# 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?

-
You're thinking backward, here. Since Green's Theorem is a special case of Stokes's Theorem, we can arrive at Green's Theorem from Stokes's Theorem. We can't go the other way, in general. – Cameron Buie Jun 19 '12 at 18:27
Can't we use similar technique?? ... I mean we could verify it for it's projection along x,y,z planes. – Santosh Linkha Jun 19 '12 at 18:28
We can, but only to obtain a rather restricted version of Stokes's Theorem. Apparently, that's all you're looking for? – Cameron Buie Jun 19 '12 at 18:30
Oh ... I didn't know that!! I thought if this is true for it's projections ... then we could construct it from it's projections – Santosh Linkha Jun 19 '12 at 18:31
Perhaps this would be more productive if you explicitly stated the version of Stokes's Theorem that you are talking about. Some statements are very general, and nowhere near Green's Theorem. Some are more specific, though, and we might actually be able to do what you wish. – Cameron Buie Jun 19 '12 at 18:40

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 $S$.