For a vector field $\vec F$ and a curve $C$ with length $c$, if $\vec F$ is tangent to $C$ at every point, then the line integral of $\vec F$ along $C$ is:

$$\int_C \vec F \cdot d\vec r = \int_C ( ||\vec F|| *||d\vec r||)= ||\vec F|| *\int_C||d\vec r|| =||\vec F|| *c$$

I use $\cdot$ to mean the dot product and $*$ to mean multiplication.

I'm not sure about this reasoning. I don't think I'm using integrals correctly.

I get to $||\vec F|| * ||d\vec r||$ because $\vec F$ is always tangent to $C$, which means $\cos\theta$ in the dot product formula is $1$. I then take $||\vec F||$ out of the integral, since it's a scalar.

The next step is the dubious one: I have an integral $\int_C||d\vec r||$. Is this technically a valid integral? I don't know if I'm allowed to play around with the differential in the integral like that.

Furthermore, would it even be correct to say that $\int_C||d\vec r|| = $ (length of the curve $C$) ?

Any help is greatly appreciated!

  • $\begingroup$ The answer given below says everything that's needed, and just to emphasise another mistake: $\vec{F}$ being tangent to the curve doesn't mean $\cos$ of the angle is $1$. It could be $-1$ (the vectors could be parallel, but pointing in opposite directions). $\endgroup$
    – peek-a-boo
    Dec 8 '19 at 14:20
  • $\begingroup$ @peek-a-boo Indeed, forgot about that, thank you! $\endgroup$ Dec 9 '19 at 14:04

$||\vec{F}||$ need not be constant, so it is invalid to move it out of the integral. Consider $\vec{F}(r,\theta, \phi) = (r,\theta, \phi)$ with a path that starts at a point and proceeds radially outward, so that $\vec{F}$ is tangent to every point on the path. Here, $||\vec{F}(r,\theta, \phi)|| = r$, which is not a constant and so does not migrate out of the integral with respect to $r$. (Also, since $||\vec{F}||$ is not constant, which of its many values are you writing in front of the integral?)

Also, you should be careful with $|\mathrm{d}\vec{r}|$. If the path $C$ starts at a point, proceeds radially outwards, stops, then proceeds radially inward back to its starting point, $\int F \cdot \mathrm{d}\vec{r} = 0$, but $\int \vec{F} \cdot |\mathrm{d}\vec{r}| = 0$ is twice the integral along the part of the path that is just radially outward.

  • $\begingroup$ That all makes sense, thank you for the help! So I guess the only way what I wrote makes sense is if $\vec F$ is a constant vector field (or would it sitll not make sense?) Also, so $\int \vec F \cdot |d\vec r|$ is a valid integral (that's equal to the distance traveled)? Isn't it a breach of notation to wrap the differential of an integral in magnitude bars like that? $\endgroup$ Dec 9 '19 at 14:03
  • $\begingroup$ Recall that $\mathrm{d}\vec{r}$ is the placeholder for the $\Delta r$ in the Riemann sum underlying the integral. One can also take $|\Delta r|$, where the magnitude of the change in $r$ is what one wants (for whatever quantity one is calculating, for instance while rectifying a curve). It is probably not what you want in a dot product since $|\mathrm{d}r|$ is a scalar and, depending on your definitions, either the dot product of a vector and a scalar is undefined or is zero. $\endgroup$ Dec 9 '19 at 23:42
  • $\begingroup$ Hm okay, I'll keep that in mind. Last question (sorry to keep bothering): I'm now not sure as to when I can move $||\vec F||$ out of the integral. If $||\vec F||$ is not constant, is it always invalid to move it out of $\int_C ||\vec F|| \cdot ||d\vec r||$? Or can you still move $||\vec F|| out in some cases, even if it's not constant? If so, which cases would those be? $\endgroup$ Dec 10 '19 at 16:32
  • $\begingroup$ Generally, never. Use explicit notation: $$\int_C ||\vec{F}(r)|| \cdot ||\mathrm{d}\vec{r}|| = ||\vec{F}(\text{what goes here?})|| \int_C ||\mathrm{d}\vec{r}|| \text{.} $$ When you suppress the arguments, it is easier to wrongly believe that a varying function is a constant, but this is a very confused thing to do. $\endgroup$ Dec 10 '19 at 16:42
  • $\begingroup$ Got it! Seeing it explicitly like that helps a lot. I was always thinking of $\vec F(x,y)$, but I realize now that $d\vec r$ is just $dxdy$, and of course you can't move $\vec F(x,y)$ out of $\int \vec F(x,y) dxdy$. Thanks again for all the help! $\endgroup$ Dec 10 '19 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.