Green's Theorem substitute?

Question

This is a more general question I was doing this question. $F$ is a vector field from r2 to r2 and is irrotational, and $D$ is the simple closed curve in the $xy$ plane now the question is asking to show that the $integral[F]$ over the boundary of D is zero. Now this is obvious by Greens theorem, but is there any other way of showing that it is zero without using the theorem and without knowing what F is?

Answer 1

Show that $F$ is the gradient of some function using it is irrotational. If $F=\nabla E$ for $E:\mathbb R^2 \to \mathbb R$ you obtain

$$\int_\gamma Fds =\int_a^b \langle F(\sigma), \sigma'\rangle = \int_a^b \langle \nabla E(\sigma) ,\sigma'\rangle = \int_a^b (E\circ \sigma)' = E(P)-E(P)$$

if $\sigma(a)=\sigma(b)=P$, and you needn't assume $\gamma$ is simple (only closed). Here $\sigma :[a,b] \to\mathbb R^2$ is some parametrization of your curve.

Add Perhaps I could have added how to find $E$. If $(F_1,F_2)=F$, then write

$$E(x,y) = \int_0^x F_1(t,0)dt + \int_0^y F_2(x,s)ds$$

Note that $\dfrac{\partial E}{\partial y}(x,y)=F_2(x,y)$ by construction, and $\dfrac{\partial E(x,0)}{\partial x} = F_1(x,0)$. Moreover

\begin{align*} \frac{\partial}{\partial y}\left( \frac{\partial E(x,y)}{\partial x} -F_1(x,y)\right) &= \frac{\partial}{\partial x} \frac{\partial E(x,y)}{\partial y} -\dfrac{\partial F_1}{\partial y}(x,y) \\ &=\frac{\partial F_2}{\partial x}(x,y) -\dfrac{\partial F_1}{\partial y}(x,y) \\ &=0\end{align*}

for any $y$. Because this is true at $y=0$, we obtain that $\nabla E= F$, as we wanted.