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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?

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Your two questions (1) and (2) are really the same thing, and (1) just asks for examples where (2) asks if such examples are possible.

I'm not sure if you know this, but I found that the whole thing made a lot more sense for me when put this way: If the field is conservative, then there must be a function $f(x, y)$ such that $\nabla f = \vec F(x, y)$. Using subscripts to denote partials, $\frac{\partial P}{\partial y} = f_{xy}$ and $\frac{\partial Q}{\partial x} = f_{yx}$. And we know thanks to Young's Theorem that if $P_y=Q_x$ then the two functions must have come from a conservative vector field as described above.

That being said, if I happen to be looking at some non-conservative force field $\vec F$, it may happen such that its $P_y$ and $Q_x$ cancel out over some R, as Tunococ mentioned. This would probably happen due to some coincidence or symmetry of the problem - but as soon as R fails to meet whatever vector field's conditions are for $P_y$ and $Q_x$ cancelling out, you'll end up with a nonzero integral.

If you really must have an example, consider $$ \vec F(x, y) = x \hat i+x y\hat j $$ Then it follows that $P_y = 0$ and $Q_x = y$. I'll pick R to be the rectangular space defined by all $(x, y): a \le x \le b $ and $-c\le y\le c$ for any $c\ge 0, a\epsilon\mathbb{R}, b\epsilon\mathbb{R},$ given $a<b$. You'll notice I have a nice symmetrical rectangle. $$ \int\int_R(Q_x-P_y)dA = \int_{x=a}^b\int_{y=-c}^c(y-0)dydx=0 $$ But as soon as you loose that symmetric nature, that integral will not be equal to 0.

EDIT: You clarified that you were looking for any two functions $P_y$ and $Q_x$ such that $$ \int\int_R(Q_x-P_y)dA = 0 $$ over ANY region R (not a specially selected one) for $P_y\ne Q_x$. If you take your focus off of Green's Theorem for a second, we can see why that is not possible. If we think of the magnitude of curl here ($Q_x-P_y$) as simply functions of $x, y$, then you must agree for the conditions you specified there is a nonzero function $D(x, y)$ which satisfies: $$ \int\int_R(Q_x-P_y)dA = \int\int_RD(x, y)dA $$ Let's assume for simplicity's sake we're dealing with well-behaved functions. The only way that an integral can equal zero is if

1) All values of $D(x, y)$ are zero, which is untrue.

2) Over the entire R our integral somehow manages to cancel out.

However, you're looking for any R.

Let $S$ be the "special region" such that $$ \int\int_SD(x, y)dA=0 $$ $S$ can either exist or not exist. If it does not, then the function $D(x, y)$ by definition cannot have any region R which makes the integral zero.

If such an $S$ does exist, then we know that within $S$ there exist some regions $S_+,S_-$ and there could exist a region $S_0$ for which the following properties are true: $$ \int\int_{S_+}D(x, y)dA>0 $$ $$ \int\int_{S_-}D(x, y)dA<0 $$ $$ \int\int_{S_0}D(x, y)dA=0 $$ Such that: $$ S_++S_-+S_0 = S $$ Do you see why?

Now, if you do, then I propose taking a region $A = S_+$. We know for a fact that $$ \int\int_AD(x, y)dA \ne 0 $$ I made no assumptions about $D(x, y)$, except for being well-behaved, so there will always be a region I can find for a nonzero function $D(x, y)$ such that the integral does not equal 0.

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I do not think (1) and (2) are the same: in (1) I am asking about cancelling out in a region (arbitrary), for which you provided a very nice answer. In (2) I am wondering why it is impossible to define two such functions over ANY region. I like your answer most but I think there is a typo after "Using subscripts to denote partials": should be P_y, not f_{xy}. –  alkamid Aug 24 '12 at 15:04
@alkamid $P_y$ and $f_{xy}$ are the same thing (if $f$ exists). The actual Young's Theorem says $f_{xy}=f{yx}$. As for the typo, it was just a missing period, right? –  VF1 Aug 24 '12 at 16:52
There. I hope my edit was what you were looking for. –  VF1 Aug 24 '12 at 17:23
I found $P_y$ and $f_{xy}$ confusing, I did not know this notation and I thought it was a typo. Now your answer is perfect, thank you! –  alkamid Aug 24 '12 at 18:22
No problem. Glad to help. FYI, subscripts, in this context, mean partials - $\frac{\partial f}{\partial x}=f_x$. –  VF1 Aug 24 '12 at 19:55
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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).

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I edited 1) as it seems I was not clear - you wrote about a situation where partials are equal, I was asking when they are not equal but the integral is 0. –  alkamid Aug 24 '12 at 7:10
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I'm not entirely sure what you mean by (2), but I think Tunococ's answer is what you're looking for.

As to (1), consider $$P(x,y) = y$$ $$Q(x,y) = x^2$$ over the region $R = [0,1]\times [0,1]$.

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