Let $\mathbf{F}(x,y)=\langle 3y,x-1 \rangle$. Find an oriented path $C$ from $(2,2)$ to $(1,1)$ such that $\int_C \mathbf{F}\cdot d\mathbf{r}=4$
Is there a general way to do this beyond trial and error? I can't see one.
I also know that $\int_{C}\mathbf{F}\cdot d\mathbf{r}=4$ where $C$ is the line segment starting at $(1,1)$ and ending at $(2,2)$.
Hint: Suppose you make two different guesses for $C$. What does Green's theorem tell you about how different their line integrals will be? (See the spoiler block below for the full answer.)
To start with, let's make some convenient guesses to get a handle on what kind of answers we get. Each path is piecewise-linear, and as the computations are obvious I'll just state the results:
\begin{align} C_1:&(1,1)\to (1,2)\to (2,2)& \implies\int_{C_1}\mathbf{F}\cdot d\mathbf{r}=6\\ C_2:&(1,1)\to (2,2)&\implies \int_{C_2}\mathbf{F}\cdot d\mathbf{r}=5\\ C_3:&(1,1)\to (2,1 )\to (2,2)&\implies \int_{C_3}\mathbf{F}\cdot d\mathbf{r}=4\end{align}
So in this case it's easy to stumble upon a contour that'll give the right answer. But let's suppose for a moment we didn't bother to do this guess; could we have predicted that it would give the correct integral? To do so, we first recall why different paths give different results. Note that the curl of $\mathbf{F}$ is given as $\nabla\times\mathbf{F}=-2\hat{z}$; consequently $\mathbf{F}$ is not the gradient of a scalar function, and so we do not have path-independence.
The difference in paths, then, is just an area integral over by $(\nabla\times \mathbf{F})_z$. But we saw above that this was just $-2$, so the difference in paths is just $-2 A(D)$. This was in fact visible in the calculations earlier: Both $C_2-C_1$ and $C_3-C_2$ enclose oriented areas of $+1/2$, and so the line integrals differ by $1$.
Thus we could have deduced $C_1$ would work from calculating either $C_2$ or $C_3$. And we can in fact do better: If we pick any curve $C$ from $(1,1)$ to $(2,2)$ such that $C-C_1$ has zero oriented area (i.e. same area enclosed clockwise as anti-clockwise) then we are guaranteed to have $\int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{C_3} \mathbf{F}\cdot d\mathbf{r}=4.$ So there are infinitely many curves that give the desired answer, with $C_3$ just being the most convenient.
