91影视

We are given$\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x鈭�y)\mathbf{j}$. We need to compute the integral ${鈭�}_{C}F(x,y)dr$.

Here, the boundary curve C is the triangle described by$x=0,y=0,y=1-x$, traversed counterclockwise as shown in the following figure

In this Figure, C is a piecewise-smooth, simple closed curve, so perform a separate integration for each of the sides of this triangle, as shown in this figure.

Hence, the required integral will be,
Its derivative will be:${鈭�}_C鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}={鈭�}_{C_1}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}+{鈭�}_{C_2}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}+{鈭�}_{C_3}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

The side $C_1$is the line segment from $\left(1,0\right)$to $\left(0,1\right)$. For the side$C_1$, use the following parametrization:

Its derivative will be

$\begin{array}{r}{\mathbf{r}}^{\mathrm{鈥�}}\left(t\right)=\frac{d}{dt}鉄�1鈭�t,t鉄�\\ =鉄�鈭�1,1鉄�.\end{array}$

For the parametrization localid="1654149212316" $\mathbf{r}\left(t\right)=鉄�1鈭�t,t鉄�$rewrite $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x鈭�y)\mathbf{j}$as follows:

localid="1654149329946" role="math"

So the first integral can be evaluated as:

The side $C_2$is the line segment from $\left(0,1\right)$to $\left(0,0\right)$. For the side $C_2$, use the following parametrization $\mathbf{r}\left(t\right)=鉄�0,1鈭�t鉄�for0鈮�t猢�1.$

Its derivative will be,

$\begin{array}{r}{\mathbf{r}}^{\mathrm{鈥�}}\left(t\right)=\frac{d}{dt}鉄�0,1鈭�t鉄�\\ =鉄�0,鈭�1鉄�.\end{array}$

For the parametrization

; rewrite $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x鈭�y)\mathbf{j}$ as follows:

So the second integral can be evaluated as: $\begin{array}{rcl}\begin{array}{r}{鈭�}_{C_2}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}\end{array}& =& {鈭�}_{C_2}鈥�\mathbf{F}\left(x\right(t),y(t\left)\right)鈰�{\mathbf{r}}^{\mathrm{鈥�}}\left(t\right)dt\\ & =& \begin{array}{r}{鈭�}_0^1鈥�鉄�2鈭�t,t鈭�1鉄�鈰�鉄�0,鈭�1鉄�dt\end{array}\\ & =& \begin{array}{r}{鈭�}_0^1鈥�\left(\right(2鈭�t)鈰�0+(t鈭�1)鈰�(鈭�1\left)\right)dt\end{array}\\ & =& \begin{array}{r}{鈭�}_0^1鈥�(1鈭�t)dt\end{array}\\ & =& \begin{array}{r}{\left[t鈭�\frac{t^2}{2}\right]}_0^1\end{array}\\ & =& \left[\begin{array}{r}\left(1鈭�\frac{1^2}{2}\right)-\left(0鈭�\frac{0^2}{2}\right)\end{array}\right]\\ & =& \frac{1}{2}鈰�鈰�\left(3\right)\end{array}$

The side $C_3$is the line segment from $\left(0,0\right)$to $\left(1,0\right)$. For the side $C_3$, use the following parametrization $\mathbf{r}\left(t\right)=鉄�t,0鉄�for0鈮�t猢�1.$

Its derivative will be,

$\begin{array}{r}{\mathbf{r}}^{\mathrm{鈥�}}\left(t\right)=\frac{d}{dt}鉄�t,0鉄�\\ =鉄�1,0鉄�.\end{array}$

For the parametrization localid="1654150157545" $\mathbf{r}\left(t\right)=鉄�t,0鉄�$; rewrite $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x鈭�y)\mathbf{j}$as follows:

localid="1654150252889"

So the third integral can be evaluated as:

Substitute values of integral from (2), (3), and (4) into (1), then the required integral is, $\begin{array}{rcl}\begin{array}{r}{鈭�}_C鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}\end{array}& =& {鈭�}_{C_1}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}+{鈭�}_{C_2}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}+{鈭�}_{C_3}鈥�\mathbf{F}(x,y)鈰�d\mathbf{r}\\ & =& \begin{array}{r}鈭�\frac{1}{\ln 鈦�2}+\frac{1}{2}+\frac{1}{\ln 鈦�2}\end{array}\\ & =& \frac{1}{2}\end{array}$

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