Line Integral to Find Work on Slope (Without Explicit Use of Vector Calculus Format) Ok, so my math class has introduced the Line Integral as a way to find the work done on a two-dimensional slope by gravity on an object traversing any distance on the slope. This is all supposed to be useful for Maxwell's equations later, but I know nothing of those right now. (P.S. We use the original texts a lot of times, with some supplementary material.)
The equation we're given, (which is actually straight out of Maxwell's "Treatise On Vectors and Vector Operations") is this: $L = \int_0^S R \space cos \space\epsilon \space \mathrm{d}x$
S is the distance on the slope. R is here the force of gravity. Epsilon is the angle between the force of gravity and the tangent of the slope. Now I understand how to do this sort of thing with a straight line. It would just be the equation without the integral. $R \space cos \space\epsilon$ is just the projection of the gravity vector onto the slope, so its the "living" force that's acting in the direction down the slope. The rest of R is countered by the normal force. dx would be an actual non-infinitesimal distance. And so we'd have force x   distance which gives us work
I don't understand how this integral is working though. Why would we try to find the area under the graph of the work done if we're trying to find the work done? I also don't understand how the change of epsilon is determined in this equation. 
The book is here: http://books.google.com/books?hl=fr&lr=&id=WtExAQAAMAAJ&oi=fnd&pg=PR1&dq=The+operation+of+integration+of+the+resolved+part+of+a+vector+quantity&ots=jD6WFkOpwj&sig=GJWbq0KEb5iHrvSnvubKuMTWvMg#v=onepage&q=The%20operation%20of%20integration%20of%20the%20resolved%20part%20of%20a%20vector%20quantity&f=false
You can use the tab at the right to navigate to page 13, where this extract is from.
I'm sorry if this doesn't make much sense. But thanks for any help!
 A: Note that the integral in the quote has $s$ as parameter and integration variable, not $x$. Written in really long form this would be
$$\int_0^S R(\vec P(s))\cos\epsilon(s)\,ds$$
where $\vec P(s)=(x(s),y(s),z(s))$ and $R(\vec P(s))=\|\vec R(\vec P(s))\|$ and 
$$\cos\epsilon(s)=\frac{\langle \vec R(\vec P(s)), \dot{\vec P}(s)\rangle}{\|\vec R(\vec P(s))\|\,\|\dot{\vec P}(s)\|}$$
and since it is assumed that $s$ corresponds to the path length, $\|\dot{\vec P}(s)\|=1$.
Thus another form to write the line integral is
$$\int_\Gamma \langle \vec R(\vec P), d\vec P\rangle$$
So you see that there is no area directly involved, it is all more complicated. 
Think about the path $\Gamma$ consisting of many small straight segments $[\vec P_{k-1},\vec P_k]$, $k=1,...,N$, and that the force $\vec R$ is nearly constant along each segment. Then you can compute the work along each small segment without integration, and summing up all small segments gives an approximation of the integral formula,
$$\sum_{k=1}^N \langle \vec R(\vec P_k), \vec P_k-\vec P_{k-1}\rangle.$$
