Given $\mathbf{F}(x,y,z) = (y-z,z-x,x-y)$, calculate the line integral $\int_C\mathbf{F}\cdot d\mathbf{l}$ where $C$ is any curve on the plane $x-z=1$.
My initial instincts to tackling this problem made me apply Stoke's theorem - calculating the curl of $\mathbf{F}$ we arrive at $-2\cdot(1,1,1)$ (ie curl is constant throughout all space and also on our plane $x-z=1$) which is most definitely nonzero and so the line integral will not be 0 either.
I then realised also that it is not specified that $C$ is a closed curve - so correct me if I'm wrong but it seems there is no way to solve this in the general case whatsoever?
Seems like a strange question I have been asked, since if it intends you to pick a curve yourself, and then integrate, you could simply pick an infinitesimal one and consequently call the whole thing 0.
Yes the question should have been more clear.
We can apply Stokes' theorem if it is a closed curve. In case of a closed curve, we have $\displaystyle \int_C \vec F \cdot dr = \iint_S (\nabla \times \vec F) \cdot \hat n \ dS \ $
The point to note is that the curve is in plane $x-z=1$ and unit normal vector to the plane is $\hat n = \frac{1}{\sqrt2} (1, 0, -1)$
So $(\nabla \times \vec F) \cdot \hat n = \frac{1}{\sqrt2}(-2, -2, -2) \cdot (1, 0, -1) = 0$
So for any closed curve in plane $x - z = 1$, the line integral will be zero.
If it is not a closed curve then yes we will have to know the curve and evaluate line integral.