Use Green’s Theorem to evaluate the line integral in Exercise 35.

We are given vector field$\mathbf{F}(x,y)=y^2\mathbf{i}-xy\mathbf{j}$.

The objective is to evaluate the integral ${\int }_C \mathbf{F}\cdot d\mathbf{r}$by using Green's Theorem, where $C$is the unit circle traversed counterclockwise.

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

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

If a vector field is defined on R, then

${\int }_C \mathbf{F}\cdot d\mathbf{r}={\iint }_R \left(\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}\right)dA\cdot "....\left(1\right)$

For the vector field$\mathbf{F}(x,y)=y^2\mathbf{i}-xy\mathbf{j}$,

Now, first, find localid="1654151631288" $\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}and\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }\mathrm{y}}$

localid="1651063303202" $\begin{array}{r}\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}(-xy)\\ =-y\end{array}$

and,

localid="1654151713367" $\begin{array}{r}\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}=\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(y^2\right)\\ =2y\end{array}$

Using the Green Theorem the integral can be evaluated as

Here, the boundary curve C is a unit circle traversed counterclockwise, so the region R is bounded by a unit disk. In polar coordinates, the region of integration is described as follows,

$R=\left\{\right(r,\theta )\mid 0\le r\le 1,0\le \theta \le 2\pi \}.$

In this case,

$x=r\cos \theta ,y=r\sin \theta ,x^2+y^2=r^{2}anddA=rdrd\theta$

Change this integral (2) to polar coordinates, and integrate it.

$\begin{array}{rcl}\begin{array}{r}{\int }_C \mathbf{F}\cdot d\mathbf{r}\end{array}& =& {\iint }_R -3ydA\\ & =& \begin{array}{r}{\int }_0^{2\pi } {\int }_0^1 (-3r\sin \theta )rdrd\theta \end{array}\\ & =& \begin{array}{r}{\int }_0^{2\pi } {\int }_0^1 \left(-3r^2\sin \theta \right)drd\theta \end{array}\\ & =& \begin{array}{r}{\int }_0^{2\pi } \left[-\sin \theta {\int }_0^1 \left(3r^2\right)dr\right]d\theta \end{array}\\ & =& \begin{array}{r}{\int }_0^{2\pi } -\sin \theta {\left[r^3\right]}_0^{1}d\theta \end{array}\\ & =& \begin{array}{r}{\int }_0^{2\pi } -\sin \theta \left[1^3-0^3\right]d\theta \end{array}\\ & =& \begin{array}{r}{\int }_0^{2\pi } -\sin \theta d\theta \end{array}\\ & =& \begin{array}{r}[\cos \theta {]}_0^{2\pi }\end{array}\\ & =& \begin{array}{r}\cos 2\pi -\cos 0\end{array}\\ & =& \begin{array}{r}1-1\end{array}\\ & =& 0\end{array}$

Therefore, the required integral is evaluated as${\int }_C \mathbf{F}\cdot d\mathbf{r}\mathbf{=}\mathbf{0}$.