I need to prove that $\int_V\nabla\times\mathbf{A}\text{d}V=\int_S\text{d}\mathbf{S}\times A$ where $\mathbf{A}=\mathbf{A}(x)$ is a vector field. I've gotten to the point where I just need to show that $\int_SA\times\text{d}\mathbf{S}=-\int_S\text{d}\mathbf{S}\times A$.
I assume this follows from the fact that the vector product is antisymmetric but I'm not sure why this is a valid expression.
Like you wouldn't say $\int f(x)\text{d}x=\int\text{d}xf(x)$, but I'm sure I'm just misunderstanding something so an explanation would be much appreciated.
$\text{d}\mathbf{S}$ is a vector. You can write it as a scalar times a unit vector in that particular direction: $$\text{d}\mathbf{S}=\text{d}S\hat s$$ Now, in your integral, moving around $\text{d}S$ does not change anything. But the cross product is antisymmetric: $$\hat s\times A=-A\times\hat s$$