\left(1,-1\right)}^{\left(0,0\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy=-6$

Hence, the Solution to the problem is

${∫}_{\left(0,0\right)}^{\left(1,1\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy+{∫}_{\left(1,-1\right)}^{\left(0,0\right)}\left(2x-3y\right)dx+\left(2y-3x\right)dy=-6$