Is there a case to be made that the topic of line integrals should only involve vector fields?

My colleagues and our textbook take the position that line integrals should only be taught from a vector field perspective (specifically for calculating "work"). [In fact, our textbook defines a line integral as $\int _C \mathbf{F} \cdot d\mathbf{r}$, where $\mathbf{F}$ is a vector field and $C$ is some parametrized curve.]

I think it makes more pedagogical sense to introduce line integrals as a way to generalize what students should have just done in their integral calculus class: integrate a function along some 1-dimensional direction. Now, that "direction" can be a path in 2-space or 3-space, so we can see an area as the result of the line integral. Then, after introducing vector fields, we can consider other, meaningful things to integrate along a path, such as a dot product of the field with the path.

My motivation here is that I would like new calculus topics to be easily connected to old topics, if possible. If we jump right into calculating work without any tie back to the "calculate the area" problem students are used to, I fear they may come away thinking that line integrals are just these weird things with their own rules.

In short: Is there a prevailing setting for introducing line integrals? If so, has there been a movement to minimize the teaching of line integrals over scalar fields, focusing primarily on work calculations?

  • $\begingroup$ I think that is a reasonable point of view (the vector field) in an E&M text but that is too limiting for line integrals in general. In other words, I agree with you. Still..if that is the text, I would roll with it. Worst thing is for the instructor to deviate from the text. Especially in a lower level service class. Better to learn the imperfect perfectly than the perfect imperfectly... $\endgroup$
    – guest
    Nov 6, 2018 at 21:00
  • $\begingroup$ This is a good question. In a way I'm glad to see I'm not the only teacher who has trouble here. Motivating the idea to non-physics students in particular is a bit challenging. I tried to introduce a more general idea of a "flow". Like imagine a large city with a dense network of streets, running N/S or E/W, wiht known traffic density (a planar vector field with N/S and E/W components). Given the city border as a parametrized curve how do we calculate the net increase of the number of cars within city limits? I'm afraid I don't think that example was a hit :-( $\endgroup$ Nov 7, 2018 at 16:46
  • $\begingroup$ Sounds like you just need to go whole hog and do it all from a differential forms perspective. <ducks> $\endgroup$
    – kcrisman
    Feb 14, 2019 at 22:14

2 Answers 2


The problem is that there isn't a direct connection with area. Sure, you can do a "$\textrm{d}s$" line integral, but that isn't quite the same thing is it?

How can we motivate $\int_\gamma P(x,y)\textrm{d}x + Q(x,y)\textrm{d}y$?

I think that it is very hard to do. Some physical problems do require this, as you mention, but that seems like a much more narrow motivation than our motivation for the single variable integral.

The only "general" motivation I have found is this:

In calculus, we use integrals to anti-differentiate. The analogue of the derivative in Calc 3 is the gradient. So it is natural to ask "If you tell me that $\nabla f = \langle P(x,y) , Q(x,y) \rangle$, then what is $f$ "? This is "anti-gradienting". In other words we want to be able to find a function $f: \mathbb{R}^2 \to \mathbb{R}$ which satisfies

$$ \begin{cases} \frac{\partial f}{\partial x} = P(x,y)\\ \frac{\partial f}{\partial y} = Q(x,y) \end{cases} $$

We cannot expect to be able to solve this system of PDEs in general, because we have equality of mixed partial derivatives. So this equation cannot have a solution if $\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$.

Say we know that there is such a primitive function $f$ though. How can I approximate it?

Say we know $f(1,1) = 5$ and $\nabla f = \langle P(x,y), Q(x,y)\rangle$, and I want to approximate $f(2,3)$. The gradient's defining feature is that it allows you to approximate small changes in the function value: $f(\vec{p}+\vec{h}) \approx f(p)+\nabla f \cdot \vec{h}$. So if we know the gradient everywhere, we can approximate $f(2,3)$ by picking a path from $(1,1)$ to $(2,3)$, chopping it up into a zillion tiny vector changes, and summing those small changes dotted with $\nabla f$. You will end up recovering the notion of a line integral of $\nabla f$ along this curve.

As noted though, not every vector field is conservative (because of Clairaut's theorem). But we can still "do a line integral" to them exactly the same way we did line integrals of gradient vector fields. So I cannot interpret that line integral of a general vector field by thinking of it as the change in values of some potential function, but it is still useful.

One reason this is useful is because it gives us a test for whether a given vector field is conservative. If $\langle P(x,y), Q(x,y)\rangle$ were conservative, then the line integral would have to be independent of path.

I still do not have really "down to earth" examples of where I would want to use a line integral of a non-conservative vector field outside of physics or pure mathematics though. I would love to find economic examples (perhaps connected to arbitrage?), engineering examples, biological examples, etc.

  • $\begingroup$ The problem is that there isn't a direct connection with area. Sure, you can do a "ds" line integral, but that isn't quite the same thing is it? Maybe that's the difference. I do see these as being the same thing, because as we move along a "line", what we calculate along the way (area? work? circulation?) and whether this path can be seen as living in a vector field (or just a scalar field), are just particular applications of the big concept. $\endgroup$
    – Nick C
    Nov 6, 2018 at 23:50
  • $\begingroup$ @NickC I would say the "big idea" common to all integrals is "chop into many small pieces, approximate each piece, and sum". This concept does apply to area, but it also applies to many many things which are far from area. $\endgroup$ Nov 7, 2018 at 0:55
  • $\begingroup$ I'm not arguing against that at all (of course students should have used integrals to add up things in other contexts besides area), but area calculations seem to be such a well-established application for them that I wondered "why not introduce line integrals from something familiar, like areas?" Then I discovered that our textbook and some instructors dont even see "area" as an application for line integrals. Then I wondered if there was some reason for this. Note: I completely agree that integration should be more for students than a way to calculate area. $\endgroup$
    – Nick C
    Nov 7, 2018 at 2:44
  • $\begingroup$ Oh ya, I didn't actually read your questions completely. Surely doing line integrals of scalar valued functions first makes sense. $\endgroup$ Nov 7, 2018 at 15:24

The integral of a scalar function on a curve is easily motivated as follows, and doing so is helpful for later motivating the definition of the integral of a vector field along an oriented curve. First, an interval on the real axis can be viewed as a very special sort of curve in the plane. Its length is the integral of the constant function $1$ along this curve. The length of a polygonal curve is the sum of the lengths of its constituent segments. Any reasonable curve in the plane or space can be approximated by polygonal curves (parameterize the curve; partition the parameter domain and take the images of the points of the parition of the parameter domain as the vertices of the polygonal curve) (that a curve be "reasonable" will be defined a posteriori to mean that the approximation works). Write down the sum expressing the length of the approximating polygonal curve and observe that it can be interpreted as a Riemann sum approximating a certain integral. Take this integral as the definition of the length of the parameterized curve and note that its value does not depend on the choice of parameterization, so depends only on the image curve. Next observe that the preceding can be modified to calculate the mass of a infinitesimally thin wire. The mass density is interpreted as a scalar field along the curve. Assuming it to be approximately constant on each segment of an approximating polygonal curve one obtains a Riemann sum approximating an integral, that can be taken as the integral of the mass density along the curve. Again, observe that the result does not depend on the choice of parameterization of the curve. Finally, observe that everything continues to make sense dropping the positivity of the mass density (one can think of charge density). All this works for plane curves or space curves equally well.

The preceding is helpful in motivating the definition of the integral of a vector field along an oriented curve. The line integral of a vector field formalizes the physical notion of the work done in moving a particle subject to a field of force represented by the vector field. This simple physical motivation requires that, unlike the integral of a scalar field, the integral of a vector field should be defined so as to depend on the orientation of the curve traversed, simply because the work done in moving against the stream is the negative of the work done in moving with the stream. Again, one can approximate the curve by an oriented polygonal curve, and suppose the vector field approximately constant on each segment of the approximating polygonal curve. The work done along each segment of the approximating polygonal curve is given by the projection of the vector field along this segment. Summing up the result gives a sum that can be viewed as a Riemann sum for an integral that is now taken as the definition of the integral of the vector field along the curve. The definition depends on the choice of a parameterization of the curve, and one checks that the number it yields does not depend on the parameterization, as along as the orientation is unchanged, and changes sign if the orientation is reversed. Directly this defines the integral of the vector field $F$ along the oriented curve $C$ as $\int_{a}^{b}F(\gamma(t))\cdot \dot{\gamma}(t)\,dt$ where $\gamma:[a, b] \to C$ is a parameterization of $C$ consistent with the orientation of $C$. The notation $\int_{C}F\cdot dr$ is shorthand that indicates the pairing of the functional argument (the vector field $D$) and the geometric argument (the oriented curve $C$); the $\cdot dr$ has no meaning, rather it should be viewed as part of the notation $\int_{\,\,} \,\,\cdot dr$ that could equally well be written $\langle\,\,,\,\,\rangle$ for the pairing of the functional and geometric argument (in fact, in more sophisticated contexts, one often sees notations such as $\langle F, C\rangle$). This notation is meant to suggest the operations that need to be performed in order to evaluate the pairing without indicating the choice that needs to be made to effect such an evaluation (and on which the result does not depend).

The preceding approaches adapt straightforwardly to the integration of scalars and vector fields over surfaces.

The idea of eliminating the integration of scalar fields along curves and surfaces from the curriculum seems to me wrongheaded.

The calculation of things as simple as the mass or center of mass of a plate, calculations all engineering students have to make, requires the notion of the integral of a scalar quantity. From a pedagogical point of view it is better to take a steppingstone approach - first extend the familiar integration of scalar quantities over intervals to the integration of scalar quantities over curves - then extend the latter to the integration of vector fields. An integral has two arguments, one functional and one geometric. First generalize the geometric argument (intervals to curves), then generalize the functional argument (scalars to vectors) - this is better than generalizing both at once simply because it compartamentalizes ideas.

  • $\begingroup$ +1 for the last paragraph which perfectly summarizes my motivation for this question. $\endgroup$
    – Nick C
    Nov 7, 2018 at 16:10

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