Question

Find the value of $\mathbf{\int }\mathit{F}\mathbf{·}\mathit{d}\mathit{r}$along the circle from (1,1) to (1,−1) if F= (2x−3y)i−(3x−2y)j.

Short answer

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

Step-by-step solution

Given Information

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

Definition of Green’s Theorem

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.

Find the solution

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

Use Green’s Theorem.

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

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

It can be seen that

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

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

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

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

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

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

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

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

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

Thus,

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

Hence, the Solution to the problem is

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