Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
How much does a curved wire weigh if its density changes along its length? Cut it into small pieces, multiply each piece's length by its density, and add. That sum, in the limit, is a line integral of a function along a curve. It is the arc length integral with a weight $f$ attached.
🔑 Idea clave
For a smooth curve $C$ given by $\mathbf{r}(t)$, $a \le t \le b$, and a continuous function $f$:
$$\int_C f\,ds = \lim \sum f(P_i^*)\,\Delta s_i = \int_a^b f(\mathbf{r}(t))\,|\mathbf{r}'(t)|\,dt.$$
The value does not depend on the parametrization or on the direction of travel. With $f = 1$ it gives the length of $C$; with a density $\delta$ it gives the mass of a wire.
💡 Por qué funciona
Why the factor $|\mathbf{r}'(t)|$: a small step $\Delta t$ in the parameter moves the point by about $\mathbf{r}'(t)\,\Delta t$, a piece of curve of length $\Delta s \approx |\mathbf{r}'(t)|\,\Delta t$. So $f\,\Delta s \approx f(\mathbf{r}(t))\,|\mathbf{r}'(t)|\,\Delta t$, and adding these gives the $t$-integral. Why direction does not matter: $\Delta s$ is a length, always positive. Tracing $C$ backward visits the same pieces with the same lengths, so the sum is unchanged.
The curve $\langle t, t^2/2 \rangle$ with its velocity $\mathbf{r}'(t) = \langle 1, t \rangle$. Near $t = 0$ the arrow is short: a step $dt$ covers little curve. Near the ends it is long. The factor $|\mathbf{r}'(t)| = \sqrt{1 + t^2}$ in $ds$ converts steps in $t$ into length.
Worked examples
✏️ Ejemplo resuelto: Along a segment
Evaluate $\int_C (x + y)\,ds$, where $C$ is the segment from $(0, 0)$ to $(3, 4)$.
$\mathbf{r}(t) = \langle 3t, 4t \rangle$, ${}0 \le t \le 1$, and $|\mathbf{r}'(t)| = |\langle 3, 4 \rangle| = 5$.
On $C$, $x + y = 7t$.
$\int_0^1 7t \cdot 5\,dt = \frac{35}{2}$.
Respuesta: $\frac{35}{2}$
✏️ Ejemplo resuelto: Along a quarter circle
Evaluate $\int_C xy\,ds$, where $C$ is the part of $x^2 + y^2 = 4$ in the first quadrant.
$\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$, ${}0 \le t \le \frac{\pi}{2}$, with $|\mathbf{r}'(t)| = 2$.
Dropping $|\mathbf{r}'(t)|$. Writing $ds$ as $dt$ treats every step in $t$ as one unit of length. In the first example that gives $\int_0^1 7t\,dt = \frac{7}{2}$, off by the factor ${}5$, the length of the segment. A quick check: with $f = 1$, your integral should give the length of the curve.
Práctica
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