# What is the difference between Green's Theorem and Stokes Theorem?

I don't quite understand the difference between Green's Theorem and Stokes Theorem. I know that Green's Theorem is in $\mathbb{R}^2$ and Stokes Theorem is in $\mathbb{R}^3$ and my lecture notes give Greens Theorem and Stokes Theorem as

$$\int \!\! \int_{\Omega} curl \, \underline{v} \, \mathrm{d}A = \int_{\partial \Omega} \underline{v} \, \mathrm{d} \underline{r}$$

and

$$\int \!\! \int_\Omega \nabla \times \underline{v} . \underline{n} \, \mathrm{d}A = \int_{\partial \Omega} \underline{v} \, \mathrm{d} \underline{r}$$

respectively. So why does being in $\mathbb{R}^3$ constitute the unit normal $\underline{n}$ to be dotted with the curl?

Thanks!

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Actually, Stoke's applies everywhere. Divergence applies to $\mathbb{R}^3$. Also, the dotting is basically the projection of the surface/area integral in the normal direction. – Inquest Mar 25 '12 at 13:52

Green's theorem in the plane is a special case of Stokes' theorem.

If we express Green's theorem as

$\oint_C M dx + N dy = \iint_R (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} ) dx dy$

we can express this in vector notation as

$Mdx + Ndy = (M \mathbf{i} + N \mathbf{j})(dx\mathbf{i}+dy\mathbf{j})= \mathbf{A}\cdot d\mathbf{r}$ in which $\mathbf{A} = M\mathbf{i}+ N\mathbf{j}$, $\mathbf{r} = x\mathbf{i}+y\mathbf{j}$.

We have

$\nabla \times \mathbf{A} = -\frac{\partial N}{\partial z}\mathbf{i}+\frac{\partial M}{\partial z}\mathbf{j}+ (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})\mathbf{k}$

and then $(\nabla \times \mathbf{A})\cdot \mathbf{k} = \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}$

So now we can re-write Green's theorem as (see * below re "dot" question)

$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_R (\nabla x \mathbf{A})\cdot \mathbf{k}$ dR

in which dR = dx dy.

This is basically a problem from Shaum's Vector Analysis (Spiegel)

Edit: the full generalization of

$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_R (\nabla x \mathbf{A})\cdot \mathbf{k}$ dR

to the usual version of Stokes' theorem

$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_S (\nabla x \mathbf{A})\cdot \mathbf{n}$ dS

is also a problem in Shaum's, which is really an exercise in extending Green's to three dimensions. As Nunoxic points out above, a slightly different treatment of Green's theorem generalizes to Gauss' divergence theorem, also known as Green's theorem in space.

(*) "Dot" question: In your first equation, "curl v" is the result of dotting with the unit normal. I think the usage "curl" is confusing because, comparing your two equations, it appears that you have done something in $R^3$ that you did not do in $R^2$. That is not the case.

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Green's Theorem is a special case of Stokes's Theorem. Since your surface is in the plane and oriented counterclockwise, then your normal vector is $n = \hat{k}$, the unit vector pointing straight up.

Similarly, if you compute $\nabla \times v$, where $v\, dr = Mdx + Ndy$, you would get $\left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \hat{k} = \text{curl}\, v \,\hat{k}$ as a result.

When you dot $(\nabla \times v) \cdot n$, you get the product $\text{curl}\, v$.

So even in $\mathbb{R}^2$ you are dotting the curl with the unit normal vector; but because the unit normal is aligned with the curl vector, the dot product is simply the magnitude of the curl. (In order words, when you look at the dot product formula $a \cdot b = |a||b| \cos \theta$, $\theta = 0$, $a$ is your curl and $|b| = 1$.)

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