I just started learning Stokes' theorem and encountered the following problem:
Suppose that $\mathbf{F}$ satisfies the conditions of Stokes’ theorem and that $S$ is a simple closed surface. Show that
$\iint_S (\nabla \times \mathbf{F}) \cdot \hat{n} \ ds = 0$
Solution
Consider a simple closed curve $C$ on $S$. This curve cuts $S$ into pieces $S_1$ and $S_2$. Now
$\iint_S (\nabla \times \mathbf{F}) \cdot \hat{n} \ ds = \iint_{S_1} (\nabla \times \mathbf{F}) \cdot \hat{n} + \iint_{S_2} (\nabla \times \mathbf{F}) \cdot \hat{n}$
$= \oint_C \mathbf{F} \cdot d\mathbf{r} + \oint_{-C} \mathbf{F} \cdot d\mathbf{r}$
$= \oint_C \mathbf{F} \cdot d\mathbf{r} - \oint_{C} \mathbf{F} \cdot d\mathbf{r} = 0$
I understand the following theorem, which seems to form the basis for this problem:
If a vector field $\mathbf{F}$ is defined on all of $\mathbb{R}^3$ and its components have continuous first partial derivatives, then $\nabla \times \mathbf{F} = \mathbf{0} \implies \mathbf{F}$ is conservative.
I do not understand the reasoning behind this solution. Specifically, I do not understand the following:
$= \oint_C \mathbf{F} \cdot d\mathbf{r} + \oint_{-C} \mathbf{F} \cdot d\mathbf{r}$
What is meant by this? (A rough sketch would be very helpful). Why is it that these curves cancel out? I do not understand why they would have opposite orientation? I would greatly appreciate it if someone could show me why this is the true.
I would greatly appreciate it if people could please take the time to clarify this.