Vector field explorer

Pick a field $\mathbf{F} = (P, Q)$ and place the circle with the sliders, or drag it across the graph. The readouts compare the two line integrals around the circle with the double integrals over the disk inside it. Green’s theorem says each pair should agree, and the vortex shows what happens when one of its conditions fails.

Vector field F

Center x0.0

Center y0.0

Radius1.0

You can also drag the circle or tap the graph to move its center. Longer arrows in fuller color mean a stronger field.

Circulation around the circle: 6.2832Equals 2π
Integral of the curl over the disk: 6.2832Equals 2π
Outward flux across the circle: 0.0000Equals 0
Integral of the divergence over the disk: 0.0000Equals 0
Curl at the center: 2
Divergence at the center: 0

Each line integral equals its double integral, as Green's theorem says.

Things to try

Start with the rotation field $(-y, x)$ and the unit circle at the origin. The circulation is $2\pi \approx 6.2832$ , and so is the integral of the curl, because the curl is $2$ everywhere and the disk has area $\pi$ . Move the center anywhere and nothing changes. Set the radius to $2$ and both become $8\pi \approx 25.1327$ .
Switch to the radial field $(x, y)$ . Now the circulation is $0$ and the outward flux is $2\pi r^2$ , which is the divergence $2$ times the area. The sink $(-x, -y)$ gives the same numbers with the sign flipped.
Try the shear field $(y, 0)$ . Every arrow points left or right, yet the circulation around the unit circle is $-\pi \approx -3.1416$ . Above the $x$ axis the arrows point right and below it they point left, so along both the top and the bottom of the circle they push against the counterclockwise direction. The curl is $-1$ everywhere.
Pick the gradient of $x^2 - y^2$ and move the circle around. The circulation stays $0$ for every center and radius, because the field is conservative and its curl is $0$ . The radial field is a gradient too, of $(x^2 + y^2)/2$ , and its circulation is also always $0$ .
With the polynomial field $(x^2, xy)$ the curl is $y$ and the divergence is $3x$ , so the numbers depend on where the circle sits. At the origin both integrals are $0$ even though the field is not. Move the center to $(1, 0.5)$ with radius $1$ . The circulation becomes $0.5\pi \approx 1.5708$ and the flux becomes $3\pi \approx 9.4248$ , which is the area times the curl and the divergence at the center.
Choose the vortex and keep the circle at the origin. The circulation is $2\pi$ , but the integral of the curl is $0$ because the curl is $0$ wherever the field is defined. Move the center to $(0.5, 0.3)$ or change the radius, and the circulation stays $2\pi$ as long as the origin is inside.
Now move the vortex circle to the center $(2, 0)$ with radius $1$ . The disk no longer contains the origin, the circulation drops to $0$ and the two sides agree again. A circle through the origin, such as center $(1, 0)$ with radius $1$ , makes the line integrals undefined.

What you are seeing

The circle $C$ with center $(a, b)$ and radius $r$ is traced counterclockwise,

\begin{gathered} x = a + r\cos t, \qquad y = b + r\sin t, \\[0.4em] 0 \le t \le 2\pi. \end{gathered}

The circulation adds up the part of $\mathbf{F}$ along the circle and the flux adds up the part pointing out of it. Green’s theorem turns each into a double integral over the disk $D$ , once with the scalar curl and once with the divergence.

\begin{aligned} \oint_C \mathbf{F} \cdot \mathbf{T} \, ds &= \oint_C P \, dx + Q \, dy \\[0.4em] &= \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \end{aligned}

\begin{aligned} \oint_C \mathbf{F} \cdot \mathbf{n} \, ds &= \oint_C P \, dy - Q \, dx \\[0.4em] &= \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) dA \end{aligned}

The theorem needs $P$ and $Q$ to have continuous partial derivatives on an open region that contains the whole disk. The vortex field

\mathbf{F} = \left( \frac{-y}{x^2 + y^2}, \ \frac{x}{x^2 + y^2} \right)

has curl $0$ at every point except the origin, where it is not defined at all. On a circle of radius $r$ around the origin it points along the circle with length $\frac{1}{r}$ , so

\oint_C \mathbf{F} \cdot \mathbf{T} \, ds = \frac{1}{r} \cdot 2\pi r = 2\pi.

When the disk contains the origin, the condition fails and the theorem says nothing, which is why the two sides can differ. The flux of the vortex is $0$ on both sides, but not because of the theorem. For this field $P \, dy - Q \, dx = -d(\ln r)$ , and $\ln r$ returns to its starting value after one trip around any circle that does not pass through the origin, even one that encloses it. When the disk misses the origin, the field is smooth on it and Green’s theorem gives a circulation of $0$ .

The explorer computes the line integrals numerically with the trapezoid rule in $t$ , using more points the closer a circle in the vortex field comes to the origin. For a smooth periodic integrand this rule is accurate to many digits. The double integrals use the exact curl and divergence formulas in polar coordinates around the center, with a Gauss rule in the radius and equally spaced angles. The note under a value writes it as a multiple of $\pi$ , except where the vortex needs a warning there instead.

Each arrow sits at the middle of a grid cell and points in the direction of $\mathbf{F}$ there. Its length and the strength of its color both grow with the magnitude $|\mathbf{F}|$ up to a cap, so short faint arrows mark a weak field. The cap sits where most arrows stay below it. The few very strong vectors, such as those close to the center of the vortex, are drawn at the cap so they do not shrink all the others.

The theorem with all its hypotheses is in the lesson on Green’s theorem, the formulas are collected on the multivariable reference sheet, and the whole course is on the Multivariable and Vector Calculus course page.