Line integral as an integral over the parameter interval

Statement

Let $\gamma \colon [a, b] \to \mathbb{R}^n$ be a $C^1$ curve with image $C$. Let $f \colon C \to \mathbb{R}$ be continuous, and let $F \colon C \to \mathbb{R}^n$ be a continuous vector field. Then $$\int_\gamma f \, ds = \int_a^b f(\gamma(t)) \, \|\gamma'(t)\| \, dt, \qquad \int_\gamma F \cdot d\mathbf{r} = \int_a^b F(\gamma(t)) \cdot \gamma'(t) \, dt.$$

Remarks

The left sides are the line integral of a scalar field and the line integral of a vector field. The right sides are ordinary integrals in one variable. The formula follows from $\Delta s_i = \int_{t_{i-1}}^{t_i} \|\gamma'(t)\| \, dt$, the mean value theorem for each component of $\gamma$, and the uniform continuity of $f \circ \gamma$ and $\gamma'$. The formulas extend to piecewise $C^1$ curves, by summing over the pieces. Example: for $F(x, y) = (-y, x)$ and $\gamma(t) = (\cos t, \sin t)$, $t \in [0, 2\pi]$, the integrand is $\sin^2 t + \cos^2 t = 1$, so $\int_\gamma F \cdot d\mathbf{r} = 2\pi$.

Explanations

L1 human

To compute a line integral, write the curve with a parameter $t$. Then $ds$ becomes $\|\gamma'(t)\| \, dt$ and $d\mathbf{r}$ becomes $\gamma'(t) \, dt$, and the line integral is an ordinary integral in $t$.