Find ${\int }_{C}F.d\mathbf{r}$, where $F\left(x,y\right)=xy\mathbf{i}+4\left(x-y\right)\mathbf{j}$and $C$is the boundary of the region bounded by the curves $x=1,x=e,y=e^x$and $y=\ln x$, traversed counterclockwise.

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${\int }_{C}F.d\mathbf{r}$.

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 :-

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

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

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

We have :-

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

Also we have :-

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

Then we have :-

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

This is the required answer.