23.${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$, where $C$is the closed curve in the plane $y=x$and formed by the curves $x=z$and $x^2=z$, traversed counterclockwise with respect to normal vector $\mathbf{n}=⟨1,-1,0⟩$, and where$\mathbf{F}(x,y,z)=7xy\mathbf{i}-z\mathbf{j}+3xyz\mathbf{k}$.

Consider the following vector field.

$\mathbf{F}(x,y,z)=7xy\mathbf{i}-z\mathbf{j}+3xyz\mathbf{k}\text{.}$

The goal is to determine the value of the line integral ${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$,of a curve$\mathrm{C}$

that is defined by its equation.

The curve is the closed curve in the plane $y=x$and is formed by the curves $x=z$and $x^2=z$traversed counterclockwise with respect to the normal vector. $\mathbf{n}=⟨1,-1,0⟩$

To evaluate the integral, Stokes' Theorem states that

if $S$an oriented, smooth or piecewise-smooth surface is bounded by a curve $C.$then there exists $\mathbf{n}$is an oriented unit normal vector of $S$and $C$has a parametrization that traverses$C$the surface in the counterclockwise direction $\mathbf{n}$.

since the 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 $S$in the field of the vector, then , ${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}={\iint }_{S}cur\mathbf{F}(x,y,z)\times \mathbf{n}dS$

Vector field of the first curl $\mathbf{F}(x,y,z)=7xy\mathbf{i}-z\mathbf{j}+3xyz\mathbf{k}$

The curl 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 as:

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

$=\left[\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right]\mathbf{i}-\left[\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right]\mathbf{j}+\left[\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right]\mathbf{k}\text{.}$

Then the curl of the vector pitch $\mathbf{F}(x,y,z)=7xy\mathbf{i}-z\mathbf{j}+3xyz\mathbf{k}$will be,

$curl\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 7xy& -z& 3xyz\end{array}\right|$

$=\left[\frac{\partial \left(3xyz\right)}{\partial y}-\frac{\partial (-z)}{\partial z}\right]\mathbf{i}-\left[\frac{\partial \left(3xyz\right)}{\partial x}-\frac{\partial \left(7xy\right)}{\partial z}\right]\mathbf{j}+\left[\frac{\partial (-z)}{\partial x}-\frac{\partial \left(7xy\right)}{\partial y}\right]\mathbf{k}$

$=\left[\right(3xz)-(-1\left)\right]\mathbf{i}-\left[\right(3yz)-0]\mathbf{j}+[0-7x]\mathbf{k}$

$=(3xz+1)\mathbf{i}-3yz\mathbf{j}-7x\mathbf{k}$

$=⟨3xz+1,-3yz,-7x⟩$

Normal vector is

$\mathbf{n}=⟨1,-1,0⟩$

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

$cur\mathbf{F}(x,y,z)\times \mathbf{n}=⟨3xz+1,-3yz,-7x⟩\times ⟨1,-1,0⟩$

$=(3xz+1)\times 1+(-3yz)(-1)+(-7x)\times 0$

$=3xz+1+3yz$

because$y=x,$so the value of$curl\mathbf{F}(x,y,z)\times \mathbf{n}$equal to, $curl\mathbf{F}(x,y,z)\times \mathbf{n}=3xz+1+3yz$

$=3xz+1+3yz$

$=6xz+1$

The curve$C$is the closed curve formed by the curves $x=z$and $x^2=z$

traversed counterclockwise. The figure is

The region of integration will be,$D=\left\{(x,z)\mid 0\le x\le 1,x^2\le z\le x\right\}.$

Now, To evaluate the integral${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$using Stokes' Theorem as follows:${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}={\iint }_{S}curlF(x,y,z)\times \mathbf{n}dS$

$={\iint }_D(6xz+1)dA$

$={\int }_0^1{\int }_0^x(6xz+1)dzdx$

$={\int }_0^1\left[{\int }_{x^2}^x(6xz+1)dz\right]dx$

$={\int }_0^1{\left[3x{z}^2+z\right]}_{x^2}^{x}dx$

$={\int }_0^1\left[\left(3x{x}^2+x\right)-\left(3x{\left(x^2\right)}^2+x^2\right)\right]dx$

$={\int }_0^1\left(-3x^5+3x^3-x^2+x\right)dx$

$={\left[-\frac{1}{2}{x}^6+\frac{3}{4}{x}^4-\frac{x^3}{3}+\frac{x^2}{2}\right]}_0^1$

$=\left(-\frac{1}{2}\times x^6+\frac{3}{4}\times x^4-\frac{x^3}{3}+\frac{x^2}{2}\right)-\left(-\frac{1}{2}\times x^6+\frac{3}{4}+x^4-\frac{x^3}{3}+\frac{x^2}{2}\right)$

$=\left(-\frac{1}{2}\times 1^6+\frac{3}{4}\times 1^4-\frac{1^3}{3}+\frac{1^2}{2}\right)-\left(-\frac{1}{2}\times 0^6+\frac{3}{4}\times 0^4-\frac{0^3}{3}+\frac{0^2}{2}\right)$

$=\frac{1}{12}.$

Therefore, ${\int }_C\mathrm{F}(x,y,z)\times d\mathrm{r}=\frac{1}{12}$is the required integral.