Why calculate the line integral for a vector field? *Without using work/physics*

For a while now I've been trying to find motivation and a good intuition behind the line integral for a vector field. This is the first time I'm learning this topic and I'm not interested in too much formal rigor but rather a strong geometric/mathematical intuition behind the concept.

• I've heard of the intuition for a line integral over a scalar field as being the "area of the fence between the curve and the surface/function." and I'm happy with this interpretation. Is there a similar idea for vector fields?
• I've seen the wikipedia GIF's for both scalar fields and vector fields.
• I've seen almost all the other related pages on MSE with a similar question but without finding a satisfying answer.
• I'm trying to find an explanation which doesn't rely on work from physics

https://en.wikipedia.org/wiki/Line_integral#Definition_2

While this vector field GIF was quite useful, I was still unable to understand the motivation behind the integral. I might not have understood the GIF completely, but from what I could tell it looked like the connection between the integral and an area being calculated. But wouldn't this be true of any integral? What connection does this integral and area have with the original setting of a field an a curve?

But what is the geometric significance of this? Why are we specifically taking the dot product of these two objects ($$\mathbf F(\mathbf r)$$ and $$d\mathbf r$$)? What is the motivation behind finding this integral? What is the significance/meaning of the output to this integral?

• I understand that in a physical setting one might be trying to find the work done by the force field. However just in a mathematical context what would be the use of finding this information provided by the integral?
• I would because I like calculating integrals. – Marra Apr 3 at 10:53
• Without the sole purpose being to extract joy from Mathematics* – user523384 Apr 3 at 10:56

Integration is, in my mind, the art of adding up many small things. Curve integrals, surface integrals, multiple integrals, they all come back to this one main idea. So the reason that most people do any integral ever is that they have lots of small things they want to add up.

The area interpretation of a line integral over a scalar field makes sense because you can add together the areas of many narrow rectangles. So when trying to find an interpretation for something like $$\int_C\vec F(\vec r)\cdot d\vec r$$, then you have to ask yourself what the little things that are added together are.

In the case of the work exerted by a force field on a particle, we divide the curve into small pieces, and for each little piece look at the work done by the field when moving the particle from one end to the other. Since the piece is small, we can consider the simplifications that

• the path is straight, and the total movement is in a straight line from some point $$\vec r$$ to a point $$\vec r + \Delta\vec r$$ close by
• the force field doesn't change, and is equal to $$\vec F(\vec r)$$ for the entire displacement

In that case, the work done is exactly the scalar product of the field and the displacement: $$\vec F(\vec r)\cdot \Delta \vec r$$.

Summing together all these little contributions, and letting the pieces of the path get smaller and smaller, the $$\Delta$$ changes into $$d$$, the Greek letter $$\Sigma$$ turns into a German long S, and we get the integral $$\int_C\vec F(\vec r)\cdot d\vec r$$.

Physical work is a very common example to use for this, probably because it's available to anyone with a good grasp of high school physics (which is luckily quite common among the people who come into multivariable calculus), and also because coming up with good examples that regular people have experience with is darned difficult. But anything which can be seen as a sum of scalar products between the vector field and many small displacement vectors will work. There is nothing inherently physical about the math in the above paragraphs. The physics only plays a part when finding the expression $$\vec F(\vec r)\cdot \Delta \vec r$$.

So a need to sum up many small $$\vec F(\vec r)\cdot \Delta \vec r$$ could arise form something else entirely. I just can't think of a good example right now.

• Thanks so much! Yes this makes sense. An example of this expression somewhere else would be really helpful to motivate the true generality of the concept. I was wondering, is there any geometric interpretation of this integral? To elaborate, what interaction between a field and a curve is the integral describing? Is it always to do with the displacement vectors of a particle moving on this curve? – user523384 Apr 3 at 11:14

Intuitively, integral is simply the sum of small, mostly microscopic, quantities to create a macroscopic quantities. For instance, the integral $$\int_{a}^{b} f(x) \, dx$$ of a function $$f: [a, b] \to \mathbb{R}$$ is often interpreted as (signed) area simply because we interpret the infinitesimal $$f(x) \, dx$$ as the area of the thin rectangle with corners $$(x, 0)$$ to $$(x+dx, f(x))$$. In principle, however, infinitesimals can assume any interpretation based on the contexts.

In the case of line integral in $$\mathbb{R}^3$$, the infinitesimal is given by the form

$$\omega = \mathbf{F}(\mathbf{r})\cdot d\mathbf{r} = F_1(\mathbf{r})dx_1 + F_2(\mathbf{r})dx_2 + F_3(\mathbf{r})dx_3$$

This $$\omega$$ is called differential 1-form in mathematics. A quintessential example of 1-form is the total differential of a scalar field $$\phi$$. If we are given a scalar field $$\phi$$ on $$\mathbb{R}^3$$, then its total differential is given by

$$d\phi = \frac{\partial \phi}{\partial x_1}dx_1 + \frac{\partial \phi}{\partial x_2}dx_2 + \frac{\partial \phi}{\partial x_3}dx_3.$$

This is an abstract notion encoding the fact that, if we are given a small displacement, from $$p$$ to $$p+\Delta p$$, then we can produce the corresponding small change in values of $$\phi$$ as

$$\phi(p+\Delta p) - \phi(p) \approx \frac{\partial \phi}{\partial x_1}\Delta p_1 + \frac{\partial \phi}{\partial x_2}\Delta p_2 + \frac{\partial \phi}{\partial x_3}\Delta p_3.$$

And if we have such notion of measuring small changes in values of $$\phi$$ at our hands, then for any given path $$\gamma$$ in the space and its polygonal approximation $$p_0, p_1, \cdots, p_n$$, we can sum these small changes along $$\gamma$$ to create a macroscopic changes

$$\phi(p_n) - \phi(p_0) = \sum_{i=1}^{n} [\phi(p_i) - \phi(p_{i-1})] \ \,\approx:\text{''} \int_{\gamma} d\phi.$$

Now, the idea of 1-form and its integral (i.e. line integral) is that, the notion of 'small changes along small displacement' need not be a by-product coming from a scalar field, but rather can be defined independently.

Now this is particularly important in physics because often the physics law is presented by equating various combinations of 'small changes', and then line integral helps us restore macroscopic changes thereby.

• Thank you for your answer! This was quite insightful to grasp more concretely what the line integral means, and particularly how it helps relate microscopic and macroscopic changes. I was wondering, is there a geometric interpretation of this as well? (There need not be, but you're aware of any it would be really nice to hear) From what I have thought, is it like "the magnitude of change in a function's value when travelling along a certain path"? Is this accurate? (Just trying to understand the interpretation of the last equation you had) – user523384 Apr 3 at 11:44