I have just watched the Green's theorem proof by Khan. At 7:40 he explains why for a conservative field, the partial differentials under the double integral:
$$\int \int_R \left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right ) dA$$
must be equal. He says:
(…) if F is conservative, which means it's the gradient of some function, it's path-independent, the closed integral around any path is equal to 0. (…) this thing right here [the difference under the integral] must be equal to 0. That's the only way that you're always going to enforce that this whole integral is going to be equal to 0 over any region. I'm sure you could think of situations where they cancel each other out, but really over any region that's the only way that this is going to be true.
1) What are the examples of two partials such that $\frac{\partial Q}{\partial x} \neq \frac{\partial P}{\partial y}$ whose double integrals are equal over a region and therefore cancel each other out?
2) If there exist examples for 1) (cancelling out locally), why is it impossible to define partials whose integrals cancel out over any region?
Best Answer
1) Both $P$ and $Q$ equal to zero will do. A little less obvious example is when $Q$ is a function of $y$ and $P$ is a function of $x$. Anyway, the most general case is when $(P, Q) = \nabla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y})$ where $F$ is some nice enough function.
2) It is possible to find $(P, Q)$ with $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \ne 0$ such that the integral over some domain is $0$. However, if you consider any regions, you can squeeze them as small as possible around the point where $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is not zero, and that integral will be non-zero (assuming $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is continuous).