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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?

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    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.2012-06-19
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    Can't we use similar technique?? ... I mean we could verify it for it's projection along x,y,z planes.2012-06-19
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    We can, but only to obtain a rather restricted version of Stokes's Theorem. Apparently, that's all you're looking for?2012-06-19
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    Oh ... I didn't know that!! I thought if this is true for it's projections ... then we could construct it from it's projections2012-06-19
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    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.2012-06-19

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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.”

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    The above answer is extracted from [a note](http://www.math.ntnu.no/~hanche/kurs/mat2/2003v/stokes.pdf) I wrote about this for a class in 2003; unfortunately, the note is in Norwegian, but perhaps you can glean some of the missing details from it anyhow.2012-06-19