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The planar divergence theorem: the flux form of Green's theorem
Statement
Let be a finite elementary Green region with its supplied decomposition, positively oriented, and let be on an open containing . Then
the right-hand integrand being the divergence of as a field on an open subset of .
Moreover, if is one of the arcs of the positive boundary chain and its derivative is nowhere zero on a piece with continuous derivative extension , then on that piece
where is the unit vector obtained from the tangent by a quarter turn clockwise.
Facts & Assumptions
Given: The finite elementary Green region with its supplied decomposition and positive orientation, and the field on the open .
The divergence of a field on an open subset of is ; for and coordinates named this is (Divergence and curl of a vector field).
For a finite elementary Green region the positive boundary integral is the finite sum over the surviving oriented arcs, written for the field (Positive orientation of elementary-region boundaries).
For a piecewise- path with admissible partition and continuous derivative extensions , and (Scalar line integrals with respect to arc length and vector-field line integrals, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).
A finite elementary Green region is a nonempty finite union of elementary Green regions with pairwise disjoint interiors and the stated shared-arc conditions, supplied as data (Type I, Type II, and elementary regions for Green's theorem).
For , and (The Euclidean inner product on ); a map is when each component is ( Euclidean maps and diffeomorphisms).
Let be a finite elementary Green region with its supplied decomposition, oriented positively, and let be on an open neighbourhood of . Then (Green's theorem for finite unions of elementary regions).
Proof
Put and on . These are on by [F5], since the components of are, so [L1] applies with the supplied decomposition of [F4] and gives .
Fix such an arc of the positive boundary chain, a piece of it carrying a continuous derivative extension with nowhere zero there, and set . By [F5] the vector has norm , since , and . Writing for the direction of is not needed: the map is the quarter turn clockwise, as it carries to and to .
By step 1.1 and [F1], on ; substituting into step 1.1 and reading the left side by [F2] gives the first asserted identity.
By [F3] the integral of over that piece of is , while by step 1.2 and [F5] . The two integrands are equal, so by [F3] the two integrals over that piece agree, which is the second asserted identity.
Remarks
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The first identity needs no regularity of the boundary arcs; the second does. The positive boundary chain of an elementary Green region is built from continuous piecewise- graphs, whose derivative may vanish, and where it vanishes there is no unit tangent and hence no . That is why the normal reading is a separate clause under an extra hypothesis, and why the identity that Green's theorem actually delivers is stated in the form.
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Outwardness of is not claimed here. For a positively oriented boundary the quarter turn clockwise of the tangent does point out of the region, but establishing that at a boundary point requires the same kind of local analysis that At interior base points, the graph faces of an adapted presentation induce the outward unit normal carries out in space, and it is not carried out for plane regions on this page. What is proved is the equality of the two integrals for the stated .
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Why this is called a divergence theorem. The right-hand integrand is the divergence of a field on an open subset of , and the left-hand side is the boundary integral of the normal component. The three-dimensional statement of The divergence theorem for finite gluings of elementary solid regions has the same shape; neither is derived from the other on this page.
Depends on
- Green's theorem for finite unions of elementary regions
- Divergence and curl of a $C^1$ vector field
- Positive orientation of elementary-region boundaries
- Type I, Type II, and elementary regions for Green's theorem
- Scalar line integrals with respect to arc length and vector-field line integrals
- The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$
- Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations
- $C^k$ Euclidean maps and diffeomorphisms
Used by
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Sources
- J. Feldman, A. Rechnitzer and E. Yeager, CLP-4 Vector Calculus (University of British Columbia), section 4.3 (standard reference, not scraped)
- M. Corral, Vector Calculus, chapter 4 (LibreTexts) (standard reference, not scraped)