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Green’s Theorem in the Plane · Tutorial 821 of 999

Green’s Theorem on Annuli and Sectors

Learn to orient and compute every boundary contribution on annuli and sectors, then check the result using Green’s Theorem.

Advanced 12 min read

What You'll Learn

  • Identify the positively oriented boundary of an annulus or annular sector
  • Set up circulation integrals on outer arcs, inner arcs, and radial sides
  • Compute outward flux on curved and radial boundary components
  • Use curl and divergence densities to check boundary calculations
  • Derive the polar-coordinate boundary and area formulas for these regions

Why annuli and sectors need careful boundary bookkeeping

An annulus has an outer boundary and an inner boundary. An annular sector adds two radial sides. Green’s Theorem applies to these regions, but a correct calculation depends on including every boundary component with the direction required by the region’s orientation.

This geometry is useful when a field is naturally described around a center, or when a boundary consists of circular arcs and straight radial segments. The same method handles both circulation and outward flux: calculate around the full oriented boundary, or integrate curl or divergence over the region, then compare the results.

Regions and positive boundary orientation

Definition: An annulus centered at the origin with inner radius \(a\) and outer radius \(b\) is the region \(a\le r\le b\), where \(0<a<b\). An annular sector also restricts the angle to \(\alpha\le\theta\le\beta\), with \(0<\beta-\alpha<2\pi\). In polar coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), and \(dA=r\,dr\,d\theta\).

A positively oriented boundary is traversed so that the region stays on the left. For an annular sector, traverse the outer arc from angle \(\alpha\) to \(\beta\), then the upper radial side inward, the inner arc from \(\beta\) back to \(\alpha\), and the lower radial side outward. The inner arc runs clockwise because the region must remain on its left there.

$$ \begin{array}{ll} \text{outer arc:} & r=b,\quad \alpha\le\theta\le\beta,\\ \text{upper radial side:} & \theta=\beta,\quad r=b\text{ to }a,\\ \text{inner arc:} & r=a,\quad \theta=\beta\text{ to }\alpha,\\ \text{lower radial side:} & \theta=\alpha,\quad r=a\text{ to }b. \end{array} $$

For a full annulus, the two radial sides are only the two descriptions of the same angular seam. Their contributions cancel when the field is single-valued and continuous there. Equivalently, one can describe the full annulus using its outer and inner circles alone.

Theorem: Let \(\mathbf{F}=(P,Q)\) have continuous first partial derivatives on an open set containing the region and its boundary. With positive boundary orientation, circulation and outward flux satisfy $$ \oint_{\partial D}P\,dx+Q\,dy =\iint_D(Q_x-P_y)\,dA, \qquad \oint_{\partial D}P\,dy-Q\,dx =\iint_D(P_x+Q_y)\,dA. $$ Here \(Q_x-P_y\) is the planar curl density and \(P_x+Q_y\) is the divergence density.

The hypotheses matter: the field and its first partial derivatives must be defined and continuous across the region, including its boundary. A field singular inside a hole is not automatically a problem if the hole is excluded and the field is smooth on a neighborhood of the region itself. The boundary must be traversed with the orientation of the region, including the clockwise inner boundary.

Polar-coordinate forms for the boundary integrals

Along a circular arc, \(r\) is fixed and \(\theta\) changes. Along a radial side, \(\theta\) is fixed and \(r\) changes. Differentiating \(x=r\cos\theta\) and \(y=r\sin\theta\) gives

$$ dx=\cos\theta\,dr-r\sin\theta\,d\theta, \qquad dy=\sin\theta\,dr+r\cos\theta\,d\theta. $$

Substituting these expressions into the circulation and flux forms produces convenient integrands. In these formulas, \(P\) and \(Q\) are evaluated at \((r\cos\theta,r\sin\theta)\).

$$ \begin{aligned} P\,dx+Q\,dy &=(P\cos\theta+Q\sin\theta)\,dr +r(-P\sin\theta+Q\cos\theta)\,d\theta,\\ P\,dy-Q\,dx &=(P\sin\theta-Q\cos\theta)\,dr +r(P\cos\theta+Q\sin\theta)\,d\theta. \end{aligned} $$

These expressions make the boundary pieces explicit: set \(dr=0\) on an arc and set \(d\theta=0\) on a radial side. The direction of travel is recorded by the integration limits; for example, the inner arc uses decreasing \(\theta\).

