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Line Integrals of the First Kind (with respect to Arc Length)

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

1. The Object: Mass of a Curved Wire

Callback — the density-to-total engine: You have already seen the slicing recipe in single-variable calculus: chop an interval into small pieces, multiply a density by a small length, sum, and take a limit. We now run the same engine on a curve instead of a straight interval.

1.1 The Physical Picture

Imagine a thin wire bent into a curve $C$ in space. At each point $(x,y,z)$ on the wire the linear density (mass per unit length) is $\rho(x,y,z)$. How do we find the total mass?

  1. Slice the wire into $n$ tiny arcs of length $\Delta s_i$.
  2. On each arc, pick a sample point $P_i^*$ and read the density $\rho(P_i^*)$.
  3. Multiply: the mass of the $i$-th piece $\approx \rho(P_i^*)\,\Delta s_i$.
  4. Sum: $\displaystyle\sum_{i=1}^{n}\rho(P_i^*)\,\Delta s_i$.
  5. Limit: let $n\to\infty$ (all $\Delta s_i\to 0$).
Definition — Line Integral of the First Kind
$$\int_C f(x,y,z)\,ds \;=\; \lim_{n\to\infty}\sum_{i=1}^{n} f(P_i^*)\,\Delta s_i$$

The measure is the arc-length element $ds$. The integrand $f$ is a scalar function evaluated on the curve.

1.2 Readings Collected at the Door

Physical / Geometric QuantityIntegrand $f$Integral
Total mass of a wire$\rho(x,y,z)$ (linear density)$m = \displaystyle\int_C \rho\,ds$
Total charge on a filament$\lambda(x,y,z)$ (charge density)$Q = \displaystyle\int_C \lambda\,ds$
Average temperature along a path$T(x,y,z)$$\bar{T} = \dfrac{1}{L}\displaystyle\int_C T\,ds$
Arc length itself$f \equiv 1$$L = \displaystyle\int_C 1\,ds$
Identity-pattern callback (third appearance): Setting $f=1$ in any integral-with-respect-to-a-measure recovers the measure of the domain. For arc length: $\int_C 1\,ds = \text{length}(C)$. You saw the same pattern with $\int_a^b 1\,dx = b-a$ and $\iint_D 1\,dA = \text{Area}(D)$.

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