${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}$, where $C$ is the boundary of the region in the plane $z=2x-y+10$ and that lies above the curves $x=1,x=2$, and $y=e^x$, and where

$\mathbf{F}(x,y,z)=(3x+y)\mathbf{i}+(y-2z)\mathbf{j}+(2+3z)\mathbf{k}.$

Consider the following vector field $\mathbf{F}(x,y,z)=(3x+y)\mathbf{i}+(y-2z)\mathbf{j}+(2+3z)\mathbf{k}$ The goal is to evaluate the line integral ${\int }_C\mathrm{F}(x,y,z)·d\mathbf{r}$,whose curve $C$ is defined as:

The curve$C$is relative to the curve in which the plane$z=2x-y+10$of the boundary region is on$x=1,x=2,y=e^x$.

The curve $C$is the boundary of the region in the plane$z=2x-y+10$and that lies above the curves. $x=1,x=2,y=e^x$. The curve$C$parameterized by the following:

for $1\le t\le 2$.

Then, the derivative of the original value will be.

${\mathbf{r}}^{\text{'}}\left(t\right)=\frac{d}{dt}\mathbf{r}\left(t\right)$

The parametrization, rewritten as a vector field $\mathbf{F}(x,y,z)=(3x+y)\mathbf{i}+(y-2z)\mathbf{j}+(2+3z)\mathbf{k}$as follows:

$\mathbf{F}(x,y,z)=(3x+y)\mathbf{i}+(y-2z)\mathbf{j}+(2+3z)\mathbf{k}$

$\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)=\left(3t+e^{\text{'}}\right)\mathbf{i}+\left(e^{\text{'}}-2\left(2t-e^{\text{'}}+10\right)\right)\mathbf{j}+\left(2+3\left(2t-e^{\text{'}}+10\right)\right)\mathbf{k}$

$=\left(3t+e^t\right)\mathbf{i}+\left(5e^t-4t-20\right)\mathbf{j}+\left(-2e^{\text{'}}+6t+22\right)\mathbf{k}$

$=\left(3t+e^{\text{'}},5e^{\text{'}}-4t-20,-2e^t+6t+22\right)$

the required integral of line will be,

${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}$

$={\int }_C\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)·{\mathbf{r}}^{\text{'}}\left(t\right)dt$

$={\int }_1^2\left(\left(3t+e^t\right)·1+\left(5e^t-4t-20\right)·e^t+\left(-2e^{\text{'}}+6t+22\right)·\left(2-e^{\text{'}}\right)\right)dt$

$={\int }_1^2\left(3t+e^{\text{'}}+5e^{2t}-4t{e}^{\text{'}}-20e^{\text{'}}-4e^t+12t+44+2e^{2t}-6t{e}^{\text{'}}-22e^t\right)dt$

$={\int }_1^2\left(7e^{2t}-5(2t+9)e^{\text{'}}+15t+44\right)dt$

$={\left[\frac{7}{2}{e}^{2t}-5e^{\text{'}}(2t+7)+\frac{15t^2}{2}+44t\right]}_1^2$

$=\left(\frac{7}{2}{e}^{22}-5e^2(2·2+7)+\frac{15·2^2}{2}+44·2\right)-\left(\frac{7}{2}{e}^{21}-5e^1(2·1+7)+\frac{15·1^2}{2}+44·1\right)$

$=\left(\frac{7}{2}{e}^4-55e^2+118\right)-\left(\frac{7}{2}{e}^2-45e+\frac{103}{2}\right)$

$=\frac{7}{2}{e}^4-\frac{117}{2}{e}^2+45e+\frac{133}{2}$

Therefore, the required integral of line is

${\int }_c\mathbf{F}(x,y,z)·d\mathbf{r}=\frac{7}{2}{e}^4-\frac{117}{2}{e}^2+45e+\frac{133}{2}$