91Ó°ÊÓ

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 ${∫}_C\mathbf{F}(x,y,z)×d\mathbf{r}$,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. $\mathbf{n}=⟨1,-1,0âŸ\copyright$

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 , ${∫}_C\mathbf{F}(x,y,z)×d\mathbf{r}={âˆ\neg }_{S}cur\mathbf{F}(x,y,z)×\mathbf{n}dS$

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:

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

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

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

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

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

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

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

$=⟨3xz+1,-3yz,-7xâŸ\copyright$

Normal vector is

$\mathbf{n}=⟨1,-1,0âŸ\copyright$

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

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

$=(3xz+1)×1+(-3yz)(-1)+(-7x)×0$

$=3xz+1+3yz$

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

$=3xz+1+3yz$

$=6xz+1$

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,$D=\left\{(x,z)âˆ\pounds 0≤x≤1,x^2≤z≤x\right\}.$

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

$={âˆ\neg }_D(6xz+1)dA$

$={∫}_0^1{∫}_0^x(6xz+1)dzdx$

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

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

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

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

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

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

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

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

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