Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $.

share|cite|improve this question
For usual definition of exactness of integral it often requires for every piecewise smooth closed curve the integral to be 0. But the weaker hypothesis can be proved to be equivalent as said in some book but without proof. – Hezudao Oct 11 '11 at 10:31

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.

share|cite|improve this answer
The hypothesis only holds for smooth closed curves, while $\gamma_1-\gamma_2$ need only be piecewise smooth. One needs to extend the hypothesis to a larger class of curves. – anon Oct 10 '11 at 17:51
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.