1. The Physical Motivation: Work Done by a Force Field
Imagine a particle sliding along a curved path through a force field — wind pushing a leaf along a winding trail, or gravity pulling a bead down a wire. At every point the field exerts a force, but only the component of that force along the direction of motion does work. The line integral of the second kind captures exactly this idea.
Recall — Vector Field: A vector field $\mathbf{F}$ assigns a vector to every point in space. You met this idea when studying the gradient: $\nabla f$ is a vector field. Now, instead of just drawing those arrows, we integrate them along a curve.
Building the Definition Step by Step
- Slice the curve into tiny segments $\Delta \mathbf{r}_i$.
- At each segment, the force is approximately $\mathbf{F}(\mathbf{r}_i)$.
- The work on that piece is $\mathbf{F}(\mathbf{r}_i) \cdot \Delta \mathbf{r}_i$ — the dot product extracts the component of force along the displacement.
- Sum over all pieces and take the limit:
Definition — Line Integral of the Second Kind
$$\int_C \mathbf{F} \cdot d\mathbf{r} \;=\; \lim_{n\to\infty} \sum_{i=1}^{n} \mathbf{F}(\mathbf{r}_i) \cdot \Delta \mathbf{r}_i$$Via a parametrisation $\mathbf{r}(t)$, $a \le t \le b$:
$$\boxed{\int_C \mathbf{F} \cdot d\mathbf{r} \;=\; \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt}$$Expanded Component Form
Write $\mathbf{F} = P\,\mathbf{i} + Q\,\mathbf{j} + R\,\mathbf{k}$ and $d\mathbf{r} = dx\,\mathbf{i} + dy\,\mathbf{j} + dz\,\mathbf{k}$. Then:
$$\int_C \mathbf{F} \cdot d\mathbf{r} \;=\; \int_C P\,dx + Q\,dy + R\,dz$$