Curl as Circulation per Unit Area

Local Circulation and the Curl Density

A vector field can make a small loop circulate counterclockwise or clockwise. The planar quantity that measures this local tendency is $Q_x-P_y$, where the field is $\mathbf{F}(x,y)=(P(x,y),Q(x,y))$. Its value describes circulation relative to area, rather than circulation alone.

This interpretation comes directly from the circulation form of Green’s theorem. For a positively oriented simple closed boundary $C$ enclosing a region $D$, the theorem relates the boundary circulation to the accumulated values of $Q_x-P_y$ across the region. Looking at increasingly small regions then explains why the integrand represents a local circulation density.

Definition: Let $\mathbf{F}(x,y)=(P(x,y),Q(x,y))$, with $P$ and $Q$ continuously differentiable on a neighborhood of a point. The planar curl density at that point is $Q_x-P_y$. It is the scalar curl associated with the counterclockwise orientation in the plane.

Here $Q_x$ means the partial derivative of the vertical component $Q$ with respect to $x$, and $P_y$ means the partial derivative of the horizontal component $P$ with respect to $y$. The order matters: the circulation form of Green’s theorem uses $Q_x-P_y$, not its negative.

For a positively oriented boundary $C$, circulation is the line integral $\oint_C P\,dx+Q\,dy$. Positive orientation means that, as the boundary is traversed, the enclosed region remains on the left. For a disk or rectangle, this is counterclockwise traversal.

Why the Density Is a Local Limit

Let $g(x,y)=Q_x(x,y)-P_y(x,y)$. Take a point $(a,b)$ and a disk $B_r$ of radius $r$ centered there. Assume $P$ and $Q$ are continuously differentiable on an open neighborhood of the disk. Green’s theorem gives the circulation around its counterclockwise boundary as the area integral of $g$.

\oint_{\partial B_r} P\,dx+Q\,dy = \iint_{B_r}g(x,y)\,dA.

Dividing by the disk’s area gives circulation per unit area. Because $g$ is continuous at the center, its values throughout a sufficiently small disk are close to $g(a,b)$. The average value of $g$ over that disk therefore approaches $g(a,b)$ as the radius shrinks.

Apply Green’s theorem.
The field is continuously differentiable near the disk, so the circulation equals the integral of $g$ over the disk.

Divide by area.
The circulation per unit area is the average value of $g$ over the disk.

Shrink the disk.
Continuity makes the average value approach the value of $g$ at the center.

This reasoning can be written as a limit. The inequality below follows because the difference between an average and $g(a,b)$ is bounded by the largest difference between $g$ and $g(a,b)$ on the disk. Continuity makes that bound tend to zero.

\left| \frac{1}{\pi r^2} \oint_{\partial B_r}P\,dx+Q\,dy-g(a,b) \right| = \left| \frac{1}{\pi r^2} \iint_{B_r}\bigl(g(x,y)-g(a,b)\bigr)\,dA \right| \leq \sup_{(x,y)\in B_r}|g(x,y)-g(a,b)| \longrightarrow 0.

Local limit: If $P$ and $Q$ are continuously differentiable near $(a,b)$, then $$ \lim_{r\to 0} \frac{1}{\pi r^2} \oint_{\partial B_r}P\,dx+Q\,dy = Q_x(a,b)-P_y(a,b), $$ where each circle is traversed counterclockwise.

The hypotheses matter: the partial derivatives must exist and be continuous in a neighborhood where Green’s theorem applies. The limit says that the curl density is a local rate. It does not say that the circulation around every finite loop equals the density at its center multiplied by the area.

Reading the Sign and Magnitude

A positive value of $Q_x-P_y$ means that sufficiently small counterclockwise loops around the point have positive circulation per unit area. A negative value means that their circulation per unit area is negative: counterclockwise traversal has a net tendency opposite to the positive orientation. A value of zero means the limiting circulation per unit area is zero; it does not guarantee zero circulation around every finite loop.

The magnitude measures how much circulation is produced locally per unit area. For a small region, multiplying the density near its center by the region’s area gives an approximation to circulation. This approximation becomes more accurate as the region shrinks, provided the density is continuous. For a larger region with variable density, the exact circulation depends on the integral of the density over the whole region.

Worked Example: Constant Positive Density on a Disk

Consider $\mathbf{F}(x,y)=(-2y,3x)$ on the disk $D$ of radius $2$ centered at the origin. Here $P=-2y$ and $Q=3x$, so

Q_x=3,\qquad P_y=-2,\qquad Q_x-P_y=3-(-2)=5.

The density is constantly $5$, so Green’s theorem gives the circulation as $5$ times the area of the disk. Its area is $\pi(2)^2=4\pi$, hence

\oint_{\partial D}P\,dx+Q\,dy = \iint_D 5\,dA = 5(4\pi) = 20\pi.

