The reasoning leading to Green's theorem in my course makes a step I don't understand, with no justification.

We have a function $P:R\to \mathbb{R}$ that has a partial derivative with respect to $y$ over $R$. We're computing $\displaystyle \iint \limits _R\dfrac{\partial P}{\partial y}(x,y)\,dA$. Say $R$ is regular with respect to the $x$-axis. We can easily compute that it's equal to $\displaystyle \int \limits _a^bP(x,f_2(x))\,dx-\int \limits _a^bP(x,f_1(x))\,dx$ for $a\leq x\leq b$. These are line integrals along the curves $y=f_1(x)$ and $y=f_2(x)$. We'll call them respectively $C_1$ and $C_2$.
So our original double integral becomes $\displaystyle \int \limits _{C_2}P(x,y)\,dx-\int \limits _{C_1}P(x,y)\,dx$.

The next step is the one I don't understand: my course states that this is equal to $\displaystyle -\oint \limits _{C^+}P(x,y)\,dx$.



1 Answer 1


For Green to hold, the boundary of $R$ must be positively oriented, and if you draw a picture, that would mean the lower half $C_1$ is positively oriented, as it's from smaller $x$ to bigger ones, while the upper half $C_2$ is not. Therefore $\oint_{C^+}=\int_{C_1} - \int_{C_2}$.

  • 1
    $\begingroup$ OMG thank you, I stumbled on this for so long and now it seems so obvious... $\endgroup$ Commented May 29, 2022 at 15:05

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