Curvilinear Integrals

Motivation: Work, Wire Mass, Flux

Three problems lead to curvilinear integrals. The first: find the mass of a thin wire with variable linear density ρ(x, y)—you cannot simply multiply length by density if the density changes. The second: find the work performed by a force field F(x, y) = (P, Q) when moving a particle along a curve—the force changes along the path and is not always directed along the movement. The third: compute the flux of a field through a curve (the two-dimensional analog of flux through a surface). All three problems lead to two types of curvilinear integrals.

The Curve and its Parameterization

A smooth curve C is defined parametrically: r(t) = (x(t), y(t)), t ∈ [a, b], where x(t), y(t) are continuously differentiable and |r'(t)| = √(x'² + y'²) ≠ 0 (no “stops”). Differential arc length element: ds = |r'(t)| dt = √(x'²(t) + y'²(t)) dt.

Example: semicircle C: x = cos t, y = sin t, t ∈ [0, π]. Length: ∫₀^π |r'(t)| dt = ∫₀^π 1 dt = π. ✓

Curvilinear Integral of the First Kind (by Arc Length)

∫_C f(x, y) ds = ∫ₐᵇ f(x(t), y(t)) |r'(t)| dt.

This is the integral of a scalar function f over the length of the curve. Does not depend on the orientation of the curve (the sign of ds is always positive).

Application — wire mass: M = ∫_C ρ(x, y) ds, where ρ is the linear density (kg/m).

Example: Find the mass of the semicircle x² + y² = R², y ≥ 0, with density ρ = y/R.

Parameterization: x = R cos t, y = R sin t, t ∈ [0, π], ds = R dt.

M = ∫₀^π (R sin t / R) · R dt = R ∫₀^π sin t dt = R · 2 = 2R.

Check: the greatest density is at the top point (t = π/2), the least at the ends (t = 0, π), so the result 2R < πR (maximum mass) is reasonable.

Curvilinear Integral of the Second Kind (by Coordinates)

∫_C P dx + Q dy = ∫ₐᵇ [P(x(t), y(t)) x'(t) + Q(x(t), y(t)) y'(t)] dt.

This is an integral of the “differential form” ω = P dx + Q dy. Depends on orientation: when traversing the curve in the reverse direction, the integral changes sign.

Physical meaning — field work: A = ∫_C F·dr = ∫_C P dx + Q dy, where F = (P, Q) is the force, dr = (dx, dy) is the path element. At each point, we take the projection of the force onto the tangent and integrate over the length.

Expanded example. Calculate the work of the field F = (−y, x) along the unit circle (counterclockwise).

Parameterization: x = cos t, y = sin t, t ∈ [0, 2π].

A = ∫₀^{2π} [(−sin t)(−sin t) + (cos t)(cos t)] dt = ∫₀^{2π} (sin²t + cos²t) dt = ∫₀^{2π} dt = 2π.

The field F = (−y, x) is a rotation (vortex) field. It is always directed perpendicular to the radius vector, i.e., along the circle. Therefore, the work is proportional to the length of the contour (2πR = 2π for R = 1).

Green's Theorem

This is the main theorem regarding curvilinear integrals in the plane. It connects the “global” (integral over the contour) with the “local” (double integral over the region) and is a special case of the generalized Stokes theorem.

Green's Theorem: If D is a region with piecewise-smooth boundary ∂D (traversed counterclockwise), P, Q are continuously differentiable in the closure of D, then:

∮_{∂D} P dx + Q dy = ∬_D (∂Q/∂x − ∂P/∂y) dA.

The quantity ∂Q/∂x − ∂P/∂y is the rotor (curl) of the field F = (P, Q) in the two-dimensional case.

Example — area calculation. For P = 0, Q = x: ∮{∂D} x dy = ∬D 1 dA = S(D). Similarly, P = −y, Q = 0: −∮{∂D} y dx = S(D). Average: S = (1/2) ∮{∂D} (x dy − y dx). This gives a method to compute area via a boundary traversal.

Example: area of ellipse x = a cos t, y = b sin t.

S = (1/2) ∫₀^{2π} (a cos t · b cos t − b sin t · (−a sin t)) dt = (1/2) ∫₀^{2π} ab dt = πab. ✓

Potential Fields and Path Independence

The field F = (P, Q) is called potential (or conservative) if there exists a function φ: ℝ² → ℝ (the potential, scalar potential) such that F = ∇φ.

Criteria for potentiality (in a simply connected region):

1. F = ∇φ for some φ.
2. ∂P/∂y = ∂Q/∂x (curl-free field, rotor = 0).
3. ∮_L F·dr = 0 for any closed contour L.
4. ∫_C F·dr does not depend on path (only on start and end).

All four conditions are equivalent in a simply connected region.

For a potential field: ∫_C F·dr = φ(r(b)) − φ(r(a)) — analogous to the Newton–Leibniz theorem.

Finding the potential. If ∂P/∂y = ∂Q/∂x, find φ from P = ∂φ/∂x, Q = ∂φ/∂y.

Example: P = 2xy³, Q = 3x²y². Check: ∂P/∂y = 6xy² = ∂Q/∂x. ✓ Potential: φ = ∫P dx = x²y³ + C(y). From Q = ∂φ/∂y = 3x²y² + C'(y) = 3x²y² → C' = 0 → φ = x²y³.

In physics: gravitational and electrostatic fields are potential. The potential φ = −U, where U is potential energy. Work in a gravitational field does not depend on the path—only on height difference.

Thought question: The field F = (−y/(x²+y²), x/(x²+y²)) for (x, y) ≠ (0, 0) has ∂P/∂y = ∂Q/∂x, but is not potential on ℝ² \ {0}. Compute ∮_C F·dr over the unit circle and explain this phenomenon.