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Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

On the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$, consider the vector fieldWith $r^2=x^2+y^2$, its partial derivatives satisfyThus the field has zero scalar curl everywhere it is defined. Nevertheless, on any counterclockwise circle ${\gamma }_R(t)=(R\cos t,R\sin t)$ with $R>0$,The radius cancels exactly.

Applying Green theorem to the enclosed disk would be invalid: $G$ is undefined at the origin, inside that disk. Zero curl on the path alone is insufficient.^[ref]

Now take the annulus $A=\{(x,y):1\le r\le 2\}$. Its positive boundary orientation is counterclockwise on the outer circle and clockwise on the inner circle. The complete boundary line integral is thereforeHere Green theorem applies: an open set containing the closed annulus can avoid the origin. The inner contribution remains $-2\pi$ even for an arbitrarily small positive inner radius. Decreasing the hole's area does not decrease this line integral.

The cancellation has a useful proof picture. Cut the annulus along two radial segments into two half-annuli. Each piece has a single simple closed curve as its boundary; along each cut, the two induced boundary orientations oppose, so those line integrals cancel when the two applications of Green theorem are added. Only the outer and inner circles remain.^[ref]

Locally, a choice of angle $\theta$ in polar coordinates has gradient $\nabla \theta =G$. Globally, following a circle once increases that angle by $2\pi$, so it cannot be a single-valued potential of a conservative vector field on the punctured plane. The nonzero closed-path line integral also rules out such a potential by the fundamental theorem for line integrals.

On a simply connected open region, a continuously differentiable planar vector field with zero scalar curl does have a global potential. The hole is what separates the local derivative condition from the global conclusion.^[ref]