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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!

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  • $\begingroup$ 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
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$||\vec{F}||$ 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.

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  • $\begingroup$ 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

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