ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Opposite parametrizations preserve area and negate flux

Example

The parametrizations $φ (u, v) = (u, v, 0)$ and $ψ (s, t) = (t, s, 0)$ of the horizontal unit square both give area $1$ . For the constant field $F = (0, 0, 1)$ , their fluxes are $1$ and $- 1$ respectively.

Facts & Assumptions

Given: The two maps on $[0, 1]^{2}$ and the coordinate swap $h (s, t) = (t, s)$ , so $ψ = φ \circ h$ .

[L1]

Regular reparametrization leaves area unchanged and negates flux when it reverses orientation (Surface area and scalar surface integrals are invariant under regular reparametrization, Flux is invariant under orientation-preserving reparametrization and changes sign under reversal).

[L2]

A regular patch has nonzero parameter cross product in the interior and no interior parameter point shares its image with a distinct point of the parameter region (Regular parametrized surface patches on compact Jordan parameter regions).

[L3]

The area of a regular patch is $\int_{D} J_{φ}$ with $J_{φ} = ∥ φ_{u} \times φ_{v} ∥_{2}$ , and the flux of a continuous field $F$ in the orientation induced by $φ$ is $\int_{D} (F \circ φ) \cdot (φ_{u} \times φ_{v})$ , the cross product being given by the coordinate formula (Surface area and scalar surface integrals on a regular patch, The surface area density is the norm of the cross product of the parameter tangents, Unit normal fields, orientations, and flux through a regular surface patch, The cross product in $R^{3}$ ).

Verification

technique · direct

1.1

The derivatives of $φ$ are $(1, 0, 0)$ and $(0, 1, 0)$ , with cross product $(0, 0, 1)$ ; those of $ψ$ are reversed, with cross product $(0, 0, - 1)$ . Both maps meet [L2], and $\det D h = - 1$ .

givenL2algebra

2.1

By [L3] the two area integrands are the constant $∥ (0, 0, \pm 1) ∥_{2} = 1$ and the two flux integrands are $(0, 0, 1) \cdot (0, 0, 1) = 1$ and $(0, 0, 1) \cdot (0, 0, - 1) = - 1$ . Integrating over the unit square gives area $1$ for both maps and fluxes $1$ and $- 1$ .

step 1.1L3algebra

3.1

The calculation agrees with [L1]: the coordinate swap preserves area and reverses the flux sign.

step 1.1step 2.1L1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

University of Toronto MAT237 notes, Section 5.3, An Invariance Property (standard reference, not scraped)