Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24
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Surface area, scalar integrals, and flux over a C1 graph

Statement

Let DR2 be a compact Jordan parameter region and let g be C1 on a neighbourhood of D. For the graph S={(x,y,g(x,y)):(x,y)D} and every continuous real-valued function q on S, Area(S)=D1+g22,SqdS=Dq(x,y,g(x,y))1+g22. For a continuous vector field F on S, upward flux is DF(x,y,g(x,y))(gx,gy,1). For the graph of g over D, the downward flux is the negative of DF(x,y,g(x,y))(gx,gy,1).

Facts & Assumptions

Proof

technique · direct
1.1

By [L1], φx×φy=(gx,gy,1), whose norm is 1+gx2+gy2=1+g22 and which never vanishes. The first two coordinates make φ injective, so it is a regular patch by [L2].

givenL1L2algebra
2.1

Substituting the norm from step 1.1 into the area and scalar-integral definitions in [L2] gives the first two formulas.

step 1.1L2
2.2

Retaining the signed vector from step 1.1 in the flux definition gives the upward formula; the downward orientation uses its negative and therefore negates the integral.

step 1.1L2
3.1

These substitutions establish all four displayed formulas, including the orientation distinction.

step 2.1step 2.2

Depends on

Used by

Dependency tree · two levels

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Sources