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Green’s Theorem in the Plane · Tutorial 816 of 999

Green’s Theorem on Regions with Holes

Apply both forms of Green’s theorem to regions with holes, keeping outer and inner boundary orientations and signs consistent.

Advanced 10 min read

What You'll Learn

  • Identify the positively oriented boundary of a region with holes
  • Apply the circulation form of Green’s theorem to every boundary component
  • Apply the flux form using the divergence over the full region
  • Convert outward flux on an oriented curve to the expression \(P\,dy-Q\,dx\)
  • Check boundary calculations against area integrals
  • Avoid the sign error of orienting hole boundaries counterclockwise

Why Holes Change the Boundary Calculation

A region with a hole has more than one boundary curve. Green’s theorem still relates a line integral around the boundary to a double integral over the region, but the boundary curves must be given the correct orientations. The outer boundary is traversed counterclockwise; each boundary around a hole is traversed clockwise.

This orientation is essential for both circulation and flux. In particular, the boundary integral over a hole contributes with the opposite sign from what it would contribute if that curve were traversed counterclockwise.

Key idea: The positively oriented boundary of a region with holes consists of the counterclockwise outer boundary together with every hole boundary traversed clockwise.

Setting and Notation

Let \(D\) be a bounded planar region whose boundary consists of one smooth outer curve \(C_0\) and finitely many smooth, disjoint hole boundaries \(C_1,\ldots,C_m\). Assume each hole lies inside the outer curve, and the holes do not overlap. Write \(C\) for the whole boundary with its required orientations: \(C_0\) counterclockwise and \(C_1,\ldots,C_m\) clockwise.

Let \(\mathbf{F}=(P,Q)\), where \(P\) and \(Q\) have continuous first partial derivatives on an open set containing the closure of \(D\). Under these conditions, both forms of Green’s theorem apply.

Theorem: For the positively oriented boundary \(C\) of \(D\), the circulation form is
$$ \oint_C P\,dx+Q\,dy = \iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA. $$
The flux form is
$$ \oint_C P\,dy-Q\,dx = \iint_D\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\right)dA. $$
The left sides mean the sum of the integrals over all boundary components, each with its specified orientation.

The quantity \(Q_x-P_y\) is the scalar curl density in the plane, and \(P_x+Q_y\) is the divergence. The double integrals are over the region \(D\) itself, so the removed hole interiors are not included.

Why the Flux Boundary Expression Has This Sign

There is a direct geometric reason for the expression \(P\,dy-Q\,dx\). Suppose a positively oriented boundary curve is parametrized by \(\mathbf{r}(t)=(x(t),y(t))\). With this orientation, the region lies to the left of the direction of travel, so an outward normal points to the right. A right-pointing normal vector with the appropriate length for a flux integral is \((y'(t),-x'(t))\).

Taking its dot product with \(\mathbf{F}=(P,Q)\) gives the outward flux contribution per unit parameter:

$$ (P,Q)\mathbin{\cdot}(y'(t),-x'(t)) = P\,y'(t)-Q\,x'(t). $$

Since \(dy=y'(t)\,dt\) and \(dx=x'(t)\,dt\), this becomes \(P\,dy-Q\,dx\). The same reasoning applies to every component of the positively oriented boundary. Along a hole boundary, the clockwise direction keeps the region on the left and the outward normal pointing out of \(D\), into the hole.

Worked Example: Circulation Around an Annulus

Worked Example: Circulation Around an Annulus

Let \(D\) be the annulus between the circles of radii \(1\) and \(3\), both centered at the origin. For \(\mathbf{F}=(-y/2,x/2)\), compute the circulation around the whole positively oriented boundary and check it using the circulation form of Green’s theorem.

Here \(P=-y/2\) and \(Q=x/2\), so

$$ Q_x=\frac12,\qquad P_y=-\frac12,\qquad Q_x-P_y=1. $$

The area of the annulus is the outer disk’s area minus the hole’s area:

$$ \operatorname{area}(D)=\pi(3^2)-\pi(1^2)=9\pi-\pi=8\pi. $$

Thus the area integral predicts circulation \(1\cdot 8\pi=8\pi\).

Check the boundary calculation directly. On a circle of radius \(R\) centered at the origin, use \(x=R\cos t\) and \(y=R\sin t\). For counterclockwise traversal, \(0\le t\le 2\pi\), so \(dx=-R\sin t\,dt\) and \(dy=R\cos t\,dt\). Then

$$ P\,dx+Q\,dy = -\frac{R\sin t}{2}(-R\sin t\,dt) +\frac{R\cos t}{2}(R\cos t\,dt) = \frac{R^2}{2}\,dt. $$

The outer circle contributes \(\int_0^{2\pi}\frac{9}{2}\,dt=9\pi\). The same counterclockwise calculation for the radius-\(1\) circle would give \(\pi\), but the hole boundary must be clockwise, so its actual contribution is \(-\pi\). Therefore

$$ \oint_C P\,dx+Q\,dy=9\pi-\pi=8\pi. $$

This agrees with the area integral.

Worked Example: Outward Flux Through a Square with a Circular Hole

Worked Example: Outward Flux Through a Square with a Circular Hole

Let \(D\) be the square \([-2,2]\times[-1,1]\) with the disk of radius \(1/2\) centered at the origin removed. For \(\mathbf{F}=(x,y)\), compute the outward flux using the flux form of Green’s theorem, then verify it by integrating around the boundary.

