Use Green’s Theorem to evaluate the integral:

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

The given vector field is $\mathbf{F}(x,y)=0\mathbf{i}+xy\mathbf{j}$

and vertices are$(\pm 3,\pm 3)$

Green Theorem states that:

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

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

Apply in given question, we get

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

For the vector field given

$F_1(x,y)=0$

and $F_2(x,y)=xy$

$\frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\left(xy\right)$

$\frac{\partial F_2}{\partial x}=y$

And

$\frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}\left(0\right)$

$\frac{\partial F_1}{\partial y}=0$

Using Green's Theorem, given integral becomes,

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

$={\iint }_R(y-0)dA$

role="math" localid="1653247702786" $\Rightarrow {\int }_C\mathbf{F}·d\mathbf{r}={\iint }_{R}ydA$

As per given conditions, the region of integration is bounded by square with vertices $(\pm 3,\pm 3)$is given by

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

The integral is evaluated as:

${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_{R}ydA$

$={\int }_{-3}^3{\int }_{-3}^{3}ydydx$

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

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

$={\int }_{-3}^3\left[\frac{9}{2}-\frac{9}{2}\right]dx$

$=0$

Hence, required integral is${\int }_C\mathbf{F}·d\mathbf{r}=0$