I know the classic line integral and is lengh of the curve and I compltetly understand the logic. But when I am trying to think about the meaning of line integral on vector field or scalar function I am not sure what are those expressesion are represent and how should I think on it.


1 Answer 1


Let $\gamma:[a,b] \rightarrow \mathbb{R}^2$ be a soft path. Let's keep things in $\mathbb{R}^2$ for simplicity.

Vector fields

In physics there is a thing by the name of "Work". Imagine that we have a block and, due to the action of a force $F$, the block moves $k$ meters to the right. Then the work is

$$W=F\cdot k\cdot \cos(\alpha)$$

where $\alpha$ is the angle between the Force vector and the direction of motion.

See this image

The problem is the following: This formula works fine only if the path is a straight line and if the force vector is constant. But what if the line is curved? If the line is curved the value of $\alpha$ is allays changing with time, and If the vector $F$ is allays changing the this becomes even more complex.

The solution is the following:

If you notice, if we thing of $\textbf{k}$ as a vector pointing from the beginning to the end of the trajectory then, if our trajectory is a line and if the force is constant:

$$W=F\cdot k\cdot \cos(\alpha)=\textbf{k} \cdot \textbf{F}$$

The key to generalize this for any trajectory is the following:

Let's divide the path $\gamma$ into a bunch of smaller paths: $\gamma_i$. This smaller path are so small that they look almost like a straight line.

Let $\textbf{F}(x,y)$ be our vector field that tells us the fore applied to the block in any position $(x,y)$.

Then we can calculate a little work done in one of the smaller paths by: $$\Delta W_i = \textbf{F} \cdot \Delta \gamma_i$$

Where $\Delta \gamma_i$ is a small vector pointing from the beginning of $\gamma_i$ to it's end.

$\Delta W_i$ is the work done in each segment $\gamma_i$ of our initial trajectory. Then, if we want to know the total work, we simply sum over all segments:

$$W = \sum_i \textbf{F} \cdot \Delta \gamma_i$$

This is just an approximation, but if we divide the initial line into smaller and smaller segments, then we get an even better approximation. So in the limit this becomes accurate and we get:

$$W = \int_\gamma \textbf{F} \cdot ds$$

This is how I think about the meaning of line integrals over vector fields and I think this is a good intuition for the motivation of their invention and their meaning.

Scalar Fields

Now let $F(x,y)$ be a scalar function. You should know that the image of this function is an surface above the $\mathbb{R}^2$ plane.

If you can recall (I think you do), the meaning of $$\int_a^b f(x)dx$$ is just the area under the curve from $a$ to $b$. This is the same meaning of the line integral over a scalar function.

When we calculate:

$$\int_\gamma f(x,y)ds$$

We are simply calculating the area over the line $\gamma$ (that lives in the $\mathbb{R}^2$ plane) and the surface $f(x,y)$ above it: See this small animation

  • $\begingroup$ that was great explantion! about the vector field part, if I understand you correctly the line integral over vector field is is the amount of work the vector field did along the path? or is that just specific case? $\endgroup$
    – Sagigever
    Commented Jun 12, 2020 at 11:49
  • $\begingroup$ Thank you. It's not a specific case. Let $\gamma$ be any path and $\textbf{F}$ be a vector field. Then the line integral over that vector field is the total work done by the vector field as something travels through that path. And this is the same meaning for any dimension, not for only $\mathbb{R}^2$ $\endgroup$ Commented Jun 12, 2020 at 11:52
  • $\begingroup$ This is why line integrals over vector fields are used a lot in physics $\endgroup$ Commented Jun 12, 2020 at 11:52
  • $\begingroup$ But although the meaning is work, mathematically it's a fundamental, It's involved in a lot of theorems and has a lot of applications within math. $\endgroup$ Commented Jun 12, 2020 at 11:55

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