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Green's theorem is the curl statement for a planar field lifted to R3

Statement

Let D be a finite elementary Green region with its supplied decomposition, positively oriented, and let P,Q be C1 on an open U⊆R2 containing D. Define the lift

F~(x,y,z):=(P(x,y), Q(x,y), 0)((x,y,z)∈U×R),

a field on the open set U×R⊆R3. Then F~ is C1, its curl has first and second coordinates identically 0 and third coordinate ∂xQ−∂yP at every point, independent of z, and the circulation of the planar field around the positive boundary chain equals the integral of the third coordinate of the curl of the lift:

∫∂D(P,Q)⋅dr=∬D(curl⁡F~)z(x,y,0) dA.

Facts & Assumptions

Given: The finite elementary Green region D with its supplied decomposition and positive orientation, the C1 functions P,Q on the open U⊇D, and the lift F~ of the Statement.

[F1]

The curl of a C1 field on an open subset of R3 is curl⁡G=(∂yGz−∂zGy, ∂zGx−∂xGz, ∂xGy−∂yGx) (Divergence and curl of a C1 vector field).

[F2]

A map is of class Ck when each component is, a scalar component being C1 when its first partial derivatives exist and are continuous (Ck Euclidean maps and diffeomorphisms).

[F3]

For a finite elementary Green region the positive boundary integral is the finite sum over the surviving oriented arcs, and ∫∂DG⋅dr and ∫∂DP dx+Q dy denote that sum for the field (P,Q) (Positive orientation of elementary-region boundaries, Scalar line integrals with respect to arc length and vector-field line integrals).

[F4]

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).

[L1]

Let D=D1∪⋯∪DN be a finite elementary Green region with its supplied decomposition, oriented positively, and let P,Q be C1 on an open neighbourhood of D. Then ∫∂DP dx+Q dy=∬D(∂xQ−∂yP) dA (Green's theorem for finite unions of elementary regions).

Proof

technique · direct
1.1givenF2F5

The three components of F~ are (x,y,z)↦P(x,y), (x,y,z)↦Q(x,y) and the constant 0. Their first partial derivatives are ∂xF~x=∂xP, ∂yF~x=∂yP, ∂zF~x=0; ∂xF~y=∂xQ, ∂yF~y=∂yQ, ∂zF~y=0; and all three of ∂xF~z, ∂yF~z, ∂zF~z are 0. Each of these exists and is continuous on U×R because P and Q are C1 on U, so F~ is C1 there by [F2].

2.1step 1.1F1

By [F1] and step 1.1 the three coordinates of curl⁡F~ are ∂yF~z−∂zF~y=0−0=0, then ∂zF~x−∂xF~z=0−0=0, and then ∂xF~y−∂yF~x=∂xQ−∂yP. All three are computed, and the third depends only on (x,y), so its value at (x,y,z) is its value at (x,y,0).

3.1step 2.1F3F4L1∎

By [F4] the region D carries its supplied decomposition and P,Q are C1 on the open neighbourhood U of D, so [L1] gives ∫∂DP dx+Q dy=∬D(∂xQ−∂yP) dA; by [F3] the left side is ∫∂D(P,Q)⋅dr, and by step 2.1 the integrand on the right is (curl⁡F~)z(x,y,0). That is the asserted identity, and step 2.1 is the assertion about the three curl coordinates.

Remarks

  • This is a dictionary, not a new theorem. Both sides are the two sides of Green's theorem, rewritten. What the corollary records is that the planar integrand ∂xQ−∂yP is a curl, so that the planar and the spatial developments on this page speak about one operator rather than two unrelated ones.

  • The route is deliberately one-way. The classical Stokes theorem for a C2 patch over a finite elementary Green region is proved from Green's theorem, so re-deriving Green's theorem from it would be circular. Nothing above uses Stokes' theorem.

Depends on

Used by

Dependency tree · two levels

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Sources