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Green’s Theorem in the Plane · Tutorial 817 of 999

Deriving Green’s Theorem for a Region with Holes

Learn how cuts reduce a region with holes to simple regions, and why the resulting Green’s theorem assigns clockwise orientation to each hole boundary.

Advanced 10 min read

What You'll Learn

  • State the hypotheses for Green’s theorem on a region with holes
  • Use cuts to divide a multiply connected region into simple pieces
  • Explain why integrals along the two sides of each cut cancel
  • Track the induced orientations of outer and hole boundaries
  • Write the theorem using counterclockwise hole curves with explicit minus signs
  • Check the orientation and area terms in concrete examples

Why the Holes Change the Boundary Formula

A region with a hole has more than one boundary curve. Green’s theorem still relates circulation around its boundary to the curl integrated over the region, but the boundary curves cannot all be given the same orientation. The outer boundary is traversed counterclockwise, while each hole boundary is traversed clockwise when the region is kept on the left.

This sign is not an arbitrary convention. It follows by cutting the region into simple pieces, applying the usual Green’s theorem to each piece, and adding the results. The artificial boundaries along the cuts cancel, while the original boundary components retain the orientations induced by their adjacent pieces.

Set-Up and Boundary Orientation

Let \(D\) be a bounded planar region formed from an outer simple region by removing the interiors of finitely many disjoint holes. Assume the hole boundaries do not touch one another or the outer boundary, and that the boundaries are sufficiently smooth for Green’s theorem to apply. Let \(C_0\) denote the outer boundary, and let \(C_j\), for \(1\leq j\leq m\), denote the boundary of the \(j\)-th hole.

For the formula below, each curve \(C_j\) is initially written with its own counterclockwise orientation, as though it bounded the hole itself. The positive orientation of the boundary of \(D\), however, traverses each hole boundary clockwise. We denote that oriented boundary collectively by \(\partial D\).

Orientation definition: A boundary is positively oriented for a planar region when the region lies to the left as the boundary is traversed. Thus \(C_0\) is counterclockwise, and the hole boundaries are clockwise.

Write the vector field as \(\mathbf{F}=(P,Q)\), so its circulation form is \(P\,dx+Q\,dy\). Assume \(P\) and \(Q\) have continuous first partial derivatives on an open set containing the closure of \(D\). The field must be defined and sufficiently smooth on \(D\); it need not be defined inside the holes.

Theorem: Under these hypotheses, Green’s theorem for a region with \(m\) holes is
$$ \oint_{\partial D} P\,dx+Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA = \oint_{C_0}P\,dx+Q\,dy -\sum_{j=1}^{m}\oint_{C_j}P\,dx+Q\,dy. $$

In the final expression, \(C_0\) and every \(C_j\) are counterclockwise. The minus signs convert the counterclockwise hole curves into the clockwise orientations required by the positive boundary of \(D\).

Derivation by Cutting the Region

Choose nonintersecting, piecewise-smooth cuts that connect the hole boundaries to the outer boundary. The cuts can be arranged to divide \(D\) into finitely many simple regions. The argument uses Green’s theorem on those pieces, where it applies in its ordinary form.

1
Apply Green’s theorem to each piece.
On a simple piece \(D_k\), the circulation around its positively oriented boundary equals the double integral of \(\partial Q/\partial x-\partial P/\partial y\) over \(D_k\).
2
Add the piecewise identities.
The pieces cover \(D\) without overlapping interiors, so their area integrals add to the integral over \(D\).
3
Cancel the cut integrals.
Each cut is shared by two adjacent pieces and is traversed in opposite directions in their positively oriented boundaries. Its two line integrals therefore sum to zero.
4
Read off the remaining boundary.
The uncancelled segments are the original outer and hole boundaries, with the positive orientation induced by the region.

Here is the justification for the cancellation and the signs. If a cut is parametrized from one endpoint to the other, the neighboring piece traverses it in the reverse direction. Reversing a parametrized curve changes the sign of a line integral, so the two contributions cancel. Along the outer boundary, keeping \(D\) on the left gives counterclockwise orientation. Along a hole boundary, keeping \(D\) on the left requires clockwise orientation: the missing hole must lie on the right. This proves the displayed formula when the hole curves are instead written counterclockwise with minus signs.

The double integrals also combine exactly. The pieces have disjoint interiors and their union is \(D\); their shared cuts have zero area. Hence the sum of their curl integrals is the curl integral over \(D\). No values of the field inside the holes are used at any step.

Worked Example: A Circular Annulus

Worked Example: A Circular Annulus

Let \(D\) be the annulus between the circles of radii \(2\) and \(1\), centered at the origin. Take \(P=-y\) and \(Q=x\). Compute the circulation on the positively oriented boundary and compare it with the curl integral.

First calculate the curl density:

$$ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} =1-(-1)=2. $$

For a counterclockwise circle of radius \(r\), use \(x=r\cos t\), \(y=r\sin t\), with \(0\leq t\leq2\pi\). Then \(dx=-r\sin t\,dt\) and \(dy=r\cos t\,dt\), so

$$ P\,dx+Q\,dy =(-r\sin t)(-r\sin t\,dt)+(r\cos t)(r\cos t\,dt) =r^2\,dt. $$

Thus the counterclockwise integral is \(\int_0^{2\pi}r^2\,dt=2\pi r^2\). The outer circle contributes \(2\pi(2^2)=8\pi\). The inner circle, if written counterclockwise, contributes \(2\pi(1^2)=2\pi\), and its actual positive orientation is clockwise. The total boundary circulation is therefore \(8\pi-2\pi=6\pi\).

