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Green’s Theorem in the Plane · Tutorial 808 of 999

Computing Circulation from the Curl Density

Use Green’s circulation form to replace boundary line integrals with area integrals, and track the sign correctly when a boundary’s orientation is reversed.

Advanced 9 min read

What You'll Learn

  • State the circulation form of Green’s Theorem and its conditions.
  • Identify the curl density \(Q_x-P_y\) for a planar vector field.
  • Set up and evaluate the area integral on a rectangle or disk.
  • Use piecewise bounds to integrate over a triangular region.
  • Determine how reversing boundary orientation changes the circulation.

From Boundary Motion to an Area Calculation

A circulation integral measures how a vector field acts along an oriented boundary. Its direct calculation can require splitting a boundary into several curves, parametrizing each one, and evaluating a line integral on every piece. Green’s Theorem offers another route: under suitable conditions, the total circulation around a positively oriented boundary is the double integral of a scalar quantity over the enclosed region.

That scalar quantity is \(Q_x-P_y\) for a field \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\). The method is especially useful when this expression is simple, even if the boundary itself has several segments.

Definition: For a planar vector field \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\), its circulation along an oriented, piecewise-smooth curve \(C\) is the line integral \(\int_C P\,dx+Q\,dy\). The scalar function \(Q_x-P_y\) is the density used in the circulation form of Green’s Theorem.

The Circulation Form of Green’s Theorem

Let \(D\) be a bounded planar region whose boundary \(C\) is a simple, closed, piecewise-smooth curve. The positive orientation traverses \(C\) counterclockwise, so that the region stays on the left as the boundary is followed. Suppose \(P\) and \(Q\), together with their first partial derivatives, are continuous on an open set containing \(D\). Then Green’s Theorem gives:

$$ \oint_C P\,dx+Q\,dy = \iint_D \left(Q_x-P_y\right)\,dA. $$

The integral on the left is circulation around the boundary; the one on the right adds the density \(Q_x-P_y\) over the region. For a region described by \(a\le x\le b\) and \(g_1(x)\le y\le g_2(x)\), the right side can be written as an iterated integral:

$$ \iint_D \left(Q_x-P_y\right)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} \left(Q_x-P_y\right)\,dy\,dx. $$

Other descriptions of \(D\), including polar coordinates for a disk, may be more convenient. The bounds must describe the entire region exactly once.

Orientation rule: Green’s Theorem in this form uses the positively oriented boundary. If the same boundary is traversed clockwise, its circulation is the negative of the double integral over \(D\).

Why the Area Integral Gives the Boundary Circulation

The theorem does more than provide a convenient formula: the sign and combination \(Q_x-P_y\) are tied to how oriented boundary pieces fit together. One way to understand the computation is to subdivide a region into small pieces. When their boundary integrals are added, each interior edge is traversed once in each direction, so those two contributions cancel. What remains is the integral around the outer boundary. Green’s Theorem identifies the net contribution accumulated over the small pieces with the area integral of \(Q_x-P_y\).

For applying the theorem, the practical consequence is direct: determine the density, integrate it over the full region, and check that the boundary orientation matches the theorem. The examples below also compare the result with direct boundary calculations.

1
Check the hypotheses and orientation.
Confirm the region has a simple, closed, piecewise-smooth boundary and that the field’s first partial derivatives are continuous on an open set containing the region. Identify whether the boundary is counterclockwise.
2
Compute the density.
Differentiate \(Q\) with respect to \(x\), differentiate \(P\) with respect to \(y\), and form \(Q_x-P_y\).
3
Integrate over the region.
Choose bounds that cover the region once, evaluate the double integral, and reverse the sign if the given traversal is clockwise.

Worked Example: Circulation Around a Rectangle

Worked Example: Circulation Around a Rectangle

Let \(\mathbf{F}(x,y)=\langle -y,x\rangle\), and let \(D\) be the rectangle \(0\le x\le2\), \(-1\le y\le1\). Find the circulation around its counterclockwise boundary \(C\).

Here \(P=-y\) and \(Q=x\). Their required partial derivatives are

$$ Q_x=1,\qquad P_y=-1, \qquad Q_x-P_y=1-(-1)=2. $$

The density is constant, and the rectangle has width \(2-0=2\) and height \(1-(-1)=2\). Therefore its area integral is

$$ \oint_C -y\,dx+x\,dy = \int_0^2\int_{-1}^1 2\,dy\,dx = \int_0^2 \left[2y\right]_{-1}^{1}\,dx = \int_0^2 4\,dx = 8. $$

As a check, traverse the rectangle counterclockwise starting at \((0,-1)\). Along the bottom edge, \(y=-1\), \(x\) runs from \(0\) to \(2\), and \(dy=0\), so the contribution is \(\int_0^2 1\,dx=2\). Along the right edge, \(x=2\), \(y\) runs from \(-1\) to \(1\), and \(dx=0\), so the contribution is \(\int_{-1}^1 2\,dy=4\). Along the top edge, \(y=1\), \(x\) runs from \(2\) to \(0\), giving \(\int_2^0 -1\,dx=2\). Along the left edge, \(x=0\), so \(Q=0\) and its contribution is \(0\). The sum is \(2+4+2+0=8\), agreeing with the area-integral result.

Worked Example: Circulation Around a Circle

Worked Example: Circulation Around a Circle

Let \(\mathbf{F}(x,y)=\langle -y/2,x/2\rangle\), and let \(D\) be the disk \(x^2+y^2\le9\). Compute the circulation around the counterclockwise boundary.

