Green's Theorem

A closed line integral runs around the edge of a region. Green's theorem says that circulation equals something accumulated inside:

Two descriptions of the same situation, and the theorem converts between them. Sometimes the boundary walk is easier; sometimes the region integral is.

The area case. Choose $P=-y$ and $Q=x$. The region integrand becomes the constant $2$, so the right side is twice the area, and

A two-dimensional quantity computed by walking only the edge. On a polygon this collapses to the shoelace formula, and it is what a mechanical planimeter does by tracing an outline.

Three hypotheses, each doing work. The curve must be simple and closed, the orientation positive, and the partials continuous on the enclosed region rather than merely on the curve. That last one is where the punctured-plane field escapes: it is perfectly smooth on the unit circle and undefined at the centre, so the theorem never applied, and its loop integral of $2\pi$ is not a counterexample to anything.

This unit covers applying the theorem in both directions, computing both sides as a genuine check, and recognising when an enclosed singularity voids the hypothesis.

Green's theorem's three hypotheses, each with a job.

• Simple closed curve: a curve crossing itself encloses regions with conflicting orientations, and the region integral has no single meaning.
• Positive orientation: reversing it negates the left side and not the right, so the identity holds with one specific sign convention.
• Continuous partials on the enclosed region: a singularity inside $R$ makes the double integral improper. This is precisely how the punctured-plane field escapes the conclusion, its closed integral is $2\pi$ rather than $0$ because the theorem never applied.

Green's theorem. For a positively oriented simple closed curve bounding a region on which $P$ and $Q$ have continuous partials, ${\oint }_{C}Pdx+Qdy={\iint }_R(Q_x-P_y)dA$: the circulation around the boundary equals the integrated local circulation inside. Choosing $P$ and $Q$ to make the integrand constant turns the boundary integral into an area.

To apply Green's theorem.

1. Confirm the curve is closed and simple, and orient it counterclockwise. If it is given clockwise, negate the result at the end.
2. Confirm $P$ and $Q$ have continuous partials everywhere inside, not merely on the curve. A singularity enclosed by the curve voids the theorem, which is exactly the punctured-plane case.
3. Compute the integrand $\partial Q/\partial x-\partial P/\partial y$ and integrate it over the enclosed region.
4. Where both sides are computable, compute both. Agreement is a genuine check; deriving one from the other checks nothing.

To compute an area from a boundary. Use $\frac{1}{2}{\oint }_{C}xdy-ydx$ counterclockwise. On a polygon, sides through the origin and sides where one coordinate is constant at zero contribute nothing, so the work is usually a fraction of what the vertex count suggests.

Checks. Confirm a Green's theorem result against a direct boundary walk whenever the boundary is polygonal, since that is inexpensive and the agreement is a genuine check on both.

Part 3: Green's theorem, both sides computed independently.

Take $\mathbf{F}=(-y,x)$ again on the unit square $[0,1]\times [0,1]$, traversed counterclockwise.

Left side, walk the boundary.

Sum: $0+1+1+0=2$.

Right side, integrate over the region. ${\displaystyle \frac{\partial Q}{\partial x}}-{\displaystyle \frac{\partial P}{\partial y}}=1-(-1)=2$, so

Both sides give $2$. Neither was derived from the other, so the agreement is a check on the work.

Note the top leg: traversing counterclockwise means going from $x=1$ to $x=0$, and ${\int }_1^0(-1)dx=+1$. Getting the orientation wrong here would give $-1$ and a total of $0$, disagreeing with the region integral, which is how the check earns its place.

Part 4: area from the boundary alone.

With $P=-y$ and $Q=x$ the region integrand is the constant $2$, so $\frac{1}{2}\oint xdy-ydx$ gives the area. Apply it to the triangle with vertices $(0,0)$, $(4,0)$, $(0,3)$.

• $(0,0)\to (4,0)$: $y=0$ and $dy=0$, so the integrand vanishes. Contribution $0$.
• $(4,0)\to (0,3)$: parametrise $(4-4t,3t)$, so $dx=-4dt$ and $dy=3dt$. Then $xdy-ydx=(4-4t)(3)-(3t)(-4)=12-12t+12t=12$, and ${\int }_0^{1}12dt=12$.
• $(0,3)\to (0,0)$: $x=0$ and $dx=0$, so the integrand vanishes. Contribution $0$.

against $\frac{1}{2}\times 4\times 3=6$ from the elementary formula.

Two of the three sides contributed nothing, and the area of a two-dimensional region came from a single integral along one edge.

Work and conservative forces. The work done by a force along a route is exactly a line integral. Gravity and electrostatic attraction are gradients of potential energy, so the work depends only on the endpoints, which is why "potential energy at a height" is a meaningful quantity at all. Friction is not a gradient: dragging an object around a loop and back does non-zero work, which is the physical reading of a closed integral that fails to vanish.

The classification in this unit is therefore the mathematical form of a physical distinction: forces that store energy against forces that dissipate it.

Circulation in a flow. For a velocity field, the closed line integral measures net circulation around the loop. Green's theorem says that total equals the accumulated local rotation inside, so a flow with zero local rotation everywhere has zero circulation around any loop that encloses nothing special, and a nonzero circulation indicates something enclosed.

The punctured-plane field is the model case: a vortex whose rotation is concentrated at a point removed from the domain, leaving every local measurement zero while every loop around it registers $2\pi$.

Winding numbers. That $2\pi$ counts how many times a path encircles the origin, and dividing by $2\pi$ gives an integer invariant. A quantity that is locally undetectable and globally nonzero is exactly what a topological invariant is, and this integral is the elementary example.

Planimeters and polygon area. The area formula $\frac{1}{2}\oint xdy-ydx$ computes a region's area from its boundary, which is what a mechanical planimeter does by tracing an outline. Applied to a polygon it collapses to the shoelace formula, and the triangle computed in this unit, area $6$ from a single non-trivial side, is that formula in action. Geographic information systems compute polygon areas this way.

Path-independent quantities in engineering. Where a quantity is known to be conservative, a value can be tabulated per point rather than per route: gravitational potential, electric potential, thermodynamic state functions. The practical gain is precisely the one this unit computes, a route integral replaced by a subtraction, and the practical risk is assuming it for a quantity that circulates.

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What decides in every case. Whether the field is a gradient, and on which region. That single question determines whether a quantity can be tabulated by position at all, and the cross-partial test answers it in two derivatives, subject to the region having no holes, which is where the careful cases live.