Can someone define a line integral?

[winbacker]

I have just begun exploring the topic of line integrals for my Calculus 3 class. Although I can perform the calculation properly, I don't understand the physical significance of what I've just calculated.

For example, I know that when I calculate the integral of a function which is defined along x, the physical significance is that I have calculated the area under the curve.

What then, would be the equivalent, in plain English, of the result that is obtained when calculating the integral of a function along a curve? It is tempting to guess that one has obtained the area between the function and the curve, but I don't think so, because that would just be the difference of two regular integrals.

I have come across certain analogies of the curve C being the path of a particle being pushed by a variable force F which is the function to be integrated over C. However, I'm still having a bit of trouble visualizing all of this. Can anyone provide some insight or a good way to conceptualize what exactly is happening when one computes a line integral?

 

[yyat]

Hi,

Integrating the force F (vector field) along a path C computes the work W. See http://en.wikipedia.org/wiki/Mechanical_work#Mathematical_calculation".

When F has a potential, there is an analogue of the fundamental theorem of calculus: the http://en.wikipedia.org/wiki/Gradient_theorem" .

 

[slider142]

Here's a http://planetmath.org/encyclopedia/PathIntegral.html , if it helps.

 

[HallsofIvy]

Once again, a mathematical calculation does NOT have a "physical significance". Specific applications may have "physical significance".

The application to work that yyat gives is probably the most common.

 

[csprof2000]

So I guess if you wanted a dumbed-down version of what's going on in plain English, then while an integration with respect to x gives you the area under the curve in the x-y plane, the line integral gives you the area under the "sheet" which you would get if you "drew" the curve on the surface you're doing the line integral on and took everything perpendicularly between that and the z = 0 plane.

So, for instance, if I had the relatively tame surface z = 1, and I asked you to find the line integral of this with respect to some curve having length 10, then you can imagine a blanket twisted into some funny shape representing the curve, the blanket being 10 units longs and (in this case) all of it being uniformly 1 unit off the ground. The line integral gives you the area of the blanket... in this case, 10.

 

[n!kofeyn]

The book Complex Variables by George Polya has a good explanation of a physical interpretation of the line integral. It starts on p. 143 if you are able to find it in the library. It is a great book, especially in building up physical applications of the theory.

Points in the complex plane can be though of vectors. Let f(z):\mathbb{C}\to\mathbb{C} be a complex valued function. Letting z=x+iy, this is the same as the function f(x,y):\mathbb{R}^2\to\mathbb{R}^2, since the complex plane is equivalent to \mathds{R}^2. This function assigns to each point in its domain a vector, so this function is a vector field.

We can consider the function either as a force, or as a current density in a two-dimensional flow. If we interpret f as a force, then we think of a curve C as a path, which a particle can move along. If we interpret f as a current density, then we think of C as a boundary, which points can move across. In the former case, the line integral gives us the work done in transporting a particle along C, and in the latter case, the line integral gives us the amount of matter crossing the curve C, i.e. the flux.

Then informally
\int_C f(z) \,dz = \text{Work} + i \text{Flux}

So restricted to real values, the line integral gives you the work, as mentioned. This is basically his discussion, but Polya describes these derivations, including a discussion of the importance of the tangent and normal vectors to the curve C at a point.