Why the polar formulas agree with Green’s Theorem

The area form of Green’s Theorem can be checked directly in polar coordinates. Write the circulation form as \(A\,dr+B\,d\theta\), where \(A=P\cos\theta+Q\sin\theta\) and \(B=r(-P\sin\theta+Q\cos\theta)\). Green’s Theorem in the \((r,\theta)\)-rectangle gives the boundary integral as the double integral of \(B_r-A_\theta\). The chain rule yields

$$ B_r-A_\theta=r(Q_x-P_y). $$

For completeness, the terms without a factor of \(r\) cancel when \(B_r-A_\theta\) is formed. The remaining terms are

$$ \begin{aligned} B_r-A_\theta &=r(-P_y\sin^2\theta-P_y\cos^2\theta +Q_x\cos^2\theta+Q_x\sin^2\theta)\\ &=r(Q_x-P_y), \end{aligned} $$

since \(\sin^2\theta+\cos^2\theta=1\). This is exactly the curl density multiplied by the polar area factor \(r\). For flux, write the boundary form as \(C\,dr+D\,d\theta\), where \(C=P\sin\theta-Q\cos\theta\) and \(D=r(P\cos\theta+Q\sin\theta)\). The same chain-rule calculation gives

$$ D_r-C_\theta=r(P_x+Q_y). $$

Thus the area integrands are the usual divergence and curl densities with \(dA=r\,dr\,d\theta\). For a sector, the parameter rectangle traces its boundary in the required order. For a full annulus, the rectangle’s two angular edges represent the same seam and cancel, leaving the outer and inner circles.

Worked Example: Circulation around a full annulus

Worked Example: Circulation around a full annulus

Let \(P=-y\), \(Q=x\), and let \(D\) be the annulus \(1\le r\le3\). Find the circulation using both the area integral and the boundary.

First compute the curl density: \(Q_x=1\) and \(P_y=-1\), so \(Q_x-P_y=1-(-1)=2\). The annulus has area \(\pi(3^2-1^2)=8\pi\). Therefore

$$ \iint_D(Q_x-P_y)\,dA =\int_0^{2\pi}\int_1^3 2r\,dr\,d\theta =\int_0^{2\pi}[r^2]_1^3\,d\theta =\int_0^{2\pi}8\,d\theta =16\pi. $$

On a circle of radius \(r\), \(x=r\cos\theta\), \(y=r\sin\theta\), \(dx=-r\sin\theta\,d\theta\), and \(dy=r\cos\theta\,d\theta\). Hence

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

The outer circle contributes \(\int_0^{2\pi}9\,d\theta=18\pi\). The inner circle is clockwise, so it contributes \(\int_{2\pi}^0 1\,d\theta=-2\pi\). Their sum is \(18\pi-2\pi=16\pi\), agreeing with the area calculation. The field is smooth everywhere, so the theorem’s regularity condition is satisfied.

Worked Example: Circulation on an annular sector

Worked Example: Circulation on an annular sector

Let \(P=-y\), \(Q=0\), and take \(1\le r\le3\), \(0\le\theta\le\pi/2\). Find the circulation, tracking all four boundary pieces.

Here \(Q_x-P_y=0-(-1)=1\). The area integral is

$$ \iint_D1\,dA =\int_0^{\pi/2}\int_1^3 r\,dr\,d\theta =\int_0^{\pi/2}\frac{9-1}{2}\,d\theta =4\left(\frac{\pi}{2}\right) =2\pi. $$

On the outer arc, \(r=3\) and \(\theta\) increases. Since \(P=-r\sin\theta\) and \(dx=-r\sin\theta\,d\theta\), the contribution is \(\int_0^{\pi/2}9\sin^2\theta\,d\theta=9\pi/4\). On the upper radial side, \(x=0\), so \(dx=0\), and the contribution is \(0\).

On the inner arc, \(r=1\) and \(\theta\) decreases. Its contribution is \(\int_{\pi/2}^0\sin^2\theta\,d\theta=-\pi/4\). On the lower radial side, \(\theta=0\), hence \(y=0\), \(P=0\), and the contribution is \(0\). Adding the pieces gives

$$ \frac{9\pi}{4}+0-\frac{\pi}{4}+0 =\frac{8\pi}{4} =2\pi. $$

This matches the area integral. In particular, a zero contribution from one boundary piece here follows from the specific field and geometry; it is not a reason to omit that piece in other problems.

