Gradient theorem

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $\phi \colon U \to \mathbb{R}$ be of class $C^1$, and let $\gamma \colon [a, b] \to U$ be a piecewise $C^1$ curve. Then $$\int_\gamma \nabla \phi \cdot d\mathbf{r} = \phi(\gamma(b)) - \phi(\gamma(a)).$$

Remarks

Proof for a $C^1$ curve: by the chain rule along a curve, $(\phi \circ \gamma)'(t) = \nabla \phi(\gamma(t)) \cdot \gamma'(t)$. Then the parametric formula and the fundamental theorem of calculus give the result. For a piecewise $C^1$ curve, add the pieces. So the line integral of a gradient field depends only on the end points. Around a closed curve it is $0$. Example: for $\phi(x, y) = x^2 y$ and any path from $(0, 0)$ to $(1, 2)$, $\int_\gamma (2 x y, x^2) \cdot d\mathbf{r} = 2$. This is the case of curves in the generalised Stokes theorem: the boundary of a curve consists of its two end points.

Explanations

L1 human

This is the fundamental theorem of calculus for curves. To integrate the gradient of $\phi$ along a path, you only need the values of $\phi$ at the end and at the start.