Necessary condition for $\oint_C Pdx + Q dy = 0$ Suppose $P, Q$ have continuous partial derivatives in simply connected region $R$ enclosed by the curve $C$. Is $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ should necessarily hold in every point for $\oint_C Pdx + Q dy = 0$. There is a problem that claims that $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ for every point in the region. Wouldn't this integral $\displaystyle \iint_S \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)dxdy $ be equal to zero if the term $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ were symmetric in region?
For example:- if $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x$ and the surface is a rectangle $x = -a $ to $x = a$ and $y = 0$ to $y=b$

 A: Perhaps your trouble is in understanding the statement of the theorem.  I will state it as clearly as I can.

A necessary condition that $\oint_C Pdx+Qdy=0$ for all closed paths $C$ in a simply connected region $\mathcal{R}$ is that $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ at all points in $\mathcal{R}$.  In other words, if $\oint_C Pdx+Qdy=0$ for all closed paths $C$ in a simply connected region $\mathcal{R}$, then $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ at all points in $\mathcal{R}$.

I think you've interpreted the statement as "If $\oint_C Pdx+Qdy=0$ for some closed path $C$, then $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ at all points in the region $\mathcal{R}$ bounded by $C$."  The difference in these two statements is subtle, but the first one is true, while you've shown in your example that the second one is false.
The point is, as you've shown in your example, that we can find a simply connected region $\mathcal{R}$ and a specific closed curve $C$ in $\mathcal{R}$ such that $\oint_C Pdx+Qdy=0$, with $\frac{\partial Q}{\partial x}\ne\frac{\partial P}{\partial y}$ at all points in $\mathcal{R}$.  However, in all of these cases, it will be possible to find another closed path $C'$ in $\mathcal{R}$ such that $\oint_{C'} Pdx+Qdy\ne0$.
As shown in the proof, this follows from the continuity of the partial derivatives and Green's theorem.  Once we have some point in $\mathcal{R}$ where $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}>0$, then we must have some neighborhood where $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}>0$ (by continuity).  Integrating over this neighborhood gives a positive line integral (by Green's theorem), contradicting the hypotheses.
