What does $\int_CF=\int_a^bF(C(t))\cdot\frac{dC}{dt}\,dt$ actually mean?

Question

Let $F$ be a continuous vector field on an open set $U$ and $C$ is a continuously differentiable curve on $U$.

We define the integral of $F$ along $C$ to be $$\int_CF=\int_a^bF(C(t))\cdot\frac{dC}{dt}\,dt$$ using the chain rule $$\int_CF=\int_{C(a)}^{C(b)}F(C)dC$$

But what does this all mean? I mean is there any geometric interpretation to the integral curve? or can you just describe what does the integral curve do? and the reason why we decompose a field with a curve?

Answer 1

If $F$ is a force field (e.g. electric or gravity) and $C$ is the motion of a particle, the integral tells you the total work the force does on the particle.

If $F$ is a velocity field (e.g. winds or currents), and $C$ is a frictionless rail filled with objects that can move freely along it (I like to imagine rollercoaster carts packed closely together on a track, or beads on a string), then the integral tells you the total force that $F$ imparts on all these objects. If $C$ is a closed loop the integral will tell you which way around the loop they will be pushed.

So those are a few physical interpretations. What about the calculation itself? Just like any integral, it's about adding together many small things. Divide $C$ into miniscule segments, calculate the contribution from each segment, and add them all up. That's what the integral does (at least the Riemann / Darboux version). In this case, it's probably easiest to think about all the segments having not the same length, but the same $\Delta t$, to see why $\frac{dC}{dt}$ pops up.

Answer 2

The natural definition of a "line integral" is actually the integral of a differential 1-form along a curve. Using upper and lower indices, a 1-form is anything that looks like

$$ \alpha = \sum_k a_k \: \text{d}x^k $$

so for example in two dimensions it would look like

$$\alpha = P \: \text{d}x + Q \: \text{d}y \text{ .}$$

With every 1-form $\alpha$, there is an associated vector field $\vec{A}$ which can help us geometrically interpret the integral. We have that

$$ \int_C \alpha = \int_C \vec{A}\: \cdot \: \vec{\text{d}s}$$

where we can appeal to our understanding of a Riemann sum. Consider a point $p\in C$. We find the component of the vector field at $p$ that points in the direction tangent to the curve, scale it by $\text{d}s$, and sum this over every point on the curve.

Answer 3

This clip gives a pretty good geometric interpretation of what's going on. You're integrating along a differentiable curve $C$ with parametrization $C(t)$, i.e. a function from $[a,b] \to C$ (usually another letter than C is used for the parametrization such as $r$ to avoid confusion) such that $C(a)$ and $C(b)$ are the endpoints of the curve. The derivative of the parametrization gives you the direction of the curve at each pont $t$ since it it is the tangential line of the curve at $t$.

Physically the integral is the work (difference in energy) required for a particle traverse the curve in this vector field, which is different for each dimension, i.e. you have a certain amount of work performed in say the $x$ and $y$ directions, if we are working in $\mathbb{R}^2$. The scalar product in your integrand ensures that the integral gives you the sum of the two directions, i.e. the total work, as it splits up the total work into its components