91Ó°ÊÓ

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

and vertices are$(Â\pm 3,Â\pm 3)$

Green Theorem states that:

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

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

Apply in given question, we get

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

For the vector field given

$F_1(x,y)=0$

and $F_2(x,y)=xy$

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

$\frac{∂F_2}{∂x}=y$

And

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

$\frac{∂F_1}{∂y}=0$

Using Green's Theorem, given integral becomes,

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

$={âˆ\neg }_R(y-0)dA$

role="math" localid="1653247702786" $⇒{∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_{R}ydA$

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

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

The integral is evaluated as:

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

$={∫}_{-3}^3{∫}_{-3}^{3}ydydx$

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

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

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

$=0$

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