Check by parametrizing the boundary counterclockwise: $x=2\cos t$, $y=2\sin t$, for $0\leq t\leq2\pi$. Then $dx=-2\sin t\,dt$, $dy=2\cos t\,dt$, $P=-4\sin t$, and $Q=6\cos t$. Thus

\oint_{\partial D}P\,dx+Q\,dy = \int_0^{2\pi} \left(8\sin^2t+12\cos^2t\right)\,dt = 8\pi+12\pi = 20\pi.

The direct boundary calculation agrees with the area calculation. The positive density means the net circulation is counterclockwise.

Variable Density: The Average Matters

When the density varies over a region, the total circulation is the area integral of all its local contributions. Positive and negative values can reinforce or cancel. The average density over a region $D$ of positive area is the total circulation divided by the area:

\text{average density on }D = \frac{1}{\operatorname{Area}(D)} \iint_D(Q_x-P_y)\,dA = \frac{1}{\operatorname{Area}(D)} \oint_{\partial D}P\,dx+Q\,dy.

This average equals the density at some point of the region when the density is continuous and the region is connected, by the integral mean value theorem. Even without using that conclusion, the integral formula shows why a single point’s density cannot generally determine circulation around a large region.

Worked Example: Variable Density on a Rectangle

Let $\mathbf{F}(x,y)=(0,x^2)$ and let $D=[1,3]\times[0,2]$, with its boundary traversed counterclockwise. We have $P=0$ and $Q=x^2$, so

Q_x=2x,\qquad P_y=0,\qquad Q_x-P_y=2x.

Integrating across the rectangle gives

\oint_{\partial D}P\,dx+Q\,dy = \int_0^2\int_1^3 2x\,dx\,dy = \int_0^2 [x^2]_1^3\,dy = \int_0^2 (9-1)\,dy = 16.

The rectangle has area $(3-1)(2-0)=4$, so its average density is $16/4=4$. The density itself ranges from $2$ at $x=1$ to $6$ at $x=3$, and its average over the rectangle is $4$.

Check directly on the four sides. On the bottom and top, $dy=0$, and $P=0$, so each contribution is zero. On the right side, $x=3$, $y$ increases from $0$ to $2$, and $Q\,dy=9\,dy$, giving $18$. On the left side, $x=1$, $y$ decreases from $2$ to $0$, and $Q\,dy=1\,dy$, giving $-2$. The total is $0+18+0-2=16$, as required.

A Disk with Density That Changes Across It

A density may vary even when it has one sign throughout a region. In that case, the sign of the total circulation is clear, while its magnitude must account for how the density varies with position.

Worked Example: Radially Varying Density on a Disk

Consider $\mathbf{F}(x,y)=\bigl(-y(x^2+y^2),\,x(x^2+y^2)\bigr)$ on the unit disk. Thus $P=-x^2y-y^3$ and $Q=x^3+xy^2$. Differentiate each component with respect to the required variable:

Q_x=3x^2+y^2,\qquad P_y=-x^2-3y^2, \qquad Q_x-P_y=4x^2+4y^2.

In polar coordinates, $x^2+y^2=r^2$, and the unit disk has $0\leq r\leq1$, $0\leq\theta\leq2\pi$, with area element $r\,dr\,d\theta$. Therefore

\oint_{\partial D}P\,dx+Q\,dy = \int_0^{2\pi}\int_0^1 4r^2(r\,dr\,d\theta) = \int_0^{2\pi}\left[r^4\right]_0^1\,d\theta = 2\pi.

Check on the boundary $r=1$, where $x=\cos t$, $y=\sin t$, $0\leq t\leq2\pi$. Along it, $P=-\sin t$, $Q=\cos t$, $dx=-\sin t\,dt$, and $dy=\cos t\,dt$. Hence

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

The density is zero at the center and positive elsewhere in the disk. Its average is $2\pi/\pi=2$, so the total circulation is positive despite the density not being constant.

A Common Sign Error

A frequent mistake is to reverse the derivatives and use $P_y-Q_x$. For the field $\mathbf{F}=(-2y,3x)$, that incorrect expression gives $-2-3=-5$, which would predict circulation $-20\pi$ around the radius-$2$ disk. But the direct boundary calculation gives $20\pi$. The correction is to use the circulation form of Green’s theorem with the positive orientation: the integrand is $Q_x-P_y=3-(-2)=5$.

Takeaway: For a counterclockwise boundary, the local circulation density is $Q_x-P_y$. Its value is circulation per unit area in the shrinking-region limit; on a finite region, integrate the density over the entire area.

Check Your Understanding

Use the definition and the circulation form of Green’s theorem to answer these questions.

For $\mathbf{F}(x,y)=(y,x)$, compute the curl density. What does its value imply about limiting circulation per unit area?
If $Q_x-P_y$ is negative throughout a disk, what is the sign of its counterclockwise boundary circulation?
Why is the circulation around a finite region not generally equal to the density at its center times the area?
For $\mathbf{F}(x,y)=(0,3x^2)$ on $1\leq x\leq2$, $0\leq y\leq1$, find the density and the net counterclockwise circulation.
State the smoothness condition needed to apply the local-limit argument using Green’s theorem.