Green's Theorem: Turning a Line Integral into a Double Integral

Green's theorem is an exchange rate between two kinds of integral. On one side is a line integral taken around a closed curve; on the other is a double integral over the flat region that curve encloses. Being able to move between them turns many line integrals that would need several parametrised pieces into one double integral — and occasionally the reverse.

The exchange the theorem makes

For a curve $C$ that is positively oriented, piecewise smooth, simple and closed, and a field $\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ whose components have continuous first partial derivatives on the enclosed region $D$,

$\oint_{C} P\,dx+Q\,dy=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$

The combination on the right is the two-dimensional curl of the field. Notice what the theorem does not require of the curve: it does not have to be a circle or a rectangle. Any simple closed curve without self-intersections will do, which is what makes the exchange useful.

Read the conditions before using it

Three of the hypotheses are easy to skip and each one can change the answer:

• Positive orientation. The curve must be traversed so that the region stays on the left. Walking the boundary of a disk that way is counterclockwise. Using the clockwise traversal gives the negative of the correct value, because the double integral is unchanged while the line integral reverses sign.
• Simple and closed. A curve that crosses itself encloses more than one region, and the theorem in this form no longer applies directly.
• Smoothness on the whole region, not just on the curve. If a component has a singularity inside $D$, the double integral is improper and may not converge, even though the line integral is perfectly well defined. This is the situation to watch for when a denominator vanishes somewhere in the interior.

The area formula as a special case

Choosing $P=-y$ and $Q=x$ gives $\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=1-(-1)=2$, so

$\oint_{C}(-y\,dx+x\,dy)=\iint_{D}2\,dA=2\cdot\text{Area}(D),\qquad\text{Area}(D)=\frac{1}{2}\oint_{C}(-y\,dx+x\,dy).$

Checking this on a circle of radius $R$ centred at the origin, parametrised counterclockwise by $x=R\cos t$, $y=R\sin t$ with $0\le t\le 2\pi$: here $dx=-R\sin t\,dt$ and $dy=R\cos t\,dt$, so

$-y\,dx+x\,dy=R^{2}\sin^{2}t\,dt+R^{2}\cos^{2}t\,dt=R^{2}\,dt,\qquad \frac{1}{2}\int_{0}^{2\pi}R^{2}dt=\pi R^{2}.$

The same combination therefore measures the area of any region whose boundary you can parametrise, including regions whose shape makes an ordinary geometric formula awkward.

Where the exchange stops

If the region has a hole, its boundary is not one curve but several, and the inner boundary has to be traversed clockwise so that the region stays on the left of the direction of travel. Applying the theorem to the outer circle alone then computes the double integral over the whole disk, not over the ring, which is a different number. The theorem also says nothing about a curve that is not closed: the double integral has to be replaced by the appropriate path-independence argument, and that only works when the field is conservative.

A habit that prevents sign errors

Before writing the double integral, do two things on paper: sketch the region and mark the traversal direction on the boundary, then write the order of integration you intend to use. Most mistakes in these problems are orientation mistakes or a reversed order of integration, not calculus mistakes in the integrand itself.