Green's Theorem

The central idea

Let $C$ be a positively oriented, piecewise smooth, simple closed curve bounding a planar region $D$, and let $P,Q$ have continuous partial derivatives on an open set containing $D$. Then $\oint_C Pdx+Qdy=\iint_D(Q_x-P_y)dA$. The flux form is $\oint_C Pdy-Qdx=\iint_D(P_x+Q_y)dA$.

Definitions, hypotheses, and notation

Multiply connected regions require every boundary component with induced orientation: the outer boundary is counterclockwise and hole boundaries clockwise, keeping the region on the left. If a field is singular in a hole excluded from $D$, the theorem may still apply on the remaining region, but not across the singular point.

Area can be computed from line integrals such as $A=\frac12\oint_C xdy-ydx$. This follows by choosing a field with scalar curl one. Different choices yield equivalent area formulas, and orientation controls whether the signed result is positive.

Conceptual meaning

Green's theorem converts circulation or flux around a boundary into accumulated local rotation or divergence across the interior. Positive orientation is counterclockwise: as the boundary is traversed, the region stays on the left.

A dependable method and decision rule

1. Verify $C$ is closed, simple or properly decomposed, and positively oriented.
2. Choose circulation or flux form from the differential expression.
3. Compute the scalar curl or planar divergence.
4. Describe and integrate over $D$ in suitable coordinates.
5. Reverse the sign if the original curve is clockwise.

Fully worked example

Graphical or geometric meaning

Tiny counterclockwise circulations around interior cells cancel on shared edges. Only the outer boundary remains. This cancellation explains both the orientation and the transition from local curl density to global circulation.

Common mistakes and why they fail

Verification and reasonableness checks

• Test orientation with the region-on-left rule.
• Compare with direct parametrization for a circle or rectangle.
• Check units: derivative density times area equals line-integral units.

Orientation and holes determine the boundary

For a positively oriented simple closed curve $C$ bounding a planar region $D$, circulation form gives $\oint_C P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA$ under the required smoothness. Positive orientation means the region stays on the left: outer boundaries are counterclockwise, while boundaries around holes are clockwise. The flux form uses a different derivative combination, so identify the requested line-integral form before substituting. A field singularity inside $D$ can invalidate the theorem even if $C$ avoids it. Green's theorem can also compute area through suitable choices of $P$ and $Q$. Check orientation explicitly; reversing every boundary component changes the sign of the line integral.

When the boundary is piecewise smooth, parameterize each segment in the induced order and verify that consecutive endpoints meet. Applying the double-integral side can then replace several separate boundary computations with one region integral.

Practice

1. Find circulation of $\langle-y/2,x/2\rangle$ around a circle of radius $2$ CCW.
2. What is positive orientation?
3. What happens for clockwise traversal?

Connections and next steps

• Line Integrals
• Curl and Divergence
• Stokes' Theorem