In this case, \(P=-y/2\) and \(Q=x/2\), so

$$ Q_x=\frac12,\qquad P_y=-\frac12, \qquad Q_x-P_y=\frac12-\left(-\frac12\right)=1. $$

The density is \(1\). In polar coordinates the disk is given by \(0\le r\le3\), \(0\le\theta\le2\pi\), and \(dA=r\,dr\,d\theta\). Thus

$$ \oint_C -\frac{y}{2}\,dx+\frac{x}{2}\,dy = \int_0^{2\pi}\int_0^3 r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^3d\theta = \int_0^{2\pi}\frac92\,d\theta = 9\pi. $$

For a direct check, parametrize the circle counterclockwise by \(x=3\cos t\), \(y=3\sin t\), for \(0\le t\le2\pi\). Then \(dx=-3\sin t\,dt\) and \(dy=3\cos t\,dt\). Substitution into the line integral gives

$$ -\frac{3\sin t}{2}(-3\sin t)\,dt + \frac{3\cos t}{2}(3\cos t)\,dt = \frac92(\sin^2t+\cos^2t)\,dt = \frac92\,dt. $$

Integrating from \(0\) to \(2\pi\) gives \(\frac92(2\pi)=9\pi\), as required. If the same circle were traversed clockwise, the circulation would be \(-9\pi\).

Worked Example: A Piecewise-Bounded Triangular Region

Worked Example: A Piecewise-Bounded Triangular Region

Let \(\mathbf{F}(x,y)=\langle 0,x^2/2+xy\rangle\). Let \(D\) be the triangle with vertices \((0,0)\), \((2,0)\), and \((0,2)\), and take its boundary counterclockwise. Find the circulation.

Here \(P=0\) and \(Q=x^2/2+xy\). Consequently,

$$ Q_x=x+y,\qquad P_y=0, \qquad Q_x-P_y=x+y. $$

For each \(x\) from \(0\) to \(2\), \(y\) ranges from the bottom edge \(y=0\) to the sloping edge \(y=2-x\). These bounds describe the triangle once. The area integral is

$$ \begin{aligned} \oint_C P\,dx+Q\,dy &=\int_0^2\int_0^{2-x}(x+y)\,dy\,dx\\ &=\int_0^2\left[xy+\frac{y^2}{2}\right]_0^{2-x}dx\\ &=\int_0^2\left(x(2-x)+\frac{(2-x)^2}{2}\right)dx\\ &=\int_0^2\left(2x-x^2+2-2x+\frac{x^2}{2}\right)dx\\ &=\int_0^2\left(2-\frac{x^2}{2}\right)dx\\ &=\left[2x-\frac{x^3}{6}\right]_0^2 =4-\frac{8}{6} =\frac83. \end{aligned} $$

Check the result directly along the three edges. The bottom edge goes from \((0,0)\) to \((2,0)\); since \(dy=0\) and \(P=0\), its contribution is \(0\). The sloping edge goes from \((2,0)\) to \((0,2)\). Set \(x=2-2t\), \(y=2t\), with \(0\le t\le1\). Then \(dy=2\,dt\), and

$$ Q=\frac{(2-2t)^2}{2}+(2-2t)(2t) =2-2t^2. $$

The contribution from this edge is therefore \(\int_0^1(2-2t^2)2\,dt=4-\frac43=\frac83\). On the final edge, \(x=0\), so \(Q=0\), and its contribution is \(0\). The total circulation is \(0+\frac83+0=\frac83\), agreeing with the double integral.

Common Sign and Setup Errors

A frequent error is to use \(P_y-Q_x\) instead of \(Q_x-P_y\). In the rectangular example, that mistaken density would be \(-1-1=-2\), producing \(\int_0^2\int_{-1}^1 -2\,dy\,dx=-8\). But the boundary was counterclockwise and its direct circulation is \(8\). The correction is to use the circulation density \(Q_x-P_y\), which is \(1-(-1)=2\), giving \(8\).

A different sign error occurs when the boundary is clockwise. The area integral in Green’s Theorem is for positive, counterclockwise orientation. Reversing the direction of traversal reverses the sign of the line integral; it does not change the region or the density. For example, the clockwise circulation around the rectangle in the first example is \(-8\), not \(8\).

Finally, careful bounds matter. For the triangle, integrating \(x+y\) over a rectangle that contains the triangle would include points outside \(D\) and compute a different quantity. Sketching the region and checking the endpoints of each bound is a useful safeguard.

When This Method Is Useful

Computing circulation from the curl density is most efficient when the region is easy to describe and \(Q_x-P_y\) is simpler to integrate than the boundary is to parametrize. The density may be constant, as in the rectangle and disk examples, or vary over the region, as in the triangle. In either case, the same checks apply: verify the theorem’s conditions, compute \(Q_x-P_y\) in the correct order, use bounds for exactly the enclosed region, and match the sign to the orientation.

Check Your Understanding

Use the circulation form of Green’s Theorem to answer these questions.

  1. For \(\mathbf{F}(x,y)=\langle y,x^2\rangle\), what is the density \(Q_x-P_y\)?
  2. If a counterclockwise boundary has circulation \(12\), what is the circulation around the same boundary traversed clockwise?
  3. For the rectangle \(1\le x\le4\), \(0\le y\le2\), and constant density \(3\), what is the circulation around its counterclockwise boundary?
  4. What are the \(x\)-then-\(y\) bounds for the triangle \(x\ge0\), \(y\ge0\), \(x+y\le3\)?
  5. Why is it incorrect to use \(P_y-Q_x\) in the circulation form of Green’s Theorem?

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