91影视

We have given that :-

$F\left(x,y\right)=xy\mathbf{i}+4\left(x-y\right)\mathbf{j}$.

Also $C$is boundary that is bounded by $x=1,x=e,y=e^x$and $y=\ln x$.

We have to find the value of${鈭�}_{C}F.d\mathbf{r}$.

We have given that :-

$F\left(x,y\right)=xy\mathbf{i}+4\left(x-y\right)\mathbf{j}$

Then $F_1=xy,F_2=4\left(x-y\right)$

So that :-

$\frac{鈭�F_1}{鈭�y}=x,\frac{鈭�F_2}{鈭�x}=4$.

Then by applying Green's Theorem, stated that :-

${鈭�}_{C}F\left(x,y\right).d\mathbf{r}=鈭�{鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA.$

We have :-

${鈭�}_{C}F\left(x,y\right).d\mathbf{r}=鈭�{鈭�}_R\left(4-x\right)dA$

Also we have :-

$x=1,x=e,y=e^x,y=\ln x$.

Then we have :-

$$\begin{gathered} {鈭�}_{C}F\left(x,y\right).d\mathbf{r}={鈭�}_1^e{鈭�}_{e^x}^{Inx}\left(4-x\right)dydx \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}={鈭�}_1^e\left({鈭�}_{e^x}^{Inx}\left(4-x\right)dy\right)dx \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}={鈭�}_1^e\left(4-x\right){\left[y\right]}_{e^x}^{Inx}dx \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}={鈭�}_1^e\left(4-x\right)\left(\ln x-e^x\right)dx \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}={鈭�}_1^e\left(4\ln \left(x\right)-4e^x-x\ln \left(x\right)+x{e}^x\right)dx \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}={\left[\left[4(x\ln \left(x\right)-x)\right]-\left[4e^x\right]-\left[\frac{x^2}{2}\ln \left(x\right)-\frac{x^2}{4}\right]+\left[x{e}^x-e^x\right]\right]}_1^e \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}=\left[4\left(e\ln \left(e\right)-e\right)-4e^e-\frac{e^2}{2}\ln \left(e\right)+\frac{e^2}{4}+e路e^e-e^e\right]-\left[4\left(\ln \left(1\right)-1\right)-4e-\frac{1}{2}\ln \left(1\right)+\frac{1}{4}+e-e\right] \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}=\left[4e-e-4e^e-\frac{e^2}{4}+e路e^e-e^e\right]-\left[4\left(0-1\right)-4e-0+\frac{1}{4}\right] \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}=3e-5e^e-\frac{e^2}{4}+e路e^e+4+4e-\frac{1}{4} \\ 鈬�{鈭�}_{C}F\left(x,y\right).d\mathbf{r}=7e-5e^e-\frac{e^2}{4}+e路e^e+\frac{15}{4} \end{gathered}$$

This is the required answer.