Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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 +∞ (Schwarz lanterns can have mesh tending to zero while their polyhedral areas diverge).

[L2]

If a<b and r:[a,b]→[0,∞) is C1 on a neighbourhood of [a,b] and positive on (a,b), the surface obtained by rotating r about the axis has area 2π∫abr(s)1+r′(s)2 ds (The surface of revolution has area 2π∫abr(s)1+r′(s)2 ds).

Refutation

technique · direct
1.1givenL1

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

2.1step 1.1L2algebra

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 2π∫0Hr1+0 ds=2πrH.

3.1step 1.1step 2.1∎

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

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources