Independence of a line integral from the path

Question

I'd like to understand why this statement is true:

If $C$ is a simple closed smooth curve in the plane $x+y+z=1$, then $$\int_C zdx-2xdy+3ydz$$ depends only on the orientation of $C$ and the area of the region enclosed by $C$ but not on the shape of $C$ or its location in the plane.

What is a criterion I can use to verify this? Am I supposed to apply Stokes' theorem? If so, how exactly?

Answer 1

We have

• $\vec F=(z,-2x,3y) \implies \nabla \times \vec F=3i+j-2k$

and

$$(\nabla \times \vec F)\cdot d\vec S=(3i+j-2k)\cdot\frac1{\sqrt3}(i+j+k) \,dS=\frac2{\sqrt3}\,dS$$

Answer 2

Yes, you use Stoke's Theorem. With Stoke's, this integral becomes $$\int_S <3,1,-2> \cdot dS$$ And as long as the curve is in the plane the integral reduces to 2 times the enclosed area.