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!
$||\vec{F}||$ 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.