What exactly is a vector line integral? I am having trouble wrapping my head around the idea and understanding what is actually happening, and I am hoping for a more intuitive explanation, or at least a better understanding of what is happening.   I know that it is supposed to be the work done by the vector field along a path, but that doesn't really help me. I had trouble with scalar line integrals too, but I think I finally have it. A scalar line integral is just like a regular integral that they teach to first year Calc students, except the function to be integrated doesn't necessarily have just one variable, and the region being integrated is not restricted to the x axis, but is a one dimensional path on the xy-plane (or whatever, depending on how many variables). Is that right? What is actually going on in a vector line integral?
 A: You should know that for a constant force and velocity, the work done by that force on an object moving in a straight path is $W = Fs$. But sometimes the force isn't constant throughout, but changes with respect to position, we have to integrate
$$ W = \int_{x_1}^{x_2} F(x) \, dx$$
Where $F$ is positive or negative depending on it's direction with respect to the $x$ axis. Since force is a vetor, we can write it like this.
$$ W = \int_{x_1}^{x_2} \textbf{F}(x) \cdot d\textbf{x} $$
You can think of $d\textbf{x}$ is an infinitesimal change in displacement as the object moves along its path. Basically we have the work is the sum of every dot product of $\textbf{F}$ and $d\textbf{x}$ at each position $x$, as $x$ goes from $x_1$ to $x_2$
In 2 dimensions this is basically the same, but force is dependent on 2 coordinates instead of 2, so it's a multi-variable function $\textbf{F}(x,y)$. $\textbf{F}$ is usually called a vector field, but you can think of it as a function that returns a different vector at each respective point on the $xy$-plane 
The integral takes a new form
$$ W = \int_{(x_1,y_1)}^{(x_2,y_2)} \textbf{F}(x,y) \cdot d\textbf{s} $$
This is the work done as the object is moving from $(x_1,y_1)$ to $(x_2,y_2)$. Here, $d\textbf{s}$ is still a displacement vector, but it signifies a change in direction with respect to both axes. However, in 2-dimensions there can be infinitely many way to travel from one point to another, so we have to specify a path. Thus, the integral is usually written as
$$ W = \int_{\gamma} \textbf{F}(x,y) \cdot d\textbf{s} $$
where $\gamma$ is the curve the object is traveling on. Since $d\textbf{s}$ is a change in displacement, it is a vector tangent to the curve at each point.
A: Ok so in one dimension you have a function $f: \mathbb{R} \rightarrow \mathbb{R}$ and the integral of this is intuitively the area under the curve. But here is the catch, it matters whether you are travelling left or right along the $x$-axis.
Usually we think of "walking right along the $x$-axis" and roughly consider positive values of $f$ to generate positive area.
Now for scalar valued functions in $2$ dimensions we have $f: \mathbb{R}^2 \rightarrow\mathbb{R}$. This time there are infinitely many possible directions to "walk" in. We can travel along any curve in the plane and each will generate different areas in general. This is what the scalar line integral is really telling you.
(the first picture here is what you should be seeing: https://www.google.co.uk/search?q=scalar+line+integral&biw=1027&bih=673&source=lnms&tbm=isch&sa=X&ei=nF9aVNf4M7Ku7AaEr4DwBw&sqi=2&ved=0CAcQ_AUoAg)
It becomes tough to visualise for higher dimensions but the same intuition works. 
As for the vector line integral in $2$ dimensions you now have a vector valued function $f: \mathbb{R}^2 \rightarrow\mathbb{R}^2$. As usual one can take any path in the plane but now you get a vector output. You can visualise the line integral as some sort of work done in travelling down that path (with respect to the "force" as defined by the function).
Again this works for higher dimensions but is hard to visualise (it is hard enough for $2$ dimensions.