Here \(P=x\) and \(Q=y\), so the divergence is

$$ P_x+Q_y=1+1=2. $$

The square has area \(4\cdot2=8\). The removed disk has area \(\pi(1/2)^2=\pi/4\). Hence

$$ \iint_D 2\,dA = 2\left(8-\frac{\pi}{4}\right) = 16-\frac{\pi}{2}. $$

To check, the flux integrand is \(P\,dy-Q\,dx=x\,dy-y\,dx\). On the square’s counterclockwise boundary, calculate one side at a time:

  • Bottom edge: \(y=-1\), \(x\) runs from \(-2\) to \(2\), and \(dy=0\). Thus \(x\,dy-y\,dx=dx\), giving \(\int_{-2}^2 dx=4\).
  • Right edge: \(x=2\), \(y\) runs from \(-1\) to \(1\), and \(dx=0\). Thus the integral is \(\int_{-1}^1 2\,dy=4\).
  • Top edge: \(y=1\), \(x\) runs from \(2\) to \(-2\), and \(dy=0\). Thus the integral is \(\int_2^{-2}-dx=4\).
  • Left edge: \(x=-2\), \(y\) runs from \(1\) to \(-1\), and \(dx=0\). Thus the integral is \(\int_1^{-1}-2\,dy=4\).

The outer boundary contributes \(4+4+4+4=16\). On the circular boundary, \(x=r\cos t,\ y=r\sin t\) gives \(x\,dy-y\,dx=r^2\,dt\). A counterclockwise traversal would contribute \(2\pi r^2\); the required clockwise traversal contributes \(-2\pi r^2\). With \(r=1/2\), that is \(-2\pi(1/2)^2=-\pi/2\). The total boundary flux is therefore \(16-\pi/2\), agreeing with the area integral.

Several Holes, Several Boundary Contributions

When there is more than one hole, each hole contributes its own boundary integral. The total is not found by choosing one representative hole or by changing the area integral’s region: include every component in the boundary sum, and integrate over the region with all holes removed.

Worked Example: Circulation with Two Holes

Let \(D\) be the disk of radius \(4\) centered at the origin with two disks of radius \(1/2\) removed. Put the hole centers at \((-1,0)\) and \((1,0)\). The holes are disjoint and lie inside the outer disk. For \(\mathbf{F}=(-y,x)\), calculate the circulation using both forms of Green’s theorem.

For this field, \(P=-y\) and \(Q=x\), so

$$ Q_x-P_y=1-(-1)=2. $$

The region’s area is

$$ \pi(4^2)-2\pi\left(\frac12\right)^2 = 16\pi-\frac{\pi}{2} = \frac{31\pi}{2}. $$

The circulation form therefore gives \(2\cdot31\pi/2=31\pi\).

For a boundary check, on any circle of radius \(r\) centered at \((a,b)\), parametrize counterclockwise by \(x=a+r\cos t,\ y=b+r\sin t\). Then

$$ -y\,dx+x\,dy = -(b+r\sin t)(-r\sin t)\,dt +(a+r\cos t)(r\cos t)\,dt = \left(r^2+br\sin t+ar\cos t\right)dt. $$

Over \(0\le t\le2\pi\), the sine and cosine terms integrate to zero, leaving \(2\pi r^2\). The outer circle, traversed counterclockwise with \(r=4\), contributes \(2\pi(16)=32\pi\). Each radius-\(1/2\) hole would contribute \(2\pi(1/4)=\pi/2\) counterclockwise, so each contributes \(-\pi/2\) with the required clockwise orientation. The full boundary integral is

$$ 32\pi-\frac{\pi}{2}-\frac{\pi}{2}=31\pi. $$

This matches the area-integral calculation.

A Common Orientation Mistake

A frequent error is to traverse every boundary curve counterclockwise. In the annulus example, doing this would give \(9\pi+\pi=10\pi\) for the circulation. That is the integral around the outer circle plus the integral around the hole in the wrong direction; it is not the positively oriented boundary integral for the annulus.

The correction is to reverse the hole’s traversal. Reversing a curve’s direction changes the sign of its line integral, so the correct total is \(9\pi-\pi=8\pi\). A useful check is geometric: while traversing every component of a positively oriented boundary, the region must remain on the left. This means counterclockwise on the outside and clockwise around each hole.

When the Two Forms Are Useful

The circulation form is convenient when \(Q_x-P_y\) is easy to integrate over the region. The flux form is convenient when the divergence \(P_x+Q_y\) is simpler. In either case, the area integral uses only the part of the plane belonging to \(D\), while the boundary integral includes every component with its prescribed direction.

Boundary calculations can also be divided into a sum of separate integrals:

$$ \oint_C(\cdots) = \oint_{C_0}(\cdots) + \sum_{j=1}^{m}\oint_{C_j}(\cdots), $$

where \(C_0\) is counterclockwise and each \(C_j\) is clockwise. This notation records orientation as part of the integral; it does not mean that all curves should be parametrized in the same direction.

Check Your Understanding

For each question, keep track of which region is integrated over and how every boundary component is oriented.

  1. For a region with one outer boundary and three holes, how many boundary integrals appear in the full Green’s theorem boundary sum?
  2. What is the required orientation of the outer boundary, and what is the required orientation of each hole boundary?
  3. For \(\mathbf{F}=(P,Q)\), what scalar density appears in the circulation form, and what density appears in the flux form?
  4. Why does a hole boundary contribute the negative of its counterclockwise integral?
  5. A region is the disk of radius \(3\) with a concentric disk of radius \(1\) removed. For \(\mathbf{F}=(-y/2,x/2)\), what circulation does the circulation form predict?

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