Worked Example: Outward flux with nonzero radial contributions

Worked Example: Outward flux with nonzero radial contributions

Let \(P=1\), \(Q=0\), and take \(1\le r\le2\), \(0\le\theta\le\pi/2\). Compute the outward flux around the sector boundary.

The divergence is \(P_x+Q_y=0\), so Green’s Theorem predicts zero flux. Check each boundary piece using \(P\,dy-Q\,dx=dy\). On the outer arc, \(y=2\sin\theta\), so \(dy=2\cos\theta\,d\theta\), giving \(\int_0^{\pi/2}2\cos\theta\,d\theta=2\).

On the upper radial side, \(y=r\) and the direction is from \(r=2\) to \(r=1\). Thus the contribution is \(\int_2^1 dr=-1\). On the inner arc, \(y=\sin\theta\) and the direction is from \(\pi/2\) to \(0\), giving \(\int_{\pi/2}^0\cos\theta\,d\theta=-1\). On the lower radial side, \(y=0\) is constant, so \(dy=0\) and the contribution is \(0\).

The total boundary flux is \(2-1-1+0=0\). The area check is \(\iint_D0\,dA=0\), so both calculations agree. This example shows that radial sides can make nonzero flux contributions even when the total flux vanishes.

Worked Example: Flux from a radial field

Worked Example: Flux from a radial field

Let \(P=x\), \(Q=y\), and take \(1\le r\le2\), \(0\le\theta\le\pi/3\). Find the outward flux by both methods.

The divergence is \(P_x+Q_y=1+1=2\). The area calculation gives

$$ \iint_D2\,dA =\int_0^{\pi/3}\int_1^2 2r\,dr\,d\theta =\int_0^{\pi/3}[r^2]_1^2\,d\theta =\int_0^{\pi/3}3\,d\theta =\pi. $$

On a circle \(r=\text{constant}\), \(P=r\cos\theta\), \(Q=r\sin\theta\), \(dx=-r\sin\theta\,d\theta\), and \(dy=r\cos\theta\,d\theta\). Therefore \(P\,dy-Q\,dx=r^2(\cos^2\theta+\sin^2\theta)\,d\theta=r^2\,d\theta\). The outer arc contributes \(\int_0^{\pi/3}4\,d\theta=4\pi/3\); the clockwise inner arc contributes \(\int_{\pi/3}^0 1\,d\theta=-\pi/3\).

On either radial side, \(dy=\sin\theta\,dr\) and \(dx=\cos\theta\,dr\). Thus \(P\,dy-Q\,dx=(r\cos\theta)(\sin\theta\,dr)-(r\sin\theta)(\cos\theta\,dr)=0\). Both radial contributions are zero. The boundary total is \(4\pi/3-\pi/3+0+0=\pi\), agreeing with the area integral.

Using the two methods effectively

The boundary method is often convenient when each arc and radial segment has a simple parametrization. The area method is often shorter when curl or divergence is constant, or depends only on \(r\). For an annular sector, the area setup is

$$ \iint_D g(x,y)\,dA =\int_\alpha^\beta\int_a^b g(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$

Here \(g\) is the curl density for circulation or the divergence density for outward flux. The factor \(r\) is essential: it is the area-scaling factor in polar coordinates. When the two methods give different answers, check the boundary directions, the inner arc’s decreasing angle, the radial sides, and the factor \(r\) in the area integral.

Key takeaway: For annuli and annular sectors, Green’s Theorem equates a sum over every oriented boundary component with an area integral of curl or divergence. Compute each component with its actual direction, then use the other form of the theorem as a check.

Check Your Understanding

For each question, identify the relevant boundary pieces and orientation before calculating.

  1. In the positive orientation of an annular sector, in which direction is the inner arc traversed?
  2. For a sector with radii \(a,b\) and angles \(\alpha,\beta\), write the polar double integral for its area.
  3. Why can a radial side contribute to circulation or outward flux even though it is not curved?
  4. For \(P=1,Q=0\), what is the divergence density, and what does Green’s Theorem predict for the total outward flux?
  5. For \(P=x,Q=y\), what are the circulation density and divergence density?

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