Line Integrals

In the module on Using Geometry we studied work and we saw that

It is often useful to think of ordinary integration as "advanced multiplication." For example, the area of a simple rectangle whose height is H and whose base goes from x = a to x = b as shown in the figure below

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is H (b - a) and the area of the figure below whose height is given by the function y = f(x)

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is

When height is a simple constant, area is height multiplied by width but when the height varies simple multiplication is replaced by integration or "advanced multiplication."

In the same way when an object moves from the point x = a to the point x = b under the influence of a force that is not constant and is given by a function f(x), then the work done by the force is given by

and the simple multiplication of our first formula for work is replaced by integration or "advanced multiplication."

In this module we look at a two-dimensional or three dimensional situation in which the force is not constant -- at different points the force may point in different directions and it may have different strangths. The figure below shows an example. Notice the force is always pointing toward the origin in this particular example.

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Not only does the force vary from point-to-point but there is another complication -- we are interested in an object traveling along a curved path as shown in the movie below.

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Because the force varies from one place to another we call it a force field. A force field is represented mathematically by a vector field. We use the notation

F(x, y) = P(x, y) i + Q(x, y) j

to describe a force field in R2 and the notation

F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k

to describe a force field in R3.

Now suppose that C is a curve in R2 or R3 like the curve shown in the figure below. This example is the path followed by the object in the movie above. We are interested in the work done by a force F acting on an object as it travels along a curve C in a particular direction. In the example below the object travels along the curve C from left to right.

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We estimate this work in the obvious way. First, we approximate the curve C by a polygonal path -- a path made up of straight line segements -- as shown in the two figures below.

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We choose points s0, s1, s2, ... sn along the path C and then connect these points as shown in the figure above. Then we estimate the work done on the i-th segment of the path by

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and the amount of work done on the whole path by

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By using a large number of small segments we can obtain a very good estimate for the amount of work done. The exact amount of work done is obtained by taking the limit of these estimates. This limit is called the line integral of the vector field F over the path C and is denoted

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In practice we usually describe the curve C by a function

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or

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and we divide the time interval [a, b] up into n subintervals. by letting

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and

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and then

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We use four examples of force fields to illustrate our work.

Before going on, you should think about the amount of work that would be done by these four force fields acting on objects traveling along various different paths.

We will look at objects traveling along various different paths. Here are some examples.


For each problem below, estimate the answer first based on your physical intuition and then estimate the answer numerically using your CAS window. The force fields F, H, L, and M are the force fields described above.


As usual, whenever we have a situation in which we can get arbitrarily good estimates for some quantity, we can find the exact value of the quantity by taking the limit of the estimates. The details are worked out in Multivariable Calculus, Linear Algebra, and Differential Equations in a Real and Complex World. The final results are.

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where we use the following notation in R2

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and the following notation in R3

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Repeat the same problems in the exercise set above using your CAS window to find the answers by integration as described above.


Diffusion Into a Region

We end this section with one long homework problem. We are interested in the way that a substance like ink or a pollutant diffuses into a two-dimensional region. Because this problem is mathematically similar to computing work, it makes a good homework problem at this time.

The way that a substance diffuses in a medium is determined by many factors -- for example, temperature, magnetic and electric fields, and the concentration of the substance itself. At each point (x, y) in a two-dimensional medium the diffusion of the substance can be described by a vector

F(x, y)

that is measured in units of mass per unit of length per unit of time. For example, if

F(x, y) = (2 mg per cm per minute) i

then the the substance is flowing in the direction of the vector i at a rate that would result in 2 milligrams of the substance crossing a vertical line whose length was 1 centimeter in one minute. Notice that if the line was horizontal -- that is, parallel to the direction of flow -- then none of the substance would cross the line.



Copyright c 1997 by Frank Wattenberg, Department of Mathematics, Montana State University, Bozeman, MT 59717