Understanding Line Integrals

  • #1

I have been trying to understand the meaning of line integrals, as opposed to the good ol' fashioned regular integral. Graphically, an integral like :

[tex] \int_{a}^{b} f(x) dx = \lim_{\|\triangle\|\rightarrow 0} \sum_{i} f(w
_i) \triangle x_i [/tex]

tells me that the area between the x-axis and the graph, from x=a to x=b is my integral. This was done by dividing that area into rectangles with width delta Xi and length f(Wi)

now how about the line integral :

[tex] \int_{C} f(x,y) ds = \lim_{\|\triangle\|\rightarrow 0} \sum_{i} f(u_i, v_i) \triangle s_i [/tex]

Now compared to my understanding of the definite integral, what does this definition exactly mean?

1. What is f(ui, vi) ?
2. "del si" is the arc length on our curve, but what does multiplying the arc length by f(ui, vi) give us?
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Answers and Replies

  • #2
1. ui and vi are the coordinates of consecutive points along the curve.
2. It's easy to see in 2 dimensions. Imagine you have the graph of a function from R2 to R, which is a 3 dimensional surface over the plane. Now imagine a curve that lies on the plane. Imagine that over each part of the curve on the plane, a surface extends straight up until it meets the graph of the function, so that something like a thin fence of variable height extends over all parts of the curve. The integral of the function over the curve is the area of that fence. The f(ui, vi) del si is the area of an infinitesimal rectangle over the curve with a base width of (del si) and a height of f(ui, vi)--an infinitesimal slice of the fence, like a picket in a picket fence.
  • #3
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 [itex] f(x,y) [/itex], then [itex] f(x,y) [/itex] values are determined by combinations of values taken from the x-y plane. If we let [itex] f(x,y)=z [/itex] then we can "see" various values of [itex] f(x,y) [/itex] 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 [itex] f(x,y)=1 [/itex] 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 [itex] z=1 [/itex]. Let's use this example to understand line integrals.

If you picture your standard x-y-z space, imagine the point [itex] (x,y,z) = (0,0,0) [/itex]. Let's use the x-y plane as our canvas to draw on (thus z=0). If you start to draw a circle from [itex] (x,y,z) = (0,0,0) [/itex]. 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: [itex] (\cos t, \sin t, 0) [/itex] Which we can also think of the notation as:

[tex] x = \cos t [/tex]
[tex] y = \sin t [/tex]
[tex] z = 0 [/tex]

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 [itex] f(x,y) [/itex] 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 [itex] \Delta s [/itex] is the differential of the length of the curve. Just like [itex] dx [/itex] is an infintisimal length of x [itex] \Delta s [/itex] is an infintisimal amount of length of the PATH.

If we go back to our example where [itex] f(x,y) = 1 [/itex] 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:

[tex] \int_C f(x,y) ds [/tex]
[tex] ds = \sqrt{\left( \frac{dx}{dy}\right)^2 + \left( \frac{dy}{dt} \right)^2} dt[/tex]

Lets use the circle as our path:
[tex] \vec r(t) = (\cos t , \sin t , 0) [/tex]

[tex] \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[/tex]

And this is what? The circumference of a circle. So the function [itex] f(x,y)=1[/itex] 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: [itex] \sin^2 x + \cos^2 x = 1 [/itex]
  • #4
http://www2.scc-fl.edu/lvosbury/CalculusIII_Folder/LineIntFiles/LineInt2.gif [Broken] is an example of a curve in the plane.

http://www2.scc-fl.edu/lvosbury/CalculusIII_Folder/LineIntFiles/LineInt3.gif [Broken] is a visualization of the line integral over that curve for a certain function. The value of the integral is the area of the "fence."
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  • #5
Slight Correction: One side of the "fence."
  • #6
thank you all very much this was indeed helpful

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