Multivariable calculus

Curl of the Rotation Field (-y, x, 0)

The curl of the field with components -y, x and 0 is the constant vector (0, 0, 2). Two components vanish; the third is the x-partial of the second component minus the y-partial of the first, which is 1 minus -1, or 2. The field spins rigidly about the z-axis, so the curl is twice its angular velocity.

P(x,y,z)=−yP(x,y,z) = -y
Q(x,y,z)=xQ(x,y,z) = x
R(x,y,z)=0R(x,y,z) = 0
∇×F=⟨0, 0, 2⟩=2k\nabla \times \mathbf{F} = \langle 0,\ 0,\ 2 \rangle = 2\mathbf{k}

Run the determinant

Curl is the cross product of the operator ∇\nabla with the field, so the setup that never lets you down is a three by three determinant: unit vectors on top, operators in the middle, components on the bottom.

∇×F=∣ijk∂∂x∂∂y∂∂zPQR∣\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}

Expanding along the top row gives the component form you will actually use. Note the middle slot: it runs PP before RR, which is the cofactor minus sign already folded in.

∇×F=⟨∂R∂y−∂Q∂z, ∂P∂z−∂R∂x, ∂Q∂x−∂P∂y⟩\nabla \times \mathbf{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z},\ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x},\ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right\rangle

Here P=−yP = -y, Q=xQ = x and R=0R = 0, so every partial is 00, 11 or −1-1. Nothing in PP or QQ mentions zz, and RR is identically zero, which kills the first two components.

∂R∂y−∂Q∂z=0−0=0,∂P∂z−∂R∂x=0−0=0\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - 0 = 0, \qquad \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - 0 = 0
∂Q∂x−∂P∂y=1−(−1)=2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2

Why the answer is a constant

Write r=⟨x,y,z⟩\mathbf{r} = \langle x, y, z \rangle and ω=⟨0,0,1⟩\boldsymbol{\omega} = \langle 0, 0, 1 \rangle. Then ω×r\boldsymbol{\omega} \times \mathbf{r} is exactly this field, so the picture is a solid body turning about the zz-axis at one radian per unit time.

ω×r=⟨−y, x, 0⟩,∇×(ω×r)=2ω\boldsymbol{\omega} \times \mathbf{r} = \langle -y,\ x,\ 0 \rangle, \qquad \nabla \times (\boldsymbol{\omega} \times \mathbf{r}) = 2\boldsymbol{\omega}

The field itself is not constant: at (3,4,z)(3, 4, z) it equals ⟨−4,3,0⟩\langle -4, 3, 0 \rangle, whose length 55 is the distance to the axis, so the flow speeds up as you move outward. The curl ignores that growth completely.

That is the right answer physically. In a rigid rotation nothing shears past anything else, so a paddle wheel dropped at the axis and one dropped a mile out turn at the same rate, and the curl records the same 2k2\mathbf{k} at both places.

The mistake: flipping the middle component

The three slots are not built from the same left to right pattern. The middle one is ∂P∂z−∂R∂x\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, with PP first, and writing ∂R∂x−∂P∂z\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} instead flips its sign.

This field is a bad place to make that error and a good place to warn about it, because the middle component is 00 either way. The mistake survives the check and shows up on the next problem.

Two habits stop it. Expand the determinant rather than recite the pattern, so the minus sign on the j\mathbf{j} cofactor does the bookkeeping for you. Then look at the geometry: this flow lives in horizontal circles, so its curl has to point along k\mathbf{k}, and ⟨0,0,2⟩\langle 0, 0, 2 \rangle does.

The three components above are recomputed on every build by numerically differentiating P, Q and R, and the build fails if any of them stops matching.

Frequently asked questions

Does a nonzero curl mean the flow lines are curved?

No. Curl measures local spin, not the shape of the trajectories. A shear flow with perfectly straight flow lines can have nonzero curl, and a vortex with circular flow lines can have zero curl away from its axis. In this field the two happen to agree, which is why it is a safe first example and a poor test of understanding.

Why is the curl twice the angular velocity rather than equal to it?

Curl collects spin from two perpendicular directions and adds them. Here ∂Q∂x=1\frac{\partial Q}{\partial x} = 1 says the yy velocity grows as you move in xx, and −∂P∂y=1-\frac{\partial P}{\partial y} = 1 says the same for the other pair. Each contributes one unit of angular speed, so the total is 22.