Vector line integrals, work and circulation

Try

A constant force F=(3,1) moves an object from A=(0,0) to B=(2,1). Its work is F·(B−A)=7 in compatible force-distance units. Only the force component along movement contributes. Moving perpendicular to the force contributes no work even when the force is large. This directional mechanism differs from a scalar wire-density integral, where every positive length piece contributes positive mass.

Predict

Predict the work when the path is traversed from B back to A. The displacement reverses, so the work is −7. For a constant force, the total depends only on endpoints, even on a curved path. For a variable field, that endpoint property requires additional structure. It is a conclusion to establish, not a default assumption based on the constant-force example.

Experiment

Compare two paths from (0,0) to (2,1) for F=(y,0). On the straight path r(t)=(2t,t), 0≤t≤1, the field is (t,0), velocity is (2,1), and work is ∫₀¹2t dt=1. On a path that first moves horizontally along y=0 and then vertically, both pieces have zero work: the first has zero field and the second has zero horizontal displacement.

Change the order of the two segments, moving up first and right second. The horizontal segment now lies at y=1 and contributes 2. Record all three results. The endpoints stayed fixed while the path changed.

Understand

For a continuous vector field F and a piecewise smooth oriented curve r, the work integral is ∫C F·dr=∫ₐᵇF(r(t))·r′(t)dt. In coordinates this is ∫P dx+Q dy+R dz. Unlike ds, the vector differential dr retains direction. A monotone orientation-preserving reparametrization leaves work unchanged; reversing orientation negates it.

Circulation is work around a closed oriented curve. For F=(−y,x) on the unit circle r=(cos t,sin t), 0≤t≤2π, the field equals the tangent r′. Their dot product is 1, so counterclockwise circulation is 2π. The field continually aligns with motion rather than canceling merely because the endpoint returns to its start.

Apply

Repair “closed path means zero work because displacement is zero.” That is true for a constant field, whose components can be pulled outside the integrals. The rotating field above is not constant and has nonzero circulation. The integral combines changing field values with changing tangents; it is not generally a single field vector dotted with total displacement.

For piecewise paths, integrate each piece with its actual orientation and add. At a corner the tangent can jump, but a finite number of corners does not prevent a piecewise smooth line integral. Reusing a parameter symbol is harmless only if each piece’s formula and interval are clearly stated.

Continue

For optional depth, interpret F·dr as tangential field component times ds. Writing T=r′/|r′| on regular pieces gives F·dr=(F·T)ds. This connects directed work with scalar integration while preserving the signed tangential component. Large field magnitude alone does not determine work; alignment matters.

Save a comparison with the same endpoints and different path integrals, including the intermediate dot products. It demonstrates path dependence for that field without requiring an abstract theorem. The next lesson asks which fields admit a potential and how such a potential makes every path integral an endpoint difference. Domain holes will matter as much as derivative calculations.