91影视

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

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

${鈭�}_C\mathbf{F}路d\mathbf{r}\text{.}$

Hence, evaluate the line integral ${鈭�}_C\mathbf{F}路d\mathbf{r}$ 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 $\mathbf{r}\left(t\right)=鉄�\left(x\right(t),y(t\left)\right)鉄�$for $a鈮�t鈮�b$.

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 $\frac{鈭�F_2}{鈭�x}$and $\frac{鈭�F_1}{鈭�y}$.

Then,

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

and,

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

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

$$\begin{gathered} {鈭�}_C\mathbf{F}路d\mathbf{r}={鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA \\ ={鈭�}_R\left(\cos x\sin y-\left(-3e^x\right)\right)dA \\ ={鈭�}_R\left(\cos x\sin y+3e^x\right)dA鈥�鈥�.\left(2\right) \end{gathered}$$

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

starting and ending at $(2,蟺)$.

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,

$R=\left\{\right(x,y)鈭�0鈮�x鈮�2,0鈮�y鈮�蟺\}$.

Then, evaluate the integral (2) as follows:

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

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