The Circulation Form of Green’s Theorem

From Boundary Circulation to Interior Rotation

A line integral around a closed curve records the accumulated tangential contribution of a vector field along that boundary. Directly evaluating it can require several separate parametrizations, especially when the boundary has many pieces.

The circulation form of Green’s theorem gives another route. It replaces the boundary integral of \(P\,dx+Q\,dy\) with a double integral over the region enclosed by the curve. The integrand is \(Q_x-P_y\), a quantity that measures the field’s local tendency to circulate counterclockwise.

The central idea: For a positively oriented, simple closed planar boundary, total counterclockwise circulation along the boundary equals the accumulated value of \(Q_x-P_y\) throughout the enclosed region.

Statement and Orientation

Let \(D\) be a bounded planar region whose boundary \(C\) is a simple, closed, piecewise-smooth curve. Suppose \(P\) and \(Q\) have continuous first partial derivatives on an open set containing \(D\) and its boundary. Give \(C\) the positive orientation: as you travel along it, the region \(D\) remains on your left. For a simple closed boundary, this is the counterclockwise direction.

Theorem (circulation form of Green’s theorem): Under these conditions,

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

Here \(Q_x=\frac{\partial Q}{\partial x}\), \(P_y=\frac{\partial P}{\partial y}\), and \(dA\) denotes area.

The left side is a line integral around the entire oriented boundary. The right side adds the scalar quantity \(Q_x-P_y\) across the region. In vector notation, if \(\mathbf{F}=\langle P,Q\rangle\), the expression \(Q_x-P_y\) is the scalar curl of \(\mathbf{F}\) in the plane.

The orientation matters. Reversing the direction of travel around \(C\) changes the sign of the line integral, while the double integral over \(D\) keeps the same value. Thus the theorem in the displayed form uses the counterclockwise orientation.

Why the Integrand Is \(Q_x-P_y\)

The sign and order of the partial derivatives can be seen by computing circulation around a small counterclockwise rectangle. Take the rectangle with corners \((x,y)\), \((x+h,y)\), \((x+h,y+k)\), and \((x,y+k)\), where \(h,k>0\). Travel along the bottom, right, top, and left edges in that order.

The bottom edge contributes \(\int_x^{x+h}P(s,y)\,ds\). The right edge contributes \(\int_y^{y+k}Q(x+h,t)\,dt\). The top edge is traversed from right to left, and the left edge from top to bottom, so the total boundary integral is

\oint P\,dx+Q\,dy = \int_x^{x+h}P(s,y)\,ds +\int_y^{y+k}Q(x+h,t)\,dt -\int_x^{x+h}P(s,y+k)\,ds -\int_y^{y+k}Q(x,t)\,dt.

Group the horizontal terms and the vertical terms. By the fundamental theorem of calculus applied to each difference,

\oint P\,dx+Q\,dy = -\int_x^{x+h}\int_y^{y+k}P_y(s,t)\,dt\,ds +\int_y^{y+k}\int_x^{x+h}Q_x(s,t)\,ds\,dt.

Since the functions being integrated are continuous, the order of integration can be exchanged in the second term. Therefore,

\oint P\,dx+Q\,dy = \int_x^{x+h}\int_y^{y+k}\left(Q_x(s,t)-P_y(s,t)\right)\,dt\,ds.

This is an exact identity for the rectangle. In particular, dividing by its area \(hk\) shows that its average circulation per unit area is the average value of \(Q_x-P_y\) on it. As the rectangle shrinks to a point where these partial derivatives are continuous, that average approaches the value of \(Q_x-P_y\) at the point. This local calculation explains both the integrand and its sign in Green’s theorem.

Applying the Theorem

To use the circulation form, first identify \(P\), \(Q\), and the enclosed region \(D\). Then verify that the boundary is positively oriented and that the needed partial derivatives are continuous. Compute \(Q_x-P_y\), set up its double integral over \(D\), and evaluate. The examples below also calculate the boundary integral directly, providing an independent check of the result.

Identify the field and region.
Write the line-integral coefficients as \(P\) and \(Q\), and specify the enclosed region \(D\).

Check the setting.
Confirm that the boundary is a simple closed curve with counterclockwise orientation and that \(P_x,P_y,Q_x,Q_y\) are continuous on a neighborhood of the region.

Form the area integral.
Calculate \(Q_x-P_y\), then integrate it over \(D\).

Check signs and arithmetic.
When practical, evaluate the boundary integral directly and compare it with the area integral.

Worked Example: Circulation Around the Unit Circle

Let \(P(x,y)=-y\) and \(Q(x,y)=x\), and let \(C\) be the unit circle oriented counterclockwise. The enclosed region is the unit disk \(D\). First,

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

In polar coordinates, the unit disk has \(0\le r\le1\), \(0\le\theta\le2\pi\), and \(dA=r\,dr\,d\theta\). Green’s theorem gives

\iint_D 2\,dA = \int_0^{2\pi}\int_0^1 2r\,dr\,d\theta = \int_0^{2\pi}\left[r^2\right]_0^1\,d\theta = \int_0^{2\pi}1\,d\theta = 2\pi.

