How to arrive at Stokes's theorem from Green's theorem? I would like to verify the identity
$$ \oint \vec F  \cdot (\hat i dx  + \hat j dy) + \oint \vec F \cdot (\hat i dx  + \hat j dy) + \oint \vec F  \cdot (\hat i dx  + \hat j dy)  = \oint \vec F \cdot (\hat i dx + \hat j dy + \hat k dz) $$
If it is incorrect then what would be the correct identity.
Green's theorem is special case of Stokes's theorem. How do we arrive at Stokes's theorem using Green's theorem?
 A: Time and space does not permit a complete answer, but here is an outline of one way to do it.
First, note that Green's theorem in the plane (applied to $f\partial g/\partial u$ and $f\partial g/\partial v$) leads to
$$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}
  \iint\limits_D\Big(\pd fu\pd gv-\pd fv\pd gu\Big)\,du\,dv
  =\oint\limits_J f\,dg$$
where $D$ is a “sufficiently nice” region in the plane and $J$ is its boundary curve.
Next, assume a three-dimensional surface $S$ is parametrized by $\mathbf{r}\colon D\to\mathbb{R}^3$, and that $\mathbf{F}$ is a vector field. Now you can prove the identity $$  (\operatorname{curl}\mathbf{F})\cdot
  \Big(\pd{\mathbf{r}}{u}\times\pd{\mathbf{r}}{v}\Big)
  =\pd{\mathbf{F}}{u}\cdot\pd{\mathbf{r}}{v}
  -\pd{\mathbf{F}}{v}\cdot\pd{\mathbf{r}}{u}$$
and discover that each component function of $\mathbf{F}$ in this equation gives rise to a term of the form of the integrand on the left in the first equation. I.e., you let $f$ be each of the components of $\mathbf{F}\circ\mathbf{r}$ in turn, with $g$ being the corresponding comonent of $\mathbf{r}$, and add the three resulting equations together. You now have Stokes's theorem as written out using the given parametrization of $S$.
“Some assembly required.”
