Find the work done by the vector field

$\mathbf{F}(x,y)=\left(\cos x-3y{e}^x\right)\mathbf{i}+\sin x\sin y\mathbf{j}$

in moving an object around the periphery of the rectangle with vertices $(0,0),(2,0),(2,\pi )$, and $(0,\pi )$, starting and ending at $(2,\pi )$.

• $\mathbf{F}(x,y)=\left(\cos x-3y{e}^x\right)\mathbf{i}+\sin x\sin y\mathbf{j}$
  
• Moving an object around the periphery of the rectangle with vertices $(0,0),(2,0),(2,\pi )$and $(0,\pi )$
• starting and ending at$(2,\pi )$
  

The work done by a vector field F moving a particle along the curve C is computed by the following line integral:

${\int }_C\mathbf{F}·d\mathbf{r}\text{.}$

Hence, evaluate the line integral ${\int }_C\mathbf{F}·d\mathbf{r}$ to evaluate the required work done.

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by $\mathbf{r}\left(t\right)=⟨\left(x\right(t),y(t\left)\right)⟩$for $a\le t\le b$.

If a vector field is defined on R, then,

For the vector field $\mathbf{F}(x,y)=\left(\cos x-3y{e}^x\right)\mathbf{i}+\sin x\sin y\mathbf{j}$,

$F_1(x,y)=\cos x-3y{e}^x$and $F_2(x,y)=\sin x\sin y$.

Now, first find $\frac{\partial F_2}{\partial x}$and $\frac{\partial F_1}{\partial y}$.

Then,

$$\begin{gathered} \frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\left(\sin x\sin y\right) \\ =\cos x\sin y \end{gathered}$$

and,

$$\begin{gathered} \frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}\left(\cos x-3y{e}^x\right) \\ =-3e^x \end{gathered}$$

Now, use Green's Theorem (1) to evaluate the integral ${\int }_C\mathbf{F}·d\mathbf{r}$as follows:

$$\begin{gathered} {\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA \\ ={\iint }_R\left(\cos x\sin y-\left(-3e^x\right)\right)dA \\ ={\iint }_R\left(\cos x\sin y+3e^x\right)dA\dots \dots .\left(2\right) \end{gathered}$$

Here, the boundary curve C is a rectangle with vertices $(0,0),(2,0),(2,\pi ),(0,\pi )$,

starting and ending at $(2,\pi )$.

So the region R bounded by this rectangle is shown in the following figure.

Viewed it as an x - or y-simple region, then the region of integration will be,

$R=\left\{\right(x,y)\mid 0\le x\le 2,0\le y\le \pi \}$.

Then, evaluate the integral (2) as follows:

$$\begin{gathered} {\int }_C\mathbf{F}·d\mathbf{r}={\int }_R\left(\cos x\sin y+3e^x\right)dA \\ ={\int }_0^{\pi }{\int }_0^2\left(\cos x\sin y+3e^x\right)dxdy \\ ={\int }_0^{\pi }\left[{\int }_0^2\left(\cos x\sin y+3e^x\right)dx\right]dy \\ ={\int }_0^{\pi }{\left[\sin x\sin y+3e^x\right]}_0^{2}dy \\ ={\int }_0^{\pi }\left[\left(\sin 2\sin y+3e^2\right)-\left(\sin 0\sin y+3e^0\right)\right]dy={\int }_0^{\pi }\left[\left(\sin 2\sin y+3e^2\right)-(0·\sin y+3·1)\right]dy \\ ={\int }_0^{\pi }\left(\sin 2\sin y+3e^2-3\right)dy \\ ={\left[-\sin 2\cos y+3e^{2}y-3y\right]}_0^{\pi } \\ =\left(-\sin 2·(-1)+3e^2\pi -3\pi \right)-\left(-\sin 2\cos 0+3e^2·0-3·0\right) \\ =\left(\sin 2+3\pi \left(e^2-1\right)\right)-(-\sin 2·1+0) \\ =\sin 2+3\pi \left(e^2-1\right)+\sin 2 \\ =2\sin 2+3\pi \left(e^2-1\right) \end{gathered}$$

Therefore, the required work done is $2\sin 2+3\pi \left(e^2-1\right)$.