Scalar line integrals and wire mass

Try

A thin wire follows the line segment from (0,0,0) to (1,2,2), with constant density 2 mass units per unit length. The segment length is 3, so mass is 6. This is a one-dimensional material embedded in three-dimensional space. Its mass should scale with length, not the area of a projection or the volume of a surrounding box. A scalar line integral generalizes this simple density-times-length calculation to curved wires and varying density.

Predict

Predict what happens if the same wire is traversed backward during the calculation. The material and density have not changed, so its mass must stay the same. This gives an early sign check: scalar arc-length accumulation is independent of traversal orientation. If a calculation reverses the mass’s sign, it has probably used a coordinate differential such as dx where an unsigned length element ds was required.

Experiment

Parameterize the segment as r(t)=(t,2t,2t), 0≤t≤1. Its derivative is (1,2,2), with magnitude 3. Therefore ds=3dt and mass is ∫₀¹2×3dt=6. Now let density at the point r(t) be 1+t. The mass becomes ∫₀¹(1+t)3dt=9/2.

Reparameterize with q(u)=r(u/2), 0≤u≤2. The density becomes 1+u/2 and speed becomes 3/2. Integrating their product again gives 9/2. Both the composed density and the speed must change; changing only the bounds would be inconsistent.

Understand

For a piecewise continuously differentiable curve r:[a,b]→R³, the scalar line integral is ∫C f ds=∫ₐᵇf(r(t))|r′(t)|dt, under the usual continuity assumptions on f. The speed factor converts parameter increments to arc length. The curve should cover the intended wire once except for harmless endpoint coincidences; repeated traversal counts repeated accumulation.

For a circle of radius R with constant density ρ, the formula gives 2πRρ. For variable density, compose its spatial formula with the curve before integrating. If f can be negative, the integral is a signed scalar accumulation, but reversing traversal still does not change its sign: the sign comes from f, not from orientation.

Apply

Take the upper semicircle r(t)=(cos t,sin t), 0≤t≤π, and density f(x,y)=y+2. Speed is 1, so mass is ∫₀π(sin t+2)dt=2+2π. A constant density equal to 2 would contribute 2π; the additional density y contributes 2. Both parts are nonnegative on this wire.

Repair “a scalar line integral is ∫f dx.” On this semicircle, dx=−sin t dt and changes according to horizontal motion, while ds=dt. A vertical segment has dx=0 but positive length. Coordinate differentials can appear in work integrals, but they measure directed coordinate change, not material length.

Continue

For optional depth, derive orientation invariance using a decreasing reparametrization. The chain rule introduces |φ′| into the speed, while the reversed substitution endpoints account for the sign of dt. The resulting accumulated length remains positive and the scalar value is unchanged. This argument assumes the same multiplicity of coverage.

Write a wire-mass setup with a diagram, parameter interval, composed density and speed on separate lines. Check units: density per length multiplied by length gives mass. Next, vector line integrals will replace density times speed with a dot product against directed velocity. That different integrand will make reversal change the sign. The physical interpretation determines the appropriate differential element.