CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Area of an elementary Green region as a boundary line integral

Statement

For a finite elementary Green region $D$ with positively oriented boundary,

$cont (D) = \frac{1}{2} \int_{\partial D} (x d y - y d x) = \int_{\partial D} x d y = - \int_{\partial D} y d x .$

Facts & Assumptions

Given: The finite elementary Green region and positive orientation in the Statement.

[L1]

Green's theorem gives $\int_{\partial D} P d x + Q d y = \iint_{D} (\partial_{x} Q - \partial_{y} P) d A$ for $C^{1}$ functions on a neighbourhood of $D$ (Green's theorem for finite unions of elementary regions).

[L2]

For a Jordan set, $\iint_{D} 1 d A = cont (D)$ (The Riemann integral of a bounded function over a bounded Jordan measurable set).

Proof

technique · direct

1.1

Choose $P (x, y) = - y / 2$ and $Q (x, y) = x / 2$ . Then $\partial_{x} Q - \partial_{y} P = 1$ , so [L1] and [L2] give $cont (D) = \frac{1}{2} \int_{\partial D} (x d y - y d x) .$

L1L2algebra

1.2

Choose $P = 0$ and $Q = x$ . Again the scalar curl is $1$ , so [L1] and [L2] give $cont (D) = \int_{\partial D} x d y$ .

L1L2algebra

1.3

Choose $P = - y$ and $Q = 0$ . Its scalar curl is $1$ , so [L1] and [L2] give $cont (D) = - \int_{\partial D} y d x$ .

L1L2algebra

2.1

Steps 1.1 to 1.3 are the three asserted formulas. They include the one-piece case because [L1] includes every nonempty finite elementary decomposition.

step 1.1step 1.2step 1.3L1∎

Depends on

Used by

The area theorem for exterior univalent functions Theorem

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, section 10.6 (standard reference, not scraped)