91Ó°ÊÓ

It is given that vector field is

$\mathbf{F}(x,y)=y\mathbf{i}-x\mathbf{j}$

Vertices are $(Â\pm 3,Â\pm 3)$traversed counterclockwise.

Green Theorem states that

Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by

$\mathbf{r}\left(t\right)=⟨\left(x\right(t),y(t\left)\right)âŸ\copyright \text{for}a≤t≤b$

Considering vector field defined on R, we get

${∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA$

Given vector field is $\mathbf{F}(x,y)=y\mathbf{i}-x\mathbf{j}$

$F_1(x,y)=y$

$F_2(x,y)=-x$

Now, $\frac{∂F_2}{∂x}=\frac{∂}{∂x}(-x)$

$\frac{∂F_2}{∂x}=-1$

And $\frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(y\right)$

$\frac{∂F_1}{∂y}=1$

Application of Green's Theorem gives integral as:

${∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA$

$={âˆ\neg }_R(-1-1)dA$

$={âˆ\neg }_R-2dA$

$=-{âˆ\neg }_{R}2dA$

As per given conditions, the region will be bounded by square with vertices $(Â\pm 3,Â\pm 3)$.

Hence, region of integration is

$R=\left\{\right(x,y)âˆ\pounds -3≤x≤3,-3≤y≤3\}$

Evaluating given integral as:

${∫}_C\mathbf{F}·d\mathbf{r}=-2{âˆ\neg }_{R}dA$

$=-2{∫}_{-3}^3{∫}_{-3}^{3}dydx$

$=-2{∫}_{-3}^3\left[{∫}_{-3}^{3}dy\right]dx$

$=-2{∫}_{-3}^3[y{]}_{-3}^{3}dx$

$=-2{∫}_{-3}^3[3-(-3\left)\right]dx$

$=-2{∫}_{-3}^{3}6dx$

$=-12{∫}_{-3}^{3}dx$

$=-12[x{]}_{-3}^3$

$=-12[3-(-3\left)\right]$

$=-12\left[6\right]$

$=-72$

Hence, required integral is${∫}_C\mathbf{F}·d\mathbf{r}=-72$