Green's theorem and oriented planar boundaries

Try

Take the field F=(0,x) around the unit square. Traverse its boundary counterclockwise: the bottom edge contributes zero because dy=0; the right edge contributes 1 because x=1 and y increases from 0 to 1; the top has dy=0; the left has x=0. Total circulation is 1. Inside the square, the planar curl Q_x−P_y is also 1. Green’s theorem relates these two calculations.

Predict

Predict what changes if the square is traversed clockwise. Each directed edge reverses, so the circulation becomes −1. The ordinary area integral of curl remains 1 unless the theorem’s orientation convention is also changed. Positive boundary orientation means the region lies on your left as you walk along each boundary component. For an outer boundary this is counterclockwise; around a hole it is clockwise.

Experiment

Compute circulation of F=(−2y,x) around a counterclockwise rectangle of width 3 and height 1. Curl is 1−(−2)=3, so the answer is 9. Verify by summing the four edges: bottom 0, right 3, top 6, left 0. On the top edge, P=−2 while dx is negative, producing positive work. This is a useful orientation check because an incorrect direction would reverse that contribution.

Change the rectangle’s height to 2. Both its area and the theorem’s prediction double. The field stays fixed; the region supplies how much curl is accumulated.

Understand

For a bounded planar region with a positively oriented piecewise smooth boundary, and P,Q with continuous first partials on an open neighborhood of the region, Green’s theorem states ∮P dx+Q dy=∫∫(Q_x−P_y)dA. Regions with finitely many suitable holes use all boundary components with their induced directions.

For a rectangle, the fundamental theorem of calculus converts the difference between values on opposite edges into an integral of a partial derivative. Combining the horizontal and vertical pairs produces Q_x−P_y. Subdivision explains the broader region theorem: interior edges appear twice in opposite directions and cancel, leaving only the external boundary and any hole boundaries. This is the mechanism, while a complete proof handles the limiting geometry carefully.

Apply

Use P=−y/2 and Q=x/2. The curl is 1, so area equals (1/2)∮(x dy−y dx) for a positively oriented simple closed boundary. This formula computes area from boundary data, but its sign changes if orientation is reversed. For a self-intersecting curve, interpretation requires signed multiplicities rather than automatically the ordinary enclosed area.

Repair an invalid application to the angular field with a singularity at the origin. Although its curl is zero away from zero, Green’s theorem cannot be applied to the full disk: the field is not C¹ on a neighborhood of that disk. On an annulus excluding the origin, the theorem is valid, and the inner clockwise circulation cancels the outer counterclockwise circulation.

Continue

For optional depth, Green’s theorem also has a flux form. For F=(P,Q), outward planar flux is ∮P dy−Q dx under positive boundary traversal; its area integral is ∫∫(P_x+Q_y)dA. This is a two-dimensional divergence statement, distinct from the circulation form’s curl.

Save a theorem application with the field domain, region, boundary arrows and derivatives visible before the final integral. Include one failed-hypothesis example and explain exactly what the theorem cannot conclude there. Next, surface parametrizations will generalize area elements to curved patches in three dimensions, preparing the corresponding circulation and flux theorems.