Line integral

created by Clobbersaurus
(idea) by 10998521 (2.2 mon) (print)   (I like it!) 2 C!s Mon May 20 2002 at 16:56:17

Definition:

The line integral (from a to b) of a vector field F(x) along a curve C(t) is defined as:

C F.ds = ab F(C(t)).C'(t)dt

Those of you reading this are now split into three groups. One, the applied mathematicians, is thinking "Well, duh!". Another, the pure mathematicians, is ripping holes in what I said and arguing over whether any parametrisation of a curve will give the same answer (it does). The third group, the non-mathematicians, is scratching its collective head and wondering how the hell it got here from nodes about lesbian ninja monkeys. What follows is for the non-mathematicians:

What the hell is that curly S thing?

That, my friend, is an integral sign. It's one of those clever little ideas that separates adding up numbers from Real Mathematics. In its most basic form, it represents the area under a curve. Look at the purty picture:

f(x)
 ^
 |
 |            _
 |         __/ \_
 |        /      \
 |     __/        \___
 |    /  |           |\_ 
 |  _/   |     X     |  \____  /
 | /     |           |       \/
 |/      |           |
0+-------+-----------+---------->x
 0       a           b

Apologies for the bad ASCII art, but that's supposed to be a graph of a function f(x). See that section with the big X in it? The area of that section is represented by ab f(x) dx (pronounced: "the integral from a to b of f of x with respect to x" or "integral from a to b fof x d x"). There are ways of calculating these things, given a, b and f, but most mathematicians (and physicists and engineers) just look the things up in a table.

What the hell do I want to do that for?

Oh, lots of things. You see, lots of things in nature (and in mathematics) turn out to be related to each other by integrals. For example, if you're looking at a function v(t) which gives your speed at time t, then your distance travelled between time a and time b is ab v(t) dt. If the current flowing through a wire at time t is I(t) then the total amount of charge transferred between time a and time b is ab I(t) dt. Of course, there are many others, but I think you get the idea by now.

OK, so what's special about line integrals?

Well, ordinary integrals are fine when your function just takes in numbers and spits out numbers. Unfortunately, most things in nature don't work like that. Instead of scalar functions, we have vector functions. These are functions that take a position and give out a direction and a magnitude. For example, the wind can be viewed as a function which takes a point in the atmosphere and gives back a wind speed and direction. Now, with a little fiddling about, one can make the standard integral work for functions which take in numbers and give out vectors (direction/magnitude pairs). However, to deal with functions which take in vectors, we need to sit down and think for a bit. When we're finished, we've got three ideas (or more, but only three of them are much use). These are line integrals, surface integrals and volume integrals.

Yes, but what is it?

The idea with a line integral is that we take a curved path, straighten it out, and integrate as though this was our x-axis. Imagine taking a piece of string, and laying it through your space. With the string still in position, measure the value of the function at each point on the string. Then straighten the string out, and create a new function, based on the values on the string. This new function can be integrated, and we refer to this as the line integral.

Alright, I think I understand that. Why do I want to do this?

Lots of reasons. For example, the energy required to move along a given path is the integral of the force opposing you, taken along that path. In general, the line integral of any conservative field along a curve is equal to the difference in potential between the endpoints.

Wow, this stuff is really interesting! Where can I learn more?

A much more rigorously mathematical treatment of this should be in any first year undergraduate mathematics, physics or engineering course. If you want to learn for yourself, pick up a book on vector calculus (or calculus of multiple variables, or multivariate calculus, or whatever they feel like calling it). For an online resource, Eric Weisstein's World of Mathematics at http://mathworld.wolfram.com/ is excellent, but it's designed for someone who knows what they're doing already, and probably isn't great for a beginner.


This has been part of the Maths for the masses project

(idea) by eien_meru (5.4 hr) (print)   (I like it!) 2 C!s Thu Nov 11 2004 at 5:42:52

The line integral is a way of extending the ordinary Riemann integral from its natural home on the Cartesian plane into the wide world of Euclidean space. Instead of integrating over a function, the line integral works over a contour, also known as a vector-valued function that is piecewise differentiable.

The new x-axis

In order to reduce something complicated like a line integral into something easier to do, like a regular integral, we parameterize the contour with a single real argument, t. If we were to think of the contour as the path a little bit of something travels along in some cheap physics experiment, then it would be convenient to choose time as the parameter.

The new y-axis

In place of the function f(x), the new integrating function becomes a scalar field, f(X). This looks like a surface, a membrane, hanging in space. In physical applications, this could be a heat gradient, or a potential energy function, or some other scalar-valued phenomena. This will serve the same purpose in the new line integral that f(x) had.

And the transition: γ: [a, b) → En

Typically, space curves are represented as a function γ from the interval [a, b) onto one of the many n-dimensional Euclidean spaces, En. This function will be our transition between the many-splendored fabric of space and the pristine flatness of the plane we've all come to know and love.

Working with work

From high school physics, we know that work is the product of force and displacement. Lets say we have a force that can easily be expressed in terms of position (like an electric or gravitational field). This force will be our scalar field, and the applied force at a position X will be given by f(X).

For displacement, we look at the path that the force field moved the object through, and attempt to parameterize it. This could be difficult, but physical phenomena in idealized conditions like these tend to follow easily calculated paths like parabolas or helices. Even if it doesn't follow an easy path, at worst we can use bezier curves to approximate it. This will give us a relationship between the position of the particle and the time it reached that position. That's our contour γ:[t0, tf). t0 is our start time, and tf is our finish time.

Now we have everything we need to employ the integral calculus method of subdividing the displacement into small pieces and adding up all the products of force and displacement:

W = ∫γ F ds

But we don't have enough yet to actually calculate the integral! There's so many problems with this: Our lower and upper bounds are with respect to time, the integral is with respect to displacement, and our force is with respect to position. It's clear we'll have to use some substitution magic to get something that can be evaluated.

First things first: we use the chain rule to perform a change in variables between displacement and time, so we can establish proper bounds.

W = ∫t0tf F ds/dt dt

But now we need to iron out our forces. We know the field (position → force) and the path (time → position), so we compose the two and obtain F(γ(t)), (time → force)!

W = ∫t0tf F(γ(t)) ds/dt dt

To make ds/dt more palatable, we resurrect the old arc length formula:

s(t) = ∫t0t √(x'12 + x'22 + ...) dt

Those little x'is are the time derivatives of the separate components of γ(t). Now, we take the derivative of both sides:

s'(t) = d/dt ∫t0t √(x'12 + x'22 + ...) dt

And unleash the fundamental theorem of calculus on the right hand side.

ds/dt = ||dγ/dt|| = √(x'12 + x'22 + ...)

Combining this with all the work we did above, we find a calculable form of the naive equation:

W = ∫t0tf F(γ(t)) ||dγ/dt|| dt

... *phew*

Vector Fields and encounters of the second kind

For all the stuff above, we used a scalar field, f(X). Well, it could be that our field was a vector field, in which case we break the line integral into component parts and do it that way:

γ F(X) • dX = ∫ab F(X) • X'(t) dt

Essentially, we find each vector component's ribbon and add all the areas together. If we unpacked the dot product, it would look like this:

γ F1(X) dX1 + &intγ F2(X) dX2 + ...

A word about parametrization

The only time the parametrization of the contour matters is when the integral is a loop integral and ends at the same point it started. In this case, there's a problem with orientation. A positively oriented curve will have its interior region on the left as its parameter increases; whereas a negatively oriented curve will have its exterior on the left.

The following identity sums up all the important features of the line integral and orientation; -γ is the path formed by following γ backwards:

γF•dX = -∫F•dX

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