Multivariable calculus

Curl of the Gradient Field (yz, xz, xy)

The curl of the field with components yz, xz and xy is the zero vector. The field is the gradient of the product xyz, and Clairaut's theorem makes the mixed second partials in each curl component agree, so all three differences cancel. Its divergence is zero as well.

P(x,y,z)=yzP(x,y,z) = yz
Q(x,y,z)=xzQ(x,y,z) = xz
R(x,y,z)=xyR(x,y,z) = xy
∇×F=⟨0, 0, 0⟩\nabla \times \mathbf{F} = \langle 0,\ 0,\ 0 \rangle

Spot the gradient before you compute anything

Set f=xyzf = xyz. Differentiating in each variable in turn gives yzyz, xzxz and xyxy, which is this field exactly. Recognising that saves the whole computation, because the curl of a gradient is always zero.

∇(xyz)=⟨yz, xz, xy⟩\nabla (xyz) = \langle yz,\ xz,\ xy \rangle

The reason is Clairaut's theorem. Each curl component is a difference of two mixed second partials of ff, taken in opposite orders, and for a function with continuous second partials the order does not matter.

∂Q∂x−∂P∂y=∂∂x(∂f∂y)−∂∂y(∂f∂x)=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}\left( \frac{\partial f}{\partial y} \right) - \frac{\partial}{\partial y}\left( \frac{\partial f}{\partial x} \right) = 0

That argument holds for any ff built from polynomials, exponentials, sines and cosines, which is every potential you will meet in this course.

Do the arithmetic anyway

The identity is worth trusting only after you have watched it work once. All six partials here are nonzero, and they cancel in matched pairs.

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

Contrast this with ⟨x2,y2,z2⟩\langle x^2, y^2, z^2 \rangle, whose curl is also zero. There every partial in the formula was zero on its own; here every partial is alive and the cancellation does the work. Both are conservative, but only one of them shows you Clairaut in action.

The mistake: swapping the two vanishing identities

Two identities get memorised together and then confused: ∇×(∇f)=0\nabla \times (\nabla f) = \mathbf{0} for any smooth ff, and ∇⋅(∇×G)=0\nabla \cdot (\nabla \times \mathbf{G}) = 0 for any smooth G\mathbf{G}. The curl of a gradient is zero; the divergence of a curl is zero. Neither says anything about the divergence of a gradient.

The divergence of a gradient is the Laplacian, ∇⋅(∇f)=∇2f\nabla \cdot (\nabla f) = \nabla^2 f, and it is usually not zero. It happens to vanish here because f=xyzf = xyz is harmonic, with all three second partials ∂2f∂x2\frac{\partial^2 f}{\partial x^2}, ∂2f∂y2\frac{\partial^2 f}{\partial y^2} and ∂2f∂z2\frac{\partial^2 f}{\partial z^2} equal to zero.

Test the difference on f=x2f = x^2. Its gradient is ⟨2x,0,0⟩\langle 2x, 0, 0 \rangle, whose curl is 0\mathbf{0} as the identity promises, but whose divergence is 22. Zero curl came free; zero divergence did not.

The identity is not taken on trust: each build differentiates P, Q and R numerically and checks that all three curl components come out zero.

Frequently asked questions

Why does Clairaut's theorem matter here?

Because the curl of a gradient is zero only when the mixed second partials agree, and that needs them to be continuous. For f=xyzf = xyz they are polynomials, so the hypothesis holds everywhere and the conclusion is safe. Functions built to have discontinuous second partials can fail it, which is why the theorem carries a hypothesis at all.

Does a zero divergence follow from the field being a gradient?

No. ∇⋅(∇f)\nabla \cdot (\nabla f) is the Laplacian ∇2f\nabla^2 f, which vanishes only for harmonic ff. Here f=xyzf = xyz is harmonic, so both operators return zero, but f=x2+y2f = x^2 + y^2 gives the gradient ⟨2x,2y,0⟩\langle 2x, 2y, 0 \rangle with zero curl and divergence 44.