Home  18.013A  Chapter 20 


We extend the idea of breaking up a curve into small pieces and summing the difference between the endpoints of each piece multiplied by some function, to define integration along a curve $C$ in any Euclidean space.
If $f$ is a scalar field, that is, a function defined on points in our space, we could sum $f\left(\stackrel{\u27f6}{r}\text{'}\right)\left({\stackrel{\u27f6}{r}}_{i}{\stackrel{\u27f6}{r}}_{i1}\right)$ where the ${\stackrel{\u27f6}{r}}_{i}$ represent subdivision points on the path, and again $\stackrel{\u27f6}{r}\text{'}$ is a point in the ith subinterval. This can be defined, but here the difference in $\stackrel{\u27f6}{r}$ 's will be a vector, and the sum will be a vector field.
A much more useful concept is obtained when we replace $f$ here by a vector field $\stackrel{\u27f6}{w}\left(\stackrel{\u27f6}{r}\right)$ and sum $\stackrel{\u27f6}{w}\left(\stackrel{\u27f6}{r}\text{'}\right)\xb7\left({\stackrel{\u27f6}{r}}_{i}{\stackrel{\u27f6}{r}}_{i1}\right)$ over the intervals on the curve $C$ that are defined by the endpoints ${\stackrel{\u27f6}{r}}_{i}$ . The result is then a number rather than a vector.
It is defined exactly as area is in the definite integral. The standard notation for it is
There are two important examples where such integrals occur.
First, suppose we integrate a unit tangent vector $\xb7d\stackrel{\u27f6}{l}$ along $C$ .
For sufficiently small intervals this unit tangent vector will be essentially a unit vector in the direction of ${\stackrel{\u27f6}{r}}_{i}{\stackrel{\u27f6}{r}}_{i1}$ namely $\frac{{\stackrel{\u27f6}{r}}_{i}{\stackrel{\u27f6}{r}}_{i1}}{\left{\stackrel{\u27f6}{r}}_{i}{\stackrel{\u27f6}{r}}_{i1}\right}$ or equivalently $\frac{d\stackrel{\u27f6}{l}}{\leftd\stackrel{\u27f6}{l}\right}$ . The result is then the integral of $\frac{d\stackrel{\u27f6}{l}\xb7d\stackrel{\u27f6}{l}}{\leftd\stackrel{\u27f6}{l}\right}$ or simply $\leftd\stackrel{\u27f6}{l}\right$ over the curve $C$ . Since this just sums the length $\leftd\stackrel{\u27f6}{l}\right$ of each of the subintervals of the curve, its sum over the entire curve will be the length of $C$ .
The length of a curve $C$ is therefore given by
where $\widehat{T}$ is a unit vector in the direction of the curve.
Second, in physics, the work done by a force $\stackrel{\u27f6}{F}$ in pushing an object along a path $C$ is given by an integral of the form
As we have noted already, we define such integrals by breaking the curve up into ever smaller pieces, taking the dot product indicated within each piece, and summing over the pieces.
So far we have generalized the notion of area or rather our approach to it, to define integration along a path in the complex plane, or in Euclidean space.
The same idea can be applied to integration over an area in two dimensions, a surface in three dimensions, a volume in three dimensions, or what you will in higher dimensions.
In every case you can break the region you are interested in into pieces whose diameters go to zero, and sum the value of an integrand multiplied by a measure of the size of the piece, over these pieces.
When integrating over a surface in three dimensions you have to take into consideration that surfaces have directions like curves do, and so we must make our definitions accordingly, as we just did for curves.
You can enter any standard vector field and any parametrized curve in the applet here and see the resulting integral unfold in the applet below.
Exercise 20.5 Integrate the vector field $x\widehat{i}+y\widehat{j}+2z\widehat{k}$ over the default helical path here using the applet, and also along the straight line, both from $z=0$ to $z=4\pi $ with the other variables 0 at the endpoints. How do the results compare?
