Stokes and Gauss' Divergence theorem on a closed smooth surface in $\Bbb R^3$

Solution 1:

While Daniel Fischer's answer probably is more direct, I find this way of showing that the circulation over a closed surface is zero (using Stokes theorem) more intuitive:

Here I split the closed surface $S$ into two surfaces $S_1$ and $S_2$ with a shared boundary, the curve $C_1$. If we apply Stokes theorem to the two surfaces separately we get:

$$\int{\int_{S_1} (\nabla \times \bar F) d \bar S} = \int_{C_1} {\bar F \cdot d \bar r}$$

and

$$\int{\int_{S_2} (\nabla \times \bar F) d \bar S} = - \int_{C_1} {\bar F \cdot d \bar r}$$

notice the negative sign before the right hand integral in the second equation. This is because the curve $C_1$ runs in the opposite direction to the normals of $S_2$ compared to $S_1$.

Combing the two surface integrals to get the integral over the entire surface, $S = S_1 \cup S_2$:

$$ \begin{eqnarray} \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int{\int_{S_1} (\nabla \times \bar F) d \bar S} + \int{\int_{S_2} (\nabla \times \bar F) d \bar S} \\ \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int_{C_1} {\bar F \cdot d \bar r} - \int_{C_1} {\bar F \cdot d \bar r} \\ &=& 0 \end{eqnarray} $$