Directly compute (i.e., without using Green’s Theorem) ${\int }_C \mathbf{F}(x,y).dr$, where $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x-y)\mathbf{j}$

and C is the triangle described by x = 0, y = 0, and y = 1 − x, traversed counterclockwise.

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

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:${\int }_C \mathbf{F}(x,y)\cdot d\mathbf{r}={\int }_{C_1} \mathbf{F}(x,y)\cdot d\mathbf{r}+{\int }_{C_2} \mathbf{F}(x,y)\cdot d\mathbf{r}+{\int }_{C_3} \mathbf{F}(x,y)\cdot d\mathbf{r}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

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

$\begin{array}{r}{\mathbf{r}}^{\mathrm{\prime }}\left(t\right)=\frac{d}{dt}⟨1-t,t⟩\\ =⟨-1,1⟩.\end{array}$

For the parametrization localid="1654149212316" $\mathbf{r}\left(t\right)=⟨1-t,t⟩$rewrite $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x-y)\mathbf{j}$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 $\mathbf{r}\left(t\right)=⟨0,1-t⟩for0\le t⩽1.$

Its derivative will be,

$\begin{array}{r}{\mathbf{r}}^{\mathrm{\prime }}\left(t\right)=\frac{d}{dt}⟨0,1-t⟩\\ =⟨0,-1⟩.\end{array}$

For the parametrization

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

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

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 $\mathbf{r}\left(t\right)=⟨t,0⟩for0\le t⩽1.$

Its derivative will be,

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

For the parametrization localid="1654150157545" $\mathbf{r}\left(t\right)=⟨t,0⟩$; rewrite $\mathbf{F}(x,y)=\left(2^x+y\right)\mathbf{i}+(2x-y)\mathbf{j}$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, $\begin{array}{rcl}\begin{array}{r}{\int }_C \mathbf{F}(x,y)\cdot d\mathbf{r}\end{array}& =& {\int }_{C_1} \mathbf{F}(x,y)\cdot d\mathbf{r}+{\int }_{C_2} \mathbf{F}(x,y)\cdot d\mathbf{r}+{\int }_{C_3} \mathbf{F}(x,y)\cdot d\mathbf{r}\\ & =& \begin{array}{r}-\frac{1}{\ln 2}+\frac{1}{2}+\frac{1}{\ln 2}\end{array}\\ & =& \frac{1}{2}\end{array}$

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