# How to use Stoke's theorem to evaluate a line integral that's not in the form $\int_{C}\vec{F}\cdot d\vec{r}$

I'm trying to compute the line integral $$\int_{C}(y-z)\,dx+(z-x)\,dy+(x-y)\,dz$$ where the curve $C$ is the intersection of the cylinder $x^2+y^2=1$ and the plane $x−z=1$. I know how to do this directly by parametrising with $\vec{r}(t)=(\cos t, \sin t, -1+\cos t)$ for $t\in [0,2\pi]$ and then substituting in for $dx$, $dy$ and $dz$, although the solution I've seen says to look at $\int_{C}\vec{F}\cdot d\vec{r}$ but how would I do this and simlilarly how would I use Stoke's theorem if it's not written as $\int_{C}\vec{F}\cdot d\vec{r}$?

From the solution, it looks like they've chosen $\vec{F}=(y-z,z-x,x-y)$ which is obviously taken from the original integrand, but I don't see why this should be true? Wouldn't I have to look at the following integral, work out the right hand side and try to match things up? $$\int_{C}(y-z)\,dx+(z-x)\,dy+(x-y)\,dz=\int_{C}\vec{F}\cdot d\vec{r}$$

Edit: Never mind, I figured it out. I didn't realise $d\vec{r}=(dx,dy,dz)$ as I'd only seen the definition as $\int_{C}\vec{F}\cdot d\vec{r}=\int_{a}^{b}\vec{F}(\vec{r}(t))\cdot \vec{r}'(t)\,dt$. My bad!

The curve $C$ bounds an elliptical disc $E$. The normal $n$ to the plane of $E$ is $(-1,0,1)$, up to sign.
Since ${\rm curl}(F)=(-2,-2,-2)$ is constant Stokes' theorem gives $$\int_C F\cdot dr=\int_E{\rm curl}(F)\cdot n\>{\rm d}\omega=\pm(-2,-2,-2)\cdot(-1,0,1)\>{\rm area}(E)=0\ ,$$ so that we don't have to bother about the orientation.