The way I picture it is like this:
Think of a standard x-y-z space. The x-y plane is a sheet of all x-y values that we can use. So if we have a function f(x,y), then f(x,y) values are determined by combinations of values taken from the x-y plane. If we let f(x,y)=z then we can "see" various values of f(x,y) inside the x-y-z space by feeding the function various x-y values from the plane.
(I promise this is getting to the line integral explanation)
For example if f(x,y)=1 then, it does not manner what values from the x-y plane are fed to the function, it is always going equal 1. So we could see this graphically in the x-y-z space as a flat plane where z=1. Let's use this example to understand line integrals.
If you picture your standard x-y-z space, imagine the point (x,y,z) = (0,0,0). Let's use the x-y plane as our canvas to draw on (thus z=0). If you start to draw a circle from (x,y,z) = (0,0,0). As you draw this circle on the x-y plane you are touching various (x,y) values right? Let's imagine (to keep this non-mathematical) that as we draw the circle we can collect all the pairs of (x,y) points in a bucket. So after we draw the circle we have a "bucket of points". If we had an empty x-y-z space, and we took every (x,y) point out of the bucket and fed it to the graph then we would see a dot appear for each of these pairs, and after an amount of time a circle would appear.
This is exactly what is being done when we parameterize a curve:
We have the vector: (\cos t, \sin t, 0) Which we can also think of the notation as:
x = \cos t
y = \sin t
z = 0
Now as we feed this vector various t values of t (one parameter) we are actually getting back 3 values (x,y,z) which we can think of as a point in the x-y-z space.
So my point is this. This vector can be anything. It does not have to be a circle. It can be a very complicated path. But IT IS A PATH. Not a limit of integration, like x=0 to x=10.
So f(x,y) can be thought of (graphically) as points above or below the x-y plane (or on it if f(x,y) =0). Oh I never explained the arclength part. The \Delta s is the differential of the length of the curve. Just like dx is an infintisimal length of x \Delta s is an infintisimal amount of length of the PATH.
If we go back to our example where f(x,y) = 1 and imagine drawing a circle on the x-y plane. We are actually feeding a lot of (x,y) values into f(x,y) and each time we are getting 1. But we are tracing a path. Each (x,y) value corresponds to an area above the x-y plane. In the case that f(x,y) = 1 the height above the x-y plane is always 1 right? So if we also track the length of the path that we take then we get something useful. We have a height going from the x-y plane to 1, and then the length of the circle. So tracking the length of the path is important.
Lets look at an example:
\int_C f(x,y) ds
ds = \sqrt{\left( \frac{dx}{dy}\right)^2 + \left( \frac{dy}{dt} \right)^2} dt
Lets use the circle as our path:
Thus:
\vec r(t) = (\cos t , \sin t , 0)
So:
\int_0^{2\pi} \,1\,\,\sqrt{\frac{d}{dt}\left( \cos t\right)^2 + \frac{d}{dt}\left \sin t \right)^2} dt = \int_0^{2\pi} = \left[ t \right]_{t=0}^{t={2\pi}}=2\pi
And this is what? The circumference of a circle. So the function f(x,y)=1 in this line integral allows us to get the circumference of a circle. We can use "any" path though. NOT just a circle. A circle is nice because that whole arc length drops down to a 1 because of the nice: \sin^2 x + \cos^2 x = 1