Check by parametrizing the counterclockwise unit circle as \(x=\cos t\), \(y=\sin t\), \(0\le t\le2\pi\). Then \(dx=-\sin t\,dt\), \(dy=\cos t\,dt\), \(P=-\sin t\), and \(Q=\cos t\). Thus

\oint_C P\,dx+Q\,dy = \int_0^{2\pi}\left(\sin^2t+\cos^2t\right)\,dt = \int_0^{2\pi}1\,dt = 2\pi.

The direct boundary calculation agrees with the double integral.

Worked Example: A Rectangle with Nonconstant Circulation Density

Let \(P(x,y)=x^2y\), \(Q(x,y)=xy^2\), and let \(C\) be the counterclockwise boundary of \(D=[0,2]\times[0,1]\). The partial derivatives and circulation density are

Q_x=y^2,\qquad P_y=x^2,\qquad Q_x-P_y=y^2-x^2.

Integrating over the rectangle,

\begin{aligned} \iint_D(y^2-x^2)\,dA &=\int_0^2\int_0^1(y^2-x^2)\,dy\,dx\\ &=\int_0^2\left(\frac13-x^2\right)\,dx\\ &=\left[\frac{x}{3}-\frac{x^3}{3}\right]_0^2\\ &=\frac23-\frac83=-2. \end{aligned}

Check directly along the four sides. On the bottom side, \(y=0\) and \(dy=0\), so the contribution is \(0\). On the right side, \(x=2\), \(dx=0\), and \(y\) runs from \(0\) to \(1\), giving

\int_0^1 Q(2,y)\,dy = \int_0^1 2y^2\,dy = \frac23.

On the top side, \(y=1\) and \(x\) runs from \(2\) to \(0\), so

\int_2^0 P(x,1)\,dx = \int_2^0 x^2\,dx = \left[\frac{x^3}{3}\right]_2^0 = -\frac83.

On the left side, \(x=0\), so \(P=Q=0\) and the contribution is \(0\). Adding the four contributions gives \(0+\frac23-\frac83+0=-2\), as obtained from Green’s theorem.

Worked Example: A Triangular Region

Let \(P(x,y)=0\), \(Q(x,y)=x^2\), and let \(C\) be the counterclockwise boundary of the triangle with vertices \((0,0)\), \((2,0)\), and \((0,1)\). The sloping side has equation \(y=1-\frac{x}{2}\). The triangle is described by \(0\le x\le2\) and \(0\le y\le1-\frac{x}{2}\). Since \(Q_x=2x\) and \(P_y=0\), the area integral is

\begin{aligned} \iint_D 2x\,dA &=\int_0^2\int_0^{1-x/2}2x\,dy\,dx\\ &=\int_0^2 2x\left(1-\frac{x}{2}\right)\,dx\\ &=\int_0^2(2x-x^2)\,dx\\ &=\left[x^2-\frac{x^3}{3}\right]_0^2\\ &=4-\frac83=\frac43. \end{aligned}

For a direct check, the bottom side has \(dy=0\), so contributes \(0\). The vertical side has \(x=0\), so \(Q=0\) and contributes \(0\). On the sloping side, travel from \((2,0)\) to \((0,1)\) using \(x=2-2t\), \(y=t\), \(0\le t\le1\). Then \(dy=dt\), and

\int_{\text{sloping side}}x^2\,dy = \int_0^1(2-2t)^2\,dt = 4\int_0^1(1-t)^2\,dt = 4\left[t-t^2+\frac{t^3}{3}\right]_0^1 = \frac43.

The boundary integral is therefore \(0+\frac43+0=\frac43\), matching the double integral.

A Common Orientation Error

A frequent mistake is to use the correct integrand but forget that the theorem’s boundary must be counterclockwise. For example, the unit-circle calculation above gives \(2\pi\) for the counterclockwise boundary. If the parametrization instead travels clockwise, such as \(x=\cos t,\ y=-\sin t\) for \(0\le t\le2\pi\), the line integral is \(-2\pi\), not \(2\pi\).

The double integral over the disk remains \(\iint_D 2\,dA=2\pi\). That does not contradict Green’s theorem: the clockwise curve is negatively oriented, so its line integral is the negative of the theorem’s positively oriented boundary integral. Either reverse the curve’s direction or include a minus sign when using the clockwise boundary.

Sign check: For the circulation form, a counterclockwise boundary uses \(Q_x-P_y\) with a positive sign. Reversing the boundary reverses the line integral’s sign.

What the Theorem Lets You Choose

The theorem is useful because it lets you choose between two descriptions of the same quantity. A complicated boundary may enclose a region with simple double-integral bounds; conversely, a difficult area integral may correspond to a boundary that is easy to parametrize. The theorem does not eliminate the need to check the region, orientation, and differentiability assumptions. Those checks are part of the calculation, not optional formalities.

The circulation density can also change sign across a region. Positive values contribute counterclockwise circulation, while negative values contribute clockwise circulation. The total boundary circulation is the signed accumulation over the entire region; positive and negative contributions can therefore cancel.

Check Your Understanding

Use the theorem’s orientation and sign convention as you answer these questions.

State the circulation form of Green’s theorem and identify the required boundary orientation.
For \(P=xy\) and \(Q=x^2\), calculate the circulation density \(Q_x-P_y\).
Why does reversing the direction around a closed curve change the line integral’s sign?
For \(P=-y\), \(Q=x\) on the unit disk, what is the circulation density and the resulting counterclockwise circulation?
Give one reason to evaluate a boundary integral directly after using Green’s theorem.