In a notable answer to [this question](https://physics.stackexchange.com/questions/31672/what-causes-a-force-field-to-be-nonconservative), Qmechanic formulates conditions for "conservative" velocity-dependent forces (e.g. the Lorentz force, but not velocity-proportional friction) that are analogous to those for traditional velocity-independent conservative forces. 

To wit, in a simply-connected domain (for the velocity-independent case, anyway), two sets of three equivalent conditions for a force to be conservative are presented: 

$$ \begin{array} {cccc}
\text{ } & \text{velocity-independent force } \boldsymbol{F}( \boldsymbol{r}(t)) & | 
 & \text{velocity-dependent force } \boldsymbol{F}(\boldsymbol{r}(t),\boldsymbol{\dot{r}}(t)) \\
1) & F\_i = - \frac{\partial U}{\partial x^i} & | & 
 F\_i = -\frac{\partial U}{\partial x^i} + \frac{d}{dt} \left( \frac{\partial U}{\partial \dot{x}\_i} \right) \\
2) & \boldsymbol{\nabla \times F} = 0 & | & 
 \frac{\delta F\_i(t)}{\delta x\_j(t')} - \frac{\delta F\_j(t')}{\delta x\_i(t)} = 0 \\
3) & \oint\_{S^1} dt \, \boldsymbol{F}(\boldsymbol{r}(t)) \boldsymbol{\cdot \, \dot{r}} (t) = 0 & | & 
 \oint\_{S^2} dt \wedge ds \, \boldsymbol{F} ( \boldsymbol{r}(t,s), 
 \boldsymbol{\dot{r}}(t,s)) \boldsymbol{\cdot \, r'}(t,s) = 0
\end {array}
$$
where $\delta$ denotes a [functional derivative](http://en.wikipedia.org/wiki/Functional-derivative), the final integral is over any "two-cycle $r: S^2 \rightarrow \mathbb{R}^3 \,$", and "a dot and a prime mean differentiation wrt. $t$ and $s$, respectively".
I have changed the formulation somewhat; I hope I didn't introduce errors.

I get the maths for the velocity-independent force conditions, but, for the velocity-dependent case, I am a bit puzzled by the functional derivatives and totally baffled by the two-cycle integral. 

My question: 

* What is this "two-cycle integral", which looks like no surface integral I've ever seen, and how is it evaluated? (and how did $\boldsymbol{r}$ acquire two arguments?). 

> 
>  - How is this functional derivative evaluated?
>  - Why are the functional derivative and two-sided integral equivalent to each other and to the potential formula for the force? 
> 
> 
> 

I suspect this is a rather large subject; references would be appreciated.