I've been studying Line Integrals over vector fields; and most books explain that "The Line Integral over Vector Fields gives us the resulting Work done by a vector field over a body/particle moving on a given curve C".

But as I understand it, that is the application/visualization from a Physics point of view. However, is there a way to visualize/understand what the Line Integral over Vector Fields is IN Mathematics? Basically: "What are we calculating when we calculate said integral?"

In conjunction to @Sammy Black's answer I found this gif very illustrative (Thanks @Mark S.)

  • 1
    $\begingroup$ In order to answer your question I need you to tell us what is the difference between Physics and Math. $\endgroup$
    – Ruy
    Jun 29, 2022 at 21:45
  • 2
    $\begingroup$ Have you already seen the award-winning image on the Wikipedia page for line integral? If so, can you elaborate on what you are not satisfied by? $\endgroup$
    – Mark S.
    Jun 30, 2022 at 3:49
  • $\begingroup$ @MarkS. I was not aware of said award winning image. It did clarify things a lot $\endgroup$ Jun 30, 2022 at 14:28

2 Answers 2


When in doubt, replace an integral expression with a corresponding Riemann sum that approximates it. $$ \int_a^b f(x) \, dx \quad\leadsto\quad \sum_{i=1}^k f(x_i^*) \, \Delta x_i^{\phantom{*}} $$ Usually, it's more clear what's being calculated in the (finite) sum. In fact, most proofs of integral expressions do just this in reverse: start with finite approximation, realize it as a Riemann sum, then pass to the limit as the partition gets finer and finer.

In the case of the line integral of a vector field along a path, the integral looks like $$ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt, $$ where the latter expression uses a parametrization $\mathbf{r}: [a, b] \to \mathbb{R}^n$ of the curve $C$. (Technical detail: we require $\mathbf{r}$ to be differentiable in the interior of the interval and never to vanish.) If we partition the interval $$ a = t_0 < t_1 < \cdots < t_i < \cdots < t_{k-1} < t_k = b, $$ then the sequence of points $\{\mathbf{r}(t_0), \mathbf{r}(t_1), \dots, \mathbf{r}(t_k)\}$ lie on the curve $C$, and by connecting them in succession by line segments, we get a polygonal (piecewise-linear) approximation of the curve $C$. Evidently, the Riemann sum looks like $$ \sum_{i=1}^k \mathbf{F}(\mathbf{r}(t_i^*)) \cdot \mathbf{r}'(t_i^*) \, \Delta t_i^{\phantom{*}}, $$ which might be easier to make geometric sense of if we normalize $$ \Delta \mathbf{r} = \mathbf{r}'(t_i^*) \, \Delta t_i^{\phantom{*}} = \frac{\mathbf{r}'(t_i^*)}{\| \mathbf{r}'(t_i^*) \|} \, \| \mathbf{r}'(t_i^*) \| \, \Delta t_i^{\phantom{*}}. $$ In this expression, the fraction is the unit tangent vector in the direction of motion along $C$ and the rest of the expression is the speed at some moment on the segment multiplied by the interval of time taken to traverse it, i.e. the length of that segment of the polygonal curve.

Thus, our Riemann sum adds up contributions from each segment the product of $$ \mathbf{F}(\mathbf{r}(t_i^*)) \cdot \frac{\mathbf{r}'(t_i^*)}{\| \mathbf{r}'(t_i^*) \|} $$ with $$ \| \mathbf{r}'(t_i^*) \| \, \Delta t_i^{\phantom{*}} $$ The former is the component of the vector field $\mathbf{F}$ along the direction of the $i$th segment, and the latter is the length of the segment. This is the most natural way to interpret this line integral. It measures the net amount that the vector field points along the curve.

Notice that if the curve $C$ is an interval on the real number line, i.e. the identity parametrization $\mathbf{r}: [a, b] \to \mathbb{R}$, $$ \mathbf{r}(t) = t, $$ and thinking of the vector field $F = \langle f \rangle$ as having a single component pointing along the line, controlled by a function $f: [a, b] \to \mathbb{R}$, then this interpretation of the line integral is precisely the standard interpretation of the integral from single-variable calculus.

I hope this helps.

  • $\begingroup$ Really good explanation. +1 $\endgroup$ Jun 30, 2022 at 1:56
  • $\begingroup$ Wow! Such a detailed explanation. I wish my teacher explained like this! $\endgroup$ Jun 30, 2022 at 14:32
  • $\begingroup$ There is a "physical" interpretation: imagine the vector field as local direction of wind : the line integral represents the total work done by the wind in helping you to walk along the path (of course, taken algebraically : wind can be against you in the average and provide a negative work... $\endgroup$
    – Jean Marie
    Jul 1, 2022 at 5:44

The line integral computes the moral equivalent of Work energy to traverse the curve. It means that the instantaneous work is the scalar product of the vector field and the curve speed vector field at a given coordinate and the integral is the resulting continuous summation.


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