The annulus has area \(\pi(2^2)-\pi(1^2)=3\pi\), so its curl integral is \(2(3\pi)=6\pi\). Both sides agree. Reversing the inner circle to clockwise changes its integral from \(2\pi\) to \(-2\pi\), confirming the orientation sign.

Worked Example: Two Square Holes

Worked Example: Two Square Holes

Let the outer region be the square \([-4,4]\times[-4,4]\). Remove the two square holes \([-2,-1]\times[-\tfrac12,\tfrac12]\) and \([1,2]\times[-\tfrac12,\tfrac12]\). For \(P=-y\), \(Q=x\), verify the theorem by computing the boundary integrals and the area integral.

As above, the curl density is \(1-(-1)=2\). The outer square has area \(8\cdot8=64\); each hole has area \(1\cdot1=1\). The area of \(D\) is \(64-1-1=62\), and the double integral is \(2\cdot62=124\).

For a counterclockwise rectangle \([a,b]\times[c,d]\), sum the integrals on its four edges. The bottom and top edges contribute \(-c(b-a)\) and \(d(b-a)\); the right and left edges contribute \(b(d-c)\) and \(-a(d-c)\). Their sum is \(2(b-a)(d-c)\). Therefore the outer square contributes \(2\cdot64=128\), and each counterclockwise hole contributes \(2\cdot1=2\).

Since the holes enter with negative signs, the positively oriented boundary integral is \(128-2-2=124\), matching the curl integral. This example shows why subtracting each hole’s counterclockwise circulation is necessary, even when there is more than one hole.

Worked Example: A Nonconstant Curl Density

Worked Example: A Nonconstant Curl Density

Let \(D\) be the rectangle \([0,4]\times[0,3]\) with the hole \([1,2]\times[1,2]\) removed. Choose \(P=0\), \(Q=x^2/2\), so the curl density is \(\partial Q/\partial x-\partial P/\partial y=x\).

Compute the outer boundary integral directly. On the horizontal edges, \(dy=0\), so the contribution is zero. On the right edge, \(x=4\) and \(y\) increases from \(0\) to \(3\), giving \(\int_0^3 (16/2)\,dy=24\). On the left edge, \(x=0\), giving zero. Thus the outer counterclockwise integral is \(24\).

For the hole’s counterclockwise boundary, the horizontal edges again contribute zero. On the right edge, \(x=2\) and \(y\) increases from \(1\) to \(2\), giving \(\int_1^2 2\,dy=2\). On the left edge, \(x=1\) and \(y\) decreases from \(2\) to \(1\), giving \(\int_2^1 \tfrac12\,dy=-\tfrac12\). The hole’s counterclockwise integral is \(2-\tfrac12=\tfrac32\). The positively oriented boundary circulation is therefore \(24-\tfrac32=\tfrac{45}{2}\).

For the area side, the integral of \(x\) over the outer rectangle is \(\int_0^4\int_0^3 x\,dy\,dx=3\cdot(4^2/2)=24\). Over the hole it is \(\int_1^2\int_1^2 x\,dy\,dx=(2^2-1^2)/2=\tfrac32\). Subtracting the hole gives \(24-\tfrac32=\tfrac{45}{2}\), which checks the boundary calculation.

A Common Orientation Mistake

A frequent error is to use counterclockwise orientation for every boundary curve and then add all the integrals. In the two-hole square example, that would give \(128+2+2=132\), not the correct value \(124\). The mistake treats the holes as if their interiors were part of the region.

Correction: Either traverse every boundary component with the region on the left, making the hole curves clockwise, or write all individual curves counterclockwise and subtract the hole integrals. Do not use counterclockwise hole curves and then add their contributions.

What the Derivation Clarifies

The cut argument explains both the scope and the structure of the theorem. The field only needs to be smooth on the region where the area integral is taken, because the proof applies Green’s theorem to pieces of \(D\) and never fills in a hole. Internal cuts do not create extra terms because their paired contributions cancel exactly.

For practical calculations, it is often convenient to keep every curve parametrized counterclockwise and use an explicit minus sign for each hole. If a curve is instead parametrized in its induced positive orientation, its direction must be clockwise around a hole. Keeping these two descriptions distinct prevents sign errors and makes the formula agree with the geometry of the region.

Check Your Understanding

Use the cut argument and boundary orientations to answer these questions.

  1. Why does the integral along an artificial cut cancel when the integrals over all the pieces are added?
  2. With the region kept on the left, which direction is used on an outer boundary, and which direction is used around a hole?
  3. If all boundary curves are written counterclockwise, what sign accompanies each hole integral?
  4. In the two-hole square example, what would the boundary integral be if the hole contributions were mistakenly added?
  5. Does the derivation require the vector field to be defined inside the holes? Explain why or why not.

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