Green's theorem and orientation

The localization proof and corner/edge estimates extend the elementary calculations below to the precisely stated finite regular boundary classes, including holes and cavities. The local proofs retained here do not assume an unproved decomposition.

Local rotation and boundary circulation

Theorem

Green's theorem, circulation form

Let $D$ be a bounded planar region whose boundary $C$ consists of finitely many disjoint piecewise smooth simple closed curves, positively oriented relative to $D$ . Suppose $P,Q$ are $C^1$ on an open neighbourhood of $\overline D$ . Then

\oint_C P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA.

Positive orientation keeps the region to the left as one traverses the boundary: counterclockwise on an outer boundary and clockwise around a hole.

Proof

Complete proof on the double-simple class

Assume first that $D$ has both descriptions $a\le x\le b$ , $g_1(x)\le y\le g_2(x)$ and $c\le y\le d$ , $h_1(y)\le x\le h_2(y)$ , with ordered continuous boundary functions whose graphs are supplied as finitely many piecewise $C^1$ arcs, and the indicated positive orientation. The fundamental theorem of calculus gives

-\iint_D P_y\,dA=\int_a^b[P(x,g_1(x))-P(x,g_2(x))]dx.

The lower boundary is traversed left to right and the upper boundary right to left, so this is precisely their $P\,dx$ contribution; vertical pieces have $dx=0$ . Similarly,

\iint_D Q_x\,dA=\int_c^d[Q(h_2(y),y)-Q(h_1(y),y)]dy.

The right boundary is traversed upwards and the left boundary downwards; horizontal pieces have $dy=0$ . Adding proves Green's formula in this class. If a specific finite decomposition into such pieces is given, with disjoint interiors and matched piecewise smooth boundary arcs, sum their identities. Each internal arc has opposite induced directions and cancels.

The boundary graphs need not have finite graph derivatives at projection endpoints, provided their continuous graphs are supplied as finitely many piecewise $C^1$ arcs. Here is the required parameter argument. For the lower graph put $H(x)=\int_a^xP(s,g_1(s))ds$ . Its integrand is continuous. Along an oriented regular arc with increasing $x=x(t)$ , the chain rule gives $dH(x(t))/dt=P(x(t),g_1(x(t)))x'(t)$ . The fundamental theorem therefore identifies the arc's $P\,dx$ integral with $\int_a^bP(x,g_1(x))dx$ , even if the inverse graph parameter has a vertical tangent at an endpoint. Reversed traversal changes the sign. The same reasoning applies to the upper graph and to the two $Q\,dy$ graphs. This supplies the boundary-integral identifications above without assuming a bounded derivative of each graph function at its endpoints.

In particular the disk $x^2+y^2\le R^2$ , $R>0$ , has continuous bounds $g_{1,2}(x)=\mp\sqrt{R^2-x^2}$ and the analogous horizontal bounds. Its semicircles have regular angle parametrizations, so this disk lies in the proved class. For the annulus $1\le x^2+y^2\le4$ , split into four quadrant pieces: in the first quadrant the vertical bounds are $0\le x\le2$ , $\sqrt{\max(0,1-x^2)}\le y\le\sqrt{4-x^2}$ , and the horizontal description is obtained by exchanging coordinates. They are continuous and their boundary arcs are piecewise $C^1$ . The reflected pieces have matching internal edges, so the stated finite-cancellation extension also covers this annulus.

This proves the formula for the double-simple class and for those given finite decompositions. The general region class in the stated theorem requires an additional decomposition or approximation result; this calculation does not by itself prove that every such region admits the required finite decomposition.

Worked example

Compute circulation through area

Let $C$ be the counterclockwise boundary of the rectangle $0\le x\le2$ , $0\le y\le1$ , and $\mathbf F=(P,Q)=(-y,x^2)$ . Then $Q_x-P_y=2x+1$ , so

\oint_C-y\,dx+x^2\,dy=\int_0^2\int_0^1(2x+1)\,dy\,dx=6.

Reversing the boundary orientation gives $-6$ .

Flux form and area

For the same positively oriented boundary, the outward flux of $(P,Q)$ is $\oint_C P\,dy-Q\,dx$ . Applying the circulation form to $(-Q,P)$ gives

\oint_C P\,dy-Q\,dx=\iint_D(P_x+Q_y)\,dA.

Thus circulation uses planar curl $Q_x-P_y$ ; outward flux uses divergence $P_x+Q_y$ . Taking $P=-y/2,Q=x/2$ in circulation form yields $\operatorname{Area}(D)=\frac12\oint_C(x\,dy-y\,dx)$ for positively oriented $C$ .

Worked example

Why a hole changes the boundary

For the annulus $1\le x^2+y^2\le4$ , the positive boundary consists of the outer circle counterclockwise and the inner circle clockwise. The field $(-y/(x^2+y^2),x/(x^2+y^2))$ is smooth on an open neighbourhood of this annulus and has zero curl there. Its outer circulation is $2\pi$ , and its inner clockwise circulation is $-2\pi$ ; the sum is zero. Applying the theorem to the entire disk would be invalid because the field is undefined at the origin.

Method comparison

Use Green's theorem when a boundary integral has an easier area integral or vice versa. First check smoothness on the whole region, include every boundary component, and determine orientation. Direct parametrization may be simpler for a short curve. A singularity inside a proposed region cannot be ignored merely because the curl vanishes away from it.

Practice

Evaluate $\oint_C(-y\,dx+x\,dy)$ for the counterclockwise circle of radius $R>0$ .
Find outward flux of $(x,y)$ across the boundary of a disk of radius $R>0$ .
State the positive orientation of both boundaries of an annulus.
Explain why Green's theorem on the full unit disk cannot be applied directly to $(-y/(x^2+y^2),x/(x^2+y^2))$ .

Full solutions

$Q_x-P_y=2$ , so the answer is $2\pi R^2$ .
The divergence is $2$ , giving outward flux $2\pi R^2$ .
Outer boundary counterclockwise, inner boundary clockwise; in each case the annular region stays to the left.
The field is undefined at the origin, which lies inside the disk, violating the smoothness hypothesis. Deleting the origin changes the topology and introduces an inner boundary if a small disk is removed.

Synthesis

Green's theorem equates a global oriented boundary integral with accumulated local curl or divergence. Orientation, regularity across the entire region, and all holes are part of the theorem—not optional bookkeeping.

Website function storage (used when needed)