$a$Use Green’s Theorem to evaluate the line integral in Exercise 37

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

The objective is to evaluate the integral ${\int }_C \mathbf{F}\cdot d\mathbf{r}$by using Green's Theorem, where $C$is the triangle described by role="math" localid="1654106106254" $x=0,y=0,y=1-x$, 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 $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x-y)\mathbf{j}$

$\begin{array}{r}{F}_1(x,y)=2^x+y\\ F_2(x,y)=2x-y.\end{array}$

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

Then,

$\begin{array}{r}\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}(2x-y)\\ =2\end{array}$

and,

$\begin{array}{r}\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }\mathrm{y}}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}(2^x+y)\\ =1\end{array}$

So the region R is bounded by the curves $x=0,y=0,y=1-x$as shown figure.

Viewed it as a y-simple region, then the region of integration will be,

$R=\left\{\right(x,y)\mid 0\le x\le 1,0\le y\le 1-x\}$

Using the Green's Theorem the line integral is evaluated as:
$\begin{array}{rcl}\begin{array}{r}{\int }_C \mathbf{F}\cdot d\mathbf{r}\end{array}& =& {\iint }_R \left(\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}\right)dA\\ & =& \begin{array}{r}{\iint }_R (2-1)dA\end{array}\\ & =& \begin{array}{r}{\iint }_R 1dA\end{array}\\ & =& \begin{array}{r}{\int }_0^1 {\int }_0^{1-x} 1dydx\end{array}\\ & =& \begin{array}{r}{\int }_0^1 \left[{\int }_0^{1-x} 1dy\right]dx\end{array}\\ & =& \begin{array}{r}{\int }_0^1 [y{]}_0^{1-x}dx\end{array}\\ & =& \begin{array}{r}{\int }_0^1 (1-x)dx\end{array}\\ & =& \begin{array}{r}{\left[x-\frac{x^2}{2}\right]}_0^1\end{array}\\ & =& \begin{array}{r}\left(1-\frac{1^2}{2}\right)-\left(0-\frac{0^2}{2}\right)\end{array}\\ & =& \frac{1}{2}\end{array}$

Therefore, the required integral is ${\int }_C \mathbf{F}\cdot d\mathbf{r}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{.}$