Intuition of Greens Theorem in the plane I'm trying to understand a special case of Greens Theorem.
Let $V: \Omega \to \mathbb{R}^2$ be a $C^1$ vector field defined an open set  $\Omega \subseteq \mathbb{R}^2$. Let $\gamma$ be a $C^1$-kurve, that is closed and has no loops and runs in the positive direction. Then
$$
\int_{\gamma} V \cdot \mathrm{d}r = \int_E \left( \dfrac{\partial V_2}{\partial x}(x,y) - \dfrac{\partial V_1}{\partial y}(x,y)\right) \mathrm{d}(x,y)
$$
where $E$ is the area enclosed by $\gamma$.
If $V$ is a closed vector field, meaning that $\dfrac{\partial V_1}{\partial y} = \dfrac{\partial V_2}{\partial x}$ then what does Greens theorem state? 
Now I can see that I'm integrating $0$ but I don't understand whats going on here. I'm very new to integrals in higher dimensions, so I'm lacking a severe amount of intuition.
 A: Sometimes it is hard to visualize and get an intuition about higher dimensional calculus, especially when one is accustomed to single variable calculus. The key here is understanding what a conservative vector field is (you call it a closed vector field). While it is true that a conservative vector field is one in which $\partial V_1 / \partial y = \partial V_2 / \partial x$, an equivalent statement is this: a conservative vector field is the gradient of some scalar function $f(x,y)$, that is 
$$
V = \langle V_1,V_2 \rangle = \nabla f = \langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y} \rangle
$$
You can see that given this definition, you can use Clairaut's Theorem to show that your definition follows:
$$
\frac{\partial V_1}{\partial y} = \frac{\partial f}{\partial y \partial x} = \frac{\partial f}{\partial x \partial y} = \frac{\partial V_2}{\partial x}
$$
Now here comes the intuition. We can think of the conservative vector field as the gradient of some function, so I think of this as a landscape. For each latitude and longitude on a certain map, we can give it an altitude as a function of those coordinates (picture a map with a bunch of hills and valleys). The gradient and thus the vector field is all the vectors that point in the direction of highest ascent. Extending the metaphor the path integral is like starting on at a point and climbing the hills and valleys, creating work as you go up a hill (proportional to the steepness and thus the dot product of your motion vector $dr$ with the gradient vector field $V$ in the path integral) and decreasing the work you put in by going down a hill. Since the path is closed, it is like you are going up and down the same amount overall, so the path integral is zero. Likewise, you can use Green's theorem and the result formulated above to conclude that this integral is zero.
Hopefully the hill metaphor is useful in visualizing these integrals.
A: This is a special case of generalized Stokes' theorem.  From 1d calculus, you're probably familiar with it in the form of the fundamental theorem of calculus:  given a function $F$ with derivative $f$, and an interval $[a,b]$ on which they $F$ is smooth, and so on, we get
$$F(b)-F(a) = \int_a^b f(x) \, dx$$
The generalization of this theorem is more or less the following:

The integral of a function over a boundary is equal to the integral of the derivative over the region enclosed by that boundary.

Your first question might be, "How is $F(b) -F(a)$ the integral of a function with respect to some boundary?"  Admittedly, this gets a lot of people--you'd need to have some notion of what it means to integrate over the endpoints of an interval.  Once you have that, you get the right answer.
But in higher dimensions, the above statement starts to make more sense geometrically.  Green's theorem is exactly it: the integral of a $V$ over some bounding curve $C$ is equal to the integral of the derivative of $V$ (in this case, the curl of $V$) over the region $D$ that is enclosed by $C$.
In any case, we're saying that, for a vector field integrated over a curve, if the curl of that vector field is zero (in any region enclosed by the curve), then the vector field integrated over the curve is zero.  This isn't as strong as the 1d case, where you could say that the antiderivative $F$ is some constant function, but it's the best you can do.
