91影视

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

The objective is to evaluate the integral ${鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}$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

$\mathbf{r}\left(t\right)=鉄�\left(x\right(t),y(t\left)\right)鉄�\text{for}a鈮�t鈮�b$

If a vector field is defined on R, then

${鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}={鈭�}_R鈥�\left(\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right)dA鈰�"....\left(1\right)$

For the vector field $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x鈭�y)\mathbf{j}$

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

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

Then,

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

and,

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

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,

$R=\left\{\right(x,y)鈭�0鈮�x鈮�1,0鈮�y鈮�1鈭�x\}$

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

Therefore, the required integral is ${鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{.}$