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

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Green's theorem for finite unions of elementary regions

Statement

Let $D = D_{1} \cup \dots \cup D_{N}$ be a finite elementary Green region with its supplied decomposition, and orient $\partial D$ positively. If $P, Q$ are $C^{1}$ on an open neighbourhood of $D$ , then

$\int_{\partial D} P d x + Q d y = \iint_{D} (\partial_{x} Q - \partial_{y} P) d A .$

Facts & Assumptions

Given: The finite elementary Green region, decomposition, orientation, and functions in the Statement.

[L1]

Every elementary piece has both a Type I and a Type II description (Type I, Type II, and elementary regions for Green's theorem).

[L2]

On a Type I piece, $\int_{\partial D_{ℓ}} P d x = - \iint_{D_{ℓ}} \partial_{y} P d A$ (The Type I boundary identity for the P dx term).

[L3]

On a Type II piece, $\int_{\partial D_{ℓ}} Q d y = \iint_{D_{ℓ}} \partial_{x} Q d A$ (The Type II boundary identity for the Q dy term).

[L4]

Boundary integrals and integrals of a continuous scalar field add from the pieces to the union, with shared arcs cancelling (Shared boundary arcs cancel when finitely many elementary regions are glued).

[L5]

The vector line integral for the field $(P, Q)$ is $\int P d x + Q d y$ (Scalar line integrals with respect to arc length and vector-field line integrals).

Proof

technique · direct

1.1

Fix a piece $D_{ℓ}$ . By [L1], [L2], and [L3], adding its Type I and Type II identities gives $\int_{\partial D_{ℓ}} P d x + Q d y = \iint_{D_{ℓ}} (\partial_{x} Q - \partial_{y} P) d A .$

givenL1L2L3algebra

2.1

Sum step 1.1 over the nonempty finite decomposition. Apply both clauses of [L4] to replace the sums by the boundary and region integrals over $D$ ; [L5] identifies the boundary integrand. This is the displayed Green identity.

step 1.1L4L5algebra

3.1

The case $N = 1$ is included in step 2.1 with no internal cancellation. The proof uses the supplied elementary decomposition and makes no assertion for an arbitrary Jordan domain.

givenstep 2.1L1∎

Depends on

Used by

Dependency tree · two levels

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Sources

J. Lebl, Basic Analysis II, Theorem 10.6.1 (standard reference, not scraped)