Line integral of a vector field

Definition

Let $\gamma \colon [a, b] \to \mathbb{R}^n$ be a rectifiable curve with image $C$, and let $F \colon C \to \mathbb{R}^n$ be a continuous vector field. For a partition $a = t_0 < t_1 < \dots < t_k = b$ and points $\tau_i \in [t_{i-1}, t_i]$, consider the sum $\sum_{i=1}^{k} F(\gamma(\tau_i)) \cdot \big( \gamma(t_i) - \gamma(t_{i-1}) \big)$. The line integral of $F$ along $\gamma$ is the limit of these sums as $\max_i (t_i - t_{i-1}) \to 0$. It is written $$\int_\gamma F \cdot d\mathbf{r} \quad \text{or} \quad \int_\gamma F_1 \, dx_1 + \dots + F_n \, dx_n.$$

Remarks

The limit exists, and it does not depend on the choice of the points $\tau_i$. For a $C^1$ curve, $\int_\gamma F \cdot d\mathbf{r} = \int_a^b F(\gamma(t)) \cdot \gamma'(t) \, dt$, by the parametric formula. For a regular curve with unit tangent vector $T$, $\int_\gamma F \cdot d\mathbf{r} = \int_\gamma (F \cdot T) \, ds$. So it is the line integral of the scalar field $F \cdot T$. The value depends on the direction of travel. Reversing the path changes its sign.

Explanations

L1 human

At each small step along the curve, take the dot product of the field with the step. This counts only the part of the field that pushes along the curve, and the line integral is the total of these contributions.