Line integral vs Arc Length I am trying to understand when do to line integral and when to do arc length. So I know the formula for arc length varies based on $dx$ or $dy$ like so:
$s=\int_a^b \sqrt{1+[f'(x)]^2} \, \mathrm{d} x$ for the arc length 
and here's a line integral equation:
$\int_c fds=\int_a^b f(r(t))\cdot r'(t)dt$
I just don't understand how the two are different/similar. Don't they both compute the same thing?
 A: Suppose you have a curve $C$ parametrized as $\mathbf{g}(t)$ for $0\le t\le 1$. Then the arc length of $C$ is defined as
$$\int_0^1\|\mathbf{g}'(t)\|\ dt$$
An intuitive way of think of the above integral is to interpret the derivative $\mathbf{g}'$ as velocity. Then the above integral is basically the statement that the net distance traveled is equal to the speed times time.
The line integral itself is also concerned with arc length. More specifically, the scalar line integral is concerned with the arc length of a curve along with a weight $f$ at each small segment of the curve.
To give a simplified example, suppose that you have an ideal elastic $C$. Further suppose that you have a function $f$ which assigns a value for each point of the elastic. Think of this $f$ as a stretch factor. If a point $p$ of the elastic is assigned a number $f(p)$, then we stretch the elastic locally around $p$ by a factor of $f(p)$. If we now add up the stretched lengths, what we have is a line integral
$$\int_C f\ ds = \int_0^1 f(\mathbf{g}(t))\,\|\mathbf{g}'(t)\|\ dt$$ 
This line integral will be larger or smaller than the actual length of the elastic depending on how the elastic is stretched or compressed as a whole. 
But if our function is $f(p) = 1$, then that corresponds to stretching the elastic at each point by a factor of $1$, i.e. leaving the elastic alone. If we add up the untouched lengths segments of the elastic, all we do is recover the actual arc length of the elastic. This is why arc-length is given by
$$\int_C 1\ ds = \int_0^1\|\mathbf{g}'(t)\|\ dt$$
an unweighted line integral.
