In the introduction to scalar line integrals, we derived the formula for $\slint{\dllp}{\dlsi}$, the line integral of a function $f$ over a curve parametrized by $\dllp(t)$ for $a \le t \le b$: \begin{align*} \slint{\dllp}{\dlsi} =\int_a^b \dlsi(\dllp(t))\| \dllp\,'(t) \| dt. \end{align*}

However, the value of this integral should not depend on the specific parametrization $\dllp$, as it is designed to capture quantities like the total mass of a wire with density $\dlsi$. The integral should only depend on the density function $\dlsi(\vc{x})$ and the image curve, denoted by $\dlc$, which is the set of points $\dllp(t)$ for all values of $t$ in the interval $[a,b]$.

For the reason, we can think of the line integral as being over the curve $\dlc$, rather than the particular parametrization given by $\dllp(t)$. To reflect this viewpoint, we could write the integral that gives the mass of the slinky as \begin{align*} \dslint = \pslint{a}{b}{\dlsi}{\dllp}, \end{align*} where the only difference from above is that we replaced $\dllp$ with $\dlc$.

The notation $\dslint$ makes it explicit that line integrals are independent of the parametrization $\dllp(t)$ (after all, the notation does not mention $\dllp(t)$). You may remember that the same curve $\dlc$ can be parametrized by many functions. For example, we gave two different parametrizations of the unit circle. Above, we parametrized the slinky by $\dllp(t) = (\cos t, \sin t, t)$, for $0 \le t \le 2\pi$. We could have equally well parametrized the same slinky by $\adllp(t) = (\cos 2t, \sin 2t, 2t)$ for $0 \le t \le \pi$. (If $\dllp(t)$ and $\adllp(t)$ were the positions at time $t$ of two particles traveling along the slinky, the particle given by $\adllp(t)$ travels twice as fast, covering the slinky in half the time, compared to the particle given by $\dllp(t)$.)

As long as the density $\dlsi(\vc{x})$ is unchanged, the mass of the slinky had better be the same, no matter which parametrization we use. Hence, it must be true that the mass is both \begin{align*} \dslint = \int_{\dllp} \dlsi \, d\als = \pslint{0}{2\pi}{\dlsi}{\dllp} \end{align*} and \begin{align*} \dslint = \int_{\adllp} \dlsi \, d\als = \pslint{0}{\pi}{\dlsi}{\adllp}. \end{align*}

Can you see why the integral is the same for both parametrizations $\dllp(t)$ and $\adllp(t)$, i.e., why \begin{align*} \pslint{0}{2\pi}{\dlsi}{\dllp} = \pslint{0}{\pi}{\dlsi}{\adllp}? \end{align*}

In the first case, you are integrating over an interval that is twice as long (from 0 to $2\pi$ versus from 0 to $\pi$), but the speed $\| \dllp'(t)\|$ is half the speed $\| \adllp'(t) \|$. The two effects cancel and the integrals are equal.

Since the line integral of a vector field $\dlvf$ over the curve is based on the line integral of a scalar function $f = \dlvf \cdot \vc{T}$, where $\vc{T}$ is the unit tangent vector of the curve, we expect that line integrals of vector fields should also be independent of the parametrization $\dllp(t)$. Indeed, this is the case, with one important exception. Since $\vc{T} = \dllp'(t)/\dllp(t)$, the unit tangent vector, and hence the integral $\dlint$, will be independent of the speed $\|\dllp'(t)\|$ of the parametrization. The unit tangent vector, as its name implies will always be of length 1. But, at any point along the curve, there are two unit tangent vectors pointing in opposite directions. If we call one of them $\vc{T}$, then the other is $-\vc{T}$. The choice of these unit tangent vectors will depend on the direction that $\dllp(t)$ transverses the curve $\dlc$ as $t$ increases.

We refer to the choice of the unit tangent vector, or equivalently, the choice of the direction to tranverse $\dlc$, as the orientation of the curve. Every simple curve has two orientations, one corresponding to one unit tangent vector $\vc{T}$ and the other corresponding to its opposite $-\vc{T}$. Scalar line integrals are independent of curve orientation, but vector line integrals will switch sign if you switch the orientation of the curve. This make sense intuitively, as the mass of the slinky shouldn't change, but the work done by a force field changes sign if you move in the opposite direction.

The examples of line integrals of scalar functions and vector fields include calculations of the same line integral with different parametrizations.