Deriving the magnetic flux from the work done by the Lorentz force, using only infinitesimals The question might sound a bit strange at first so allow me to give some context. My prof has given a derivation of the magnetic flux where he used both $\Delta$ and $d$ for elementary line and surface elements. I am confused by the simultaneous usage of both infinite and finite elements, and their meaning compared to each other. In my own version of the proof I wanted to use only infinitesimals. I will post the derivation according to my prof first, afterwards I'll give my own version (which was intended to be completely analogous), where I tried to use only infinitesimals.
The general idea is to start from the work done by the Lorentz force, define the magnetic flux and link it back to the emf.
My prof's derivation:
$$ \Delta W = \bar{F}.\Delta\bar{l} = q(\bar{v} \times \bar{B}).\Delta \bar{l} = q(\Delta \bar{l} \times \bar{v}).\bar{B} \quad \text{(permutation rule)} \\
\bar{v} = \frac{d\bar{s}}{dt} \Rightarrow \Delta W = \frac{q}{dt} (\Delta \bar{l}  \times d\bar{s}).\bar{B} \\
\Delta \bar{l} \times d\bar{s} = d\Delta \bar{S} \Rightarrow q \frac{d}{dt}(\Delta \bar{S}.\bar{B}) \\
d\phi := \Delta \bar{S}.\bar{B} \Rightarrow \Delta W = q \frac{d\Delta\phi}{dt} \\
W = \varepsilon = -\frac{d\phi}{dt} \\ $$
The last part somehow follows from integration, the - sign will be explained later. I understand most parts, save for the mixed usage of $\Delta$ and $d$, and the last integration of $\Delta \phi$ (?). Maybe it's something stupid that I'm missing but this derivation confuses me.
My version:
$$
dW = \bar{F}.d\bar{l} = q(\bar{v} \times \bar{B}).d \bar{l} = q(d \bar{l} \times \bar{v}).\bar{B} \\
\bar{v} = \frac{d\bar{s}}{dt} \Rightarrow d W = \frac{q}{dt} (d \bar{l}  \times d\bar{s}).\bar{B} \\
d \bar{l} \times d\bar{s} = d\bar{S} ^{(*)} \Rightarrow  \frac{q}{dt}(d \bar{S}.\bar{B}) \\
d\phi := d\bar{S}.\bar{B} \Rightarrow dW = q \frac{d\phi}{dt} \\
\text{Error}
$$
That's how far I can get. I can't somehow integrate $dW$ and keep the $\frac{d\phi}{dt}$. The source of this problem seems to come from $(*)$. My prof somehow summons an extra d, where in my version it is missing and I don't know what a mathematical correct way would be of fixing this.
I wasted a few hours on trying to find a solution but to no avail. I don't immediately see any mistake on my part. I haven't had any differential topology yet, so it could very well be that my understanding of differentials etc. is wrong. Any suggestions or tips on whether how to fix it or if my version is even correct are welcome.
I know of different derivations that use Faraday's law and Maxwell's 3rd equation, but I want to understand this one first.

Edit The velocity in the Lorentz force is the velocity of the conductor. I thought of the charges as being quasi-stationary in the conductor so the only velocity they have is the conductor's velocity. (If that makes any sense)
Edit 2 I should have done this sooner but here is a short explanation of the used terms: $\Delta W$ is the difference in work and the finite version of $dW$. $\Delta \bar{l}$ is the difference between two lengths of the conducting wire and the finite version of $d\bar{l}$. $d\bar{s}$ is an infinitesimal displacement of the conducting wire due to the Lorentz force. $\bar{B}$ is the magnetic induction vector, perpendicular on $\bar{l}$ and $\bar{s}$. $\Delta \phi$ is the finite difference in the magnetic flux and $d\phi$ is the infinitesimal difference. $d\bar{S}$ is the an infinitesimal surface vector, $\Delta \bar{S}$ would be the finite limit of it. No idea what is meant by $d\Delta \bar{S}$, infinitesimal difference in of a finite difference of the surface? $\varepsilon$ is the emf. q is a charge. I hope this clears things up a bit.

I was not sure whether to post the question at physics or at math. My apologies if I made the wrong choice.
 A: Here is one way I can make sense of it all. The following schematic describes the system I am going to work with:

There is a wire $\newcommand{\wire}{\mathcal{L}}\wire$ translated by an infinitesimal amount $\renewcommand{\vec}[1]{\boldsymbol{#1}}d\vec{L}$ (I will denote vectors with bold letters). The infinitesimal element of length of the wire is $d\vec{l}$. The dashed lines are fixed wires to close the circuit, through which flows an intensity $I=\lambda v$ where $\lambda$ is the density of moving charges per unit of length and $v$ is the speed of those charges. The infinitesimal work of the Lorentz force on the whole wire $\wire$ is then
$$dW = \int_\wire \lambda dl (\vec{v}\times \vec{B})\cdot d\vec{L}.$$
But $dl\,\vec{v}=v\,d\vec{l}$ as the charges are constrained to flow along the wire. Then using the formula for the current $I$,
$$dW = \int_\wire I (d\vec{l}\times \vec{B})\cdot d\vec{L}=\int_\wire I \vec{B}\cdot(d\vec{L}\times d\vec{l}).$$
But then $I$ is uniform along the wire, so we can take it out. Furthermore,  we recognise in $d\vec{L}\times d\vec{l}$ the infinitesimal oriented area $d\vec{S}$ of the surface $\newcommand{\swiped}{\mathcal{S}}\delta\swiped$ delineated by the two positions of the wire and by the fixed wires represented by dashed lines. So
$$dW = I\iint_{\delta\swiped} \vec{B}\cdot d\vec{S}.$$
We recognise the flux $d\Phi$ of $\vec{B}$ through $\delta\swiped$. It is an infinitesimal flux because $\delta\swiped$ is infinitesimal in one direction. Thus we finally get
$$dW = I\,d\Phi.$$
But then we should also have 
$$\frac{dW}{dt} = \mathcal{E}I$$ 
where $\mathcal{E}$ is the e.m.f., and therefore 
$$\mathcal{E}=\frac{d\Phi}{dt}.$$
