There is no such thing as magnetism? Here's an interesting "proof" that there is no such thing as magnetism. I know the answer but I love this so much I had to ask it here. It's a great way to confuse people!
As we all know, $$\nabla \cdot\vec{B} =0$$
Using the divergence theorem, we find $$ \iint_S \vec{B} \cdot \hat{n} \, dS = \iiint_V \nabla \cdot \vec{B} \, dV = 0$$
Since $\vec{B}$ has zero divergence, there exist a vector function $\vec{A}$ such that $$\vec{B} = \nabla \times \vec{A}$$
Combining the last two equations, we get $$\iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$
Applying Stokes' theorem, we find $$\oint_C \vec{A} \cdot \hat{t} \, ds = \iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$
Therefore, $\vec{A}$ is path independent and we can write $\vec{A} = \nabla \psi$ for some scalar function $\psi$.
Since the curl of the gradient of a function is zero, we arrive at:
$$\vec{B} = \nabla \times \nabla \psi = 0,$$
which means that all magnetic fields are zero, but that can't be!
Can you see where we went wrong?
 A: Adding a closed boundary $C$ around the volume generally splits $S$ into two surfaces $S_1$ and $S_2$ for which Stoke's theorem holds:
\begin{align*}
\oint_C \vec{A} \cdot \hat{t} \, ds &= \iint_{S_1} \hat{n} \cdot \nabla \times \vec{A} \, dS\tag{1}\\ 
\oint_C \vec{A} \cdot \hat{t} \, ds &= \iint_{S_2} \hat{n} \cdot \nabla \times \vec{A} \, dS\tag{2}
\end{align*}
I can use these two ways to show where your argument is wrong:



*

*The left hand sides of (1) and (2) must be the negative of one another because the right hand sides sum to zero, and therefore
$$\oint_C (\vec{A} - \vec{A}) \cdot \hat{t} \, ds = \iint_{S_1} \hat{n} \cdot \nabla \times \vec{A} \, dS + \iint_{S_2} \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$
So you should have used $\vec{A}-\vec{A}$ in place of $\vec{A}$ in your application of Stokes theorem and concluding arguments, to give the trivial result $\vec B - \vec B = 0$

*Let $S_1\rightarrow S,\,S_2\rightarrow 0$ as $C\rightarrow 0$.
Then $(1)$ reduces to your application of Stokes theorem which reduces to $\vec{A}\cdot\vec 0 = 0$, making your following arguments irrelevant.

A: Note that $\partial V=S$, so that 
$$\tag{1} C~=~\partial S~=~\partial^2V~=~\emptyset$$ 
is the empty set. (Topologically, the boundary of a boundary is empty, or equivalently, the boundary operator $\partial^2=0$ squares to zero.) On the other hand, the circulation 
$$\tag{2} \Gamma~=~\oint_{C=\emptyset}\vec{A}\cdot d\vec{r}~=~0$$ 
along the empty curve $C=\emptyset$ vanishes identically for any vector field $\vec{A}$. In particular, one can not conclude from (2) that the magnetic potential $\vec{A}$ should be a gradient field.
A: Here's something from wikipedia:
"This classical Kelvin–Stokes theorem relates the surface integral of the curl of a vector field F over a surface Σ in Euclidean three-space to the line integral of the vector field over its boundary ∂Σ."
The catch in the statement of Stokes theorem is "boundary".So if V is a 3D volume,then it's boundary will be it's 2D surface which will be closed as V is a 3D volume.Now here's the catch,the boundary of a closed surface will not exist or will be null.(This has been pointed out mathematically by Qmechanic as $C=\partial S=\partial^2 V=\phi$).
So in extremely simplified terms it's like concluding $a=0$ from $a \times 0=0$.
