Give an example of a vector field whose orientation does not affect the outcome of Stokes’ Theorem.

There is an example of a vector field whose orientation does not affect the outcome of Stokes’ Theorem.

"Let $S$be an oriented, smooth or piecewise-smooth surface bounded by a curve $C$. Suppose that $\mathit{n}$is an oriented unit normal vector of $S$and $C$has a parametrization that traverses $C$in the counterclockwise direction with respect to$\mathit{n}$.
If a vector field $\mathit{F}(x,y,z)={\mathit{F}}_{\mathbf{1}}(x,y,z)\mathit{i}+{\mathit{F}}_{\mathbf{2}}(x,y,z)\mathit{j}+{\mathit{F}}_{\mathbf{3}}(x,y,z)\mathit{k}$is define on $S$, then ${\int }_c\mathit{F}(x,y,z)·d\mathit{r}=\int {\int }_{S}curl\mathit{F}(x,y,z)·\mathit{n}dS$".

If$\mathit{F}(x,y,z)$ is a conservative field, then,
$curl\mathit{F}(x,y,z)=0$

Then, the integral will be
${\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS={\iint }_S\mathbf{0}·\mathbf{n}dS$
$={\iint }_{S}0dS$
$=0$
Then, by Stokes' Theorem
${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}={\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS$
$=0$
In this case, the orientation does not come out as Stokes' Theorem, because the outcome for each orientation is $0$.

So, to write the required vector field $\mathit{F}(x,y,z),$write the conservative field.

A vector field is conservative, if and only if, the following conditions are satisfied;
$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z}.$
Consider the following vector field:
$\mathbf{F}(x,y,z)=y\ln z\mathbf{i}+x\ln z\mathbf{j}+\frac{xy}{z}\mathbf{k}$

For this vector field,

role="math" localid="1650350262262" width="43" height="20">=lnz,
$\frac{\partial F_1}{\partial z}=\frac{\partial }{\partial z}\left(y\ln z\right)$

$\frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\left(x\ln z\right)$

role="math" localid="1650350268353" $=\ln z$,
$\frac{\partial F_2}{\partial z}=\frac{\partial }{\partial z}\left(x\ln z\right)$
$=\frac{x}{z}$,
$\frac{\partial F_3}{\partial x}=\frac{\partial }{\partial x}\left(\frac{xy}{z}\right)$
$=\frac{y}{z}$,
$\frac{\partial F_3}{\partial y}=\frac{\partial }{\partial y}\left(\frac{xy}{z}\right)$
$=\frac{x}{z}$.
Hence, in this case
$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z}$.
Therefore, the vector fieldrole="math" localid="1650350473890" width="208" height="41">F(x,y,z)=ylnzi+xlnzj+xyzkis conservative.

So, for this vector field, the orientation does not affect the outcome of Stokes' Theorem.
Hence, an example for a vector field $\mathit{F}(x,y)$such that integral of$\mathit{F}(x,y)$ around the unit circle is zero.
Hence, an example of a vector field whose orientation does not affect the outcome of Stokes' Theorem is $\mathit{F}(x,y,z)=y\ln z\mathit{i}+x\ln z\mathit{j}+\frac{xy}{z}\mathit{k}$.