I was trying to solve the problem to find $\int_{C}F\cdot dr$ where $C$ is the curve of intersection of cone $z^2=x^2+y^2$ and plane $z=1$. Also $F(x,y,z)=y\vec{i}+z\vec{j}+x\vec{k}$
So here we have two possible surfaces $S_1$ and $S_2$ with same boundary $C$ oriented positively.
I tried verifying for above $F$ which is divergence free and successfully showed that $$\iint_{S1}\text{curl}(F) \cdot dS_1=\iint_{S_2}\text{curl}(F)\cdot dS_2$$
But in this link:
https://mathinsight.org/stokes_theorem_examples
they have taken $F=\left(\sin x-\frac{y^3}{3}, \cos y+\frac{x^3}{3}, xyz\right)$ and its been told that
$$\iint_{S_1}\text{curl}(F)\cdot dS_1=\iint_{S_2}\text{curl}(F)\cdot dS_2$$
any clarification?
Best Answer
The field $F$ in your first example is "by coincidence" divergence free. This has no avail for the line integral $\int_C F\cdot dr$.
By Stokes' theorem this line integral is equal to the surface integral $\int_S{\rm curl}(F)\cdot dS$ for any surface $S$ with $\partial S=C$, in particular for the two surfaces $S_1$, $S_2$ in your figures: $$\int_{\partial S} F\cdot dr=\int_S{\rm curl}(F)\cdot dS\ .$$ There is no divergence appearing here.
On the other hand the curl of any field $F$ is divergence free: $${\rm div}\,{\rm curl}(F)\equiv0\ .$$ This plays a rôle in your second example: The two surfaces $S_1$, $S_2$ together bound a three-dimensional cone $K$, whereby $\partial K=S_1-S_2$. By Gauss' theorem we therefore have $$\int_{S_1}{\rm curl}(F)\cdot dS-\int_{S_2}{\rm curl}(F)\cdot dS=\int_{\partial K}{\rm curl}(F)\cdot dS=\int_K {\rm div}\,{\rm curl}(F)\>dV=0\ .$$