A question about path-independent integral Prove that for a domain $\Omega $ in ${R^2}$ and a differential form $\omega  = P\;dx + Q\;dy$, if $\oint_\gamma  \omega   = 0$ for every smooth closed curve in $\Omega $, then the path-integral of $\omega $ is path-independent, in the sense that if $\gamma _1$ and $\gamma_2$ are two piecewise-smooth curves from $p$ to $q$, then $\oint_{\gamma _1} \omega= \oint_{\gamma _2} \omega $.
 A: Consider the curve $\gamma:= \gamma_1 - \gamma_2$. This is a closed curve by definition. The result follows easily from your hypotesis $\int_\gamma\omega = 0$.
In other words, you're saying that $\omega$ is exact.
A: I think the following answer can be applied to smooth manifold, but for the time we restrict ourselves to $\mathbb{R}^2$.
First, we prove that $\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}$ in $\Omega$. For any $p=(x,y)$ in the $\Omega$, we form a circle $\gamma$ around $p$, then by the Green formula and condition, we have $\int\limits_C  {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} dxdy = 0$ where $C$ is the ball formed by the circle $\gamma$. Then when we shrink $\gamma$, we have $\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}$ at $p$.
To prove the conclusion, we have only to prove for arbitray piecewise smooth closed curve $\gamma$, $\int\limits_\gamma  \omega   = 0$. For a sigular point $a$ in $\gamma$, first we find a small ball $B$ in $\Omega$ of $a$ and break the curve into $\gamma_1$ and $\gamma_2$, and we can find a smooth curve $\gamma_3$ linking $\gamma_1$ and $\gamma_2$ and differing from them just in $B$(like here). If there are $n$ singular points in $\gamma$, then we have $n$ small balls around them named $B_1, B_2,...B_n$, and $n$ piecewise smooth closed curves $c_i$ each in one $B_i$. So we have to prove that in $B_i$, $c_i$, $\int\limits_{c_i}  \omega   = 0$. Since  $\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}$, we can consruct a function $u(x,y)$ such that ${u_x} = P$ and ${u_y} = Q$. That is, $\omega$ has a primitive in $B_i$, then $\int\limits_{c_i}  \omega   = 0$. Then we conclude the proof.
