${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}$, where C is the curve in the plane $x-y+z=20$ and that lies above the curves $y=4$and $y=x^2$ in the$xy$-plane, traversed counterclockwise with respect to$\mathbf{n}=⟨1,-1,1⟩$, and where $\mathbf{F}(x,y,z)=$ $(2x-3y+4z)\mathbf{i}+(5x+y-z)\mathbf{j}+(x+4y+2z)\mathbf{k}$.

$\mathbf{F}(x,y,z)=(2x-3y+4z)\mathbf{i}+(5x+y-z)\mathbf{j}+(x+4y+2z)\mathbf{k}$

The goal is to calculate the line integral.localid="1650767287662" ${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$, where the curve C is defined as follows:

The curve C is the plane curve.localid="1650706544139" $x-y+z=20$and that lies above the curves $y=4$and $y=x^2$in the$xy$-plane, traversed counterclockwise direction with respect to $\mathbf{n}=⟨1,-1,1⟩$.

To calculate this integral, use Stokes' Theorem. According to Stokes' Theorem,

"Assume S is an oriented, smooth or piecewise-smooth surface enclosed by a curve C. Assume is an oriented unit normal vector of S and C has a parametrization that traverses C counterclockwise with respect to.

If a vector field $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined on S, then,

localid="1650767307394" ${\int }_C \mathbf{F}(x,y,z)\times d\mathbf{r}={\iint }_S \mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}dS=$

$\mathbf{F}=(2x-3y+4z)\mathbf{i}+(5x+y-z)\mathbf{j}+(x+4y+2z)\mathbf{k}$

The vector field's curl is then applied.$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows

$\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }x}& \frac{\mathrm{\partial }}{\mathrm{\partial }y}& \frac{\mathrm{\partial }}{\mathrm{\partial }z}\\ F_1(x,y,z)& F_2(x,y,z)& F_3(x,y,z)\end{array}\right|$

$=\left[\frac{\mathrm{\partial }{F}_3}{\mathrm{\partial }y}-\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }z}\right]\mathbf{i}-\left[\frac{\mathrm{\partial }{F}_3}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }z}\right]\mathbf{j}+\left[\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}\right]\mathbf{k}$

The vector field's curl is then applied.$\mathbf{F}=(2x-3y+4z)\mathbf{i}+(5x+y-z)\mathbf{j}+(x+4y+2z)\mathbf{k}$will be,

$\mathrm{curl}\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }x}& \frac{\mathrm{\partial }}{\mathrm{\partial }y}& \frac{\mathrm{\partial }}{\mathrm{\partial }z}\\ (2x-3y+4z)& (5x+y-z)& (x+4y+2z)\end{array}\right|$

$=\left[\frac{\mathrm{\partial }}{\mathrm{\partial }y}(x+4y+2z)-\frac{\mathrm{\partial }}{\mathrm{\partial }z}(5x+y-z)\right]\mathbf{i}$

$-\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}(x+4y+2z)-\frac{\mathrm{\partial }}{\mathrm{\partial }z}(2x-3y+4z)\right]\mathbf{j}$

$+\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}(5x+y-z)-\frac{\mathrm{\partial }}{\mathrm{\partial }y}(2x-3y+4z)\right]\mathbf{k}$

$=[4-(-1\left)\right]\mathbf{i}-[1-4]\mathbf{j}+[5-(-3\left)\right]\mathbf{k}$

role="math" localid="1650708411537" $=5\mathbf{i}+3\mathbf{j}+8\mathbf{k}$

$=⟨5,3,8⟩$

The normal vector is $\mathbf{n}=⟨1,-1,1⟩$

Then, the value of curl $\mathbf{F}(x,y,z)$- $\mathbf{n}$will be,

localid="1650765726819" $\mathrm{curl}\mathbf{F}(x,y,z)\cdot \mathbf{n}=⟨5,3,8⟩\times ⟨1,-1,1⟩$

localid="1650765645431" $$\begin{gathered} =5\times 1+3\times (-1)+8\times 1 \\ =5-3+8 \\ =10 \end{gathered}$$

The curve C is located above the curves.$y=4$and $y=x^2$in the $xy$-plane, as shown in the figure below, traversed counterclockwise:

Now, apply Stokes' Theorem (1) to determine the integral.${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$as follows

${\int }_C \mathbf{F}(x,y,z)\cdot d\mathbf{r}={\iint }_S \mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}dS$

$={\iint }_D 10dA$

$={\int }_{-2}^2 {\int }_{x^2}^4 10dydx$

$={\int }_{-2}^2 \left[{\int }_{x^2}^4 10dy\right]dx$

$={\int }_{-2}^2 10[y{]}_{x^2}^{4}dx$

$={\int }_{-2}^2 10\left(4-x^2\right)dx$

$=10{\left[4x-\frac{x^3}{3}\right]}_{-2}^2$

localid="1650765592856" $=10\left[\left(4\times 2-\frac{2^3}{3}\right)-\left(4(-2)-\frac{(-2{)}^3}{3}\right)\right]$

$=10\left(\frac{32}{3}\right)$

$=\frac{320}{3}.$