### Path integral of a scalar-valued function

Recall the " >procedure for determining the length of a path, where we calculated the length of a slinky. We could use a similar procedure to calculate the mass of a slinky. In fact, if the density of the slinky (in mass per unit length, such as grams per centimeter) were constant, the total mass of the slinky would simply be its length times its density.

What if the density of the slinky varied? It might be thicker in some parts and thinner in others. This possibility is illustrated below. To make the picture easier to visualize (and to decrease the time it takes to download the " >CVT), I’m showing just one coil of the slinky here. As before, the path of the slinky is given by c(t) = (cos t, sin t,t), but this time just for 0 t 2π. The density of the slinky at the point (x,y,z) is given by some function f(x,y,z). For a given value of t, the point on the slinky is c(t), and the density of the slinky is f(c(t)).

To calculate the mass of the slinky, we repeat the procedure we used to " >calculate its length. We pretend the slinky (shown below in red) is composed by a bunch of straight line segments (shown in green). Moreover, we assume each line segment has constant density, so the mass of the line segment is simply its length times its density. The density of each line segment is given by the density of the slinky at the point where the upper end of the line segment touches the slinky. We estimate the mass of the slinky as the total mass of the line segments. The mass of the line segments is shown by the green vertical slider and is labeled by mass(dt). For comparison, the actual mass of the slinky is shown by the red mark on the green slider.

We can increase the number of line segments (decreasing the length of each line segment) so that the total mass of the line segments becomes a better estimate of the slinky mass. As in the " >slinky length CVT, the length of the line segments is controlled by the slider labeled dt (where dt should be displayed as Δt). As you drag Δt toward zero, the length of each line segment shrinks toward zero, the number of line segments increases, and the density of each line segment more closely approximates the density of the slinky. Since the total length of the line segments approaches the slinky length, the total mass of the line segments approaches the mass of the slinky (notice that mass(dt) on the green slider approaches the red mark labeled “actual mass”).

Remember that the " >length of the ith line segment can be written as ||c(ti) -c(ti-1)||. The density of the line segment is simply f(c(ti)) since c(ti) is the point where the upper end of the line segment touches the slinky. Therefore, the mass of the line segment is its density times its length: f(c(ti))||c(ti) - c(ti-1)||. The total mass of the line segments is simply the sum over all n line segments:

 ∑ inf(c(t i))||c(ti) - c(ti-1)||.
The only difference from the " >path length calculation is the f(c(ti)) factor. Hence, if you believe the result we obtained for the path length, you won’t have trouble believing that the total mass of the slinky is
 ∫ abf(c(t))||c'(t)||dt,
where a = 0 and b = 2π. This is called the path integral of f over the path parametrized by c.

We often use s(t) to be the length of the slinky from the beginning point c(a) up to the point c(t). If you think of ds as being the length of a tiny line segment approximating the slinky, then the mass of that tiny line segment is f times ds. For this reason, we often denote the integral representing the mass of the slinky as

 ∫ cfds = ∫ abf(c(t))||c'(t)||dt.

#### Path integrals do not depend on parametrization

If C is the curve parametrized by c(t) for a t b, we can think of the integral being over the curve C, rather than the particular parametrization given by c(t). To reflect this viewpoint, we could write the integral that gives the mass of the slinky as

 ∫ Cfds = ∫ abf(c(t))||c'(t)||dt,
(where the only difference from above is that we replaced c with C).

One important reason to remember the notation Cfds is that it indicates that path integrals are independent of the parametrization c(t) (after all, the notation does not mention c(t)). You may remember that the same curve C can be parametrized by many functions. For example, we gave " >two different parametrizations of the unit circle. Above, we parametrized the slinky by c(t) = (cos t, sin t,t), for 0 t 2π. We could have equally well parametrized the same slinky by p(t) = (cos 2t, sin 2t, 2t) for 0 t π. (If c(t) and p(t) were the positions at time t of two particles traveling along the slinky, the particle given by p(t) travels twice as fast, covering the slinky in half the time, compared to the particle given by c(t).) As long as the density f(x,y,z) 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

 ∫ Cfds = ∫ cfds = ∫ 02πf(c(t))||c'(t)||dt
and
 ∫ Cfds = ∫ pfds = ∫ 0πf(p(t))||p'(t)||dt.

The fact that mass of the slinky should depend only on the actual curve C of the slinky and its density f is summed up by the notation that the mass is

 ∫ Cfds.

Can you see why the integral is the same for both parametrizations c(t) and p(t), i.e., why

 ∫ 02πf(c(t))||c'(t)||dt = ∫ 0πf(p(t))||p'(t)||dt?

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

You can find some path integral examples " >here.