False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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FALSE: surface area is the supremum of inscribed polyhedral areas

Statement

The surface area of a smooth surface equals the supremum of the areas of its inscribed triangulated polyhedral surfaces.

Facts & Assumptions

Given: A fixed circular cylinder of radius $r > 0$ and height $H > 0$ .

[L1]

The cylinder admits inscribed Schwarz lanterns whose mesh tends to zero while their polyhedral areas diverge to $+ \infty$ (Schwarz lanterns can have mesh tending to zero while their polyhedral areas diverge).

[L2]

If $a < b$ and $r : [a, b] \to [0, \infty)$ is $C^{1}$ on a neighbourhood of $[a, b]$ and positive on $(a, b)$ , the surface obtained by rotating $r$ about the axis has area $2 π \int_{a}^{b} r (s) \sqrt{1 + r^{'} (s)^{2}} d s$ (The surface of revolution has area $2 π \int_{a}^{b} r (s) \sqrt{1 + r^{'} (s)^{2}} d s$ ).

Refutation

technique · direct

1.1

By [L1], the set of areas of inscribed triangulated polyhedral surfaces for this fixed cylinder is unbounded above.

givenL1

2.1

Its supremum is therefore not a finite real number. The lateral cylinder is the surface of revolution of the constant profile $r$ on $[0, H]$ , which is smooth on all of $R$ and positive, so [L2] gives it the finite surface-integral area $2 π \int_{0}^{H} r \sqrt{1 + 0} d s = 2 π r H$ .

step 1.1L2algebra

3.1

Consequently the proposed supremum does not equal surface area, even when arbitrarily small mesh is imposed.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

L. Brewin, Curvature corrected estimates for geodesic arc-length, Section 3.2 (standard reference, not scraped)