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!

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 $|b| = 1$.)
