Stokes Theorem tells us that when $S$ is a piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve $C$ with positive orientation and $\mathbf{F}$ is a vector field whose components have continuous partial derivatives on an open region in $\mathbb{R}^3$ containing $S$, then we have $$\int_C\mathbf{F }\cdot d\mathbf{r}=\iint_S \mathrm{curl }\ \mathbf{F}\cdot d\mathbf{S}.$$
The Divergence Theorem states that when $E$ is a simple solid region, $S$ is the boundary of $E$ with positive (outward) orientation, and $\mathbf{F}$ is a vector field whose component functions have continuous partial derivatives on an open region that contains $E$, then we have $$\iint_S\mathbf{F}\cdot d\mathbf{S}=\iiint_V\mathrm{div }\ \mathbf{F}dV .$$
Since $$\mathrm{div}\ \mathrm{curl}\ \mathbf{F}=0,$$ then why doesn't $\int_C\mathbf{F }\cdot d\mathbf{r}=\iint_S \mathrm{curl }\ \mathbf{F}\cdot d\mathbf{S}$ always vanish? For by the divergence theorem we have $$\int_C\mathbf{F }\cdot d\mathbf{r}=\iint_S \mathrm{curl }\ \mathbf{F}\cdot d\mathbf{S}=\iiint_V\mathrm{div }\ \mathrm{curl}\ \mathbf{F}dV =0.$$ But clearly this is not always true.
Where am I going wrong here?
- Is this true only when $S$ encloses another region $V$?
- Does $V$ need to be bounded to apply the divergence theorem? For some reason I feel like this is not the case, but I am not sure.
- Is there an obvious physical interpretation to the above result?
Many thanks for your consideration. This has stumped me for quite some time and I would be very grateful to anyone who could shed some insight.