Multivariable calculus

Curl of the Vector Field (xy, yz, zx)

The curl of the field with components xy, yz and zx is the vector with components -y, -z and -x. In each of the three differences the first partial is zero and the second is not, so nothing cancels: the y-partial of the third component is 0 and the z-partial of the second is y, which gives -y.

P(x,y,z)=xyP(x,y,z) = xy
Q(x,y,z)=yzQ(x,y,z) = yz
R(x,y,z)=zxR(x,y,z) = zx
∇×F=⟨−y, −z, −x⟩\nabla \times \mathbf{F} = \langle -y,\ -z,\ -x \rangle

Take the three differences one at a time

Every component of this field is a product of two variables and skips the third, so half of the six partials in the curl formula are derivatives of an expression that does not contain the variable. Those are the zeros.

∂R∂y−∂Q∂z=0−y=−y\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - y = -y
∂P∂z−∂R∂x=0−z=−z\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - z = -z
∂Q∂x−∂P∂y=0−x=−x\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - x = -x

Read the first line slowly, since it is the one people rush. R=zxR = zx has no yy in it, so ∂R∂y=0\frac{\partial R}{\partial y} = 0. Q=yzQ = yz differentiated in zz gives yy. The difference is 0−y0 - y, not y−0y - 0.

Use the symmetry as a check, not as a method

The field is unchanged by the cyclic relabelling x→y→z→xx \to y \to z \to x: it sends xy→yzxy \to yz, yz→zxyz \to zx and zx→xyzx \to xy. Curl respects that relabelling, so the answer must be cyclic too, and ⟨−y,−z,−x⟩\langle -y, -z, -x \rangle is.

That is worth one glance at the end, because a sign or index slip almost always breaks the pattern. It is not a way to get the answer, though: the moment a field is not cyclic the shortcut says nothing, and most fields are not.

The curl is nonzero everywhere except the origin, so this field is not conservative and no potential exists. Vanishing at a single point is not vanishing on a region.

The mistake: dropping the minus signs

Because the first term of every difference is zero, the arithmetic feels finished before the sign is applied, and the answer gets written as ⟨y,z,x⟩\langle y, z, x \rangle. That is the single most common wrong answer for this field, and it is off by a factor of −1-1 in every component.

Write the zero down. Filling in ∂R∂y=0\frac{\partial R}{\partial y} = 0 on the page, rather than skipping it, keeps the subtraction visible and the minus sign attached.

Then test one point. At (1,1,1)(1,1,1) the flow is ⟨1,1,1⟩\langle 1,1,1 \rangle and the curl is ⟨−1,−1,−1⟩\langle -1,-1,-1 \rangle, so the spin axis points back along the flow there. The wrong answer would have it pointing forward.

Each component is verified numerically on every build by differentiating P, Q and R at seeded sample points.

Frequently asked questions

Is this field conservative?

No. A conservative field has zero curl throughout its domain, and here the curl is ⟨−y,−z,−x⟩\langle -y, -z, -x \rangle, which is the zero vector only at the origin. A single point is not an open region, so no potential exists on any ball, and a search for one would break down at the first matching step.

Is the curl always antiparallel to the field?

No, that is a coincidence of the point (1,1,1)(1,1,1), where F=⟨1,1,1⟩\mathbf{F} = \langle 1,1,1 \rangle and ∇×F=⟨−1,−1,−1⟩\nabla \times \mathbf{F} = \langle -1,-1,-1 \rangle. At (1,2,3)(1,2,3) the field is ⟨2,6,3⟩\langle 2,6,3 \rangle while the curl is ⟨−2,−3,−1⟩\langle -2,-3,-1 \rangle, and those two are not parallel in either direction.