= Green theorem with holes
{c}
On the <punctured plane> $\mathbb R^2\setminus\{(0,0)\}$, consider the <vector field>
$$
G(x,y)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right).
$$
With $r^2=x^2+y^2$, its <partial derivatives> satisfy
$$
Q_x=\frac{y^2-x^2}{r^4},\qquad
P_y=\frac{y^2-x^2}{r^4},\qquad Q_x-P_y=0.
$$
Thus 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 $R>0$,
$$
G(\gamma_R(t))=\frac1R(-\sin t,\cos t),\qquad
\gamma_R'(t)=R(-\sin t,\cos t),\qquad
\oint_{\gamma_R}G\cdot d\mathbf r=\int_0^{2\pi}1\,dt=2\pi.
$$
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.<18.02SC Multivariable Calculus, Fall 2010/V5. Simply-Connected Regions>[18.02SC Multivariable Calculus, Fall 2010; V5. Simply-Connected Regions, pp. 1–2, §2: excluded origin prevents filling a closed curve within the field's domain]{ref}
Now take the <annulus> $A=\{(x,y):1\leq r\leq2\}$. Its positive <boundary orientation> is counterclockwise on the outer <circle> and clockwise on the inner <circle>. The complete boundary <line integral> is therefore
$$
\oint_{\partial A}G\cdot d\mathbf r
=\underbrace{2\pi}_{\text{outer boundary}}
+\underbrace{(-2\pi)}_{\text{inner boundary}}
=0
=\iint_A(Q_x-P_y)\,dA.
$$
Here <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>.
\Image[/green-theorem-annulus-boundary-cancellation.png]
{title=Zero curl away from an excluded origin: the outer circle contributes two pi, the clockwise inner circle minus two pi, and the annulus boundary sum is zero}
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.<18.02SC Multivariable Calculus, Fall 2010/V5. Simply-Connected Regions>[18.02SC Multivariable Calculus, Fall 2010; V5. Simply-Connected Regions, p. 1, §1, subdivision proof: integrals over cuts cancel in opposite directions]{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.<18.02SC Multivariable Calculus, Fall 2010/V5. Simply-Connected Regions>[18.02SC Multivariable Calculus, Fall 2010; V5. Simply-Connected Regions, pp. 2–3, theorem (3) and Curl Theorem: simply connected domain hypothesis for a global gradient field]{ref}
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