Use Green’s Theorem to evaluate the integral:

$\mathbf{F}(x,y)=y\mathbf{i}-x\mathbf{j}$and C is the square with vertices$(\pm 3,\pm 3)$ traversed counterclockwise.

It is given that vector field is

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

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

Green Theorem states that

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

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

Considering vector field defined on R, we get

${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA$

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

$\frac{\partial F_2}{\partial x}=-1$

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

$\frac{\partial F_1}{\partial y}=1$

Application of Green's Theorem gives integral as:

${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA$

$={\iint }_R(-1-1)dA$

$={\iint }_R-2dA$

$=-{\iint }_{R}2dA$

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

Hence, region of integration is

$R=\left\{\right(x,y)\mid -3\le x\le 3,-3\le y\le 3\}$

Evaluating given integral as:

${\int }_C\mathbf{F}·d\mathbf{r}=-2{\iint }_{R}dA$

$=-2{\int }_{-3}^3{\int }_{-3}^{3}dydx$

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

$=-2{\int }_{-3}^3[y{]}_{-3}^{3}dx$

$=-2{\int }_{-3}^3[3-(-3\left)\right]dx$

$=-2{\int }_{-3}^{3}6dx$

$=-12{\int }_{-3}^{3}dx$

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

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

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

$=-72$

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