Multivariable calculus

Curl of the Vector Field (x^2, y^2, z^2)

Q: If the curl is zero, is the field conservative?

Yes here. The field is smooth on all of \mathbb{R}^3, which is simply connected, so zero curl guarantees a potential, and f = \frac{x^3 + y^3 + z^3}{3} works. Check by differentiating: \frac{\partial f}{\partial x} = x^2, and the other two match the same way.

Q: Can a field have zero curl and zero divergence at once?

Yes. \langle yz,\ xz,\ xy \rangle is one: it is the gradient of xyz, so its curl vanishes, and no component contains its own variable, so its divergence vanishes too. Such fields are the gradients of harmonic functions, and they are the quiet case rather than the typical one.

The curl of the field with components x squared, y squared and z squared is the zero vector. Every cross derivative in the formula differentiates a component with respect to a variable it does not contain, so all six of them are zero. The divergence, by contrast, is 2x plus 2y plus 2z.

P(x,y,z) = x^2

Q(x,y,z) = y^2

R(x,y,z) = z^2

\nabla \times \mathbf{F} = \langle 0,\ 0,\ 0 \rangle

Every partial in the formula is a derivative of a constant

The curl formula uses six partials, and not one of them is a derivative of a component with respect to its own variable. $P$ is differentiated in $y$ and $z$ , $Q$ in $z$ and $x$ , $R$ in $x$ and $y$ .

Here $P = x^2$ contains no $y$ and no $z$ , so both of its partials are zero, and the same is true of $Q$ and $R$ . Six zeros make three zero differences.

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

So the answer is $\mathbf{0}$ without any cancellation happening at all. That is a different situation from a gradient field, where the terms are nonzero and cancel in pairs, and it is worth knowing which of the two you are looking at.

Zero curl does not mean nothing is happening

This flow is pushing hard, and never in a circle. Along the $x$ -axis every particle moves in the $+x$ direction, since $x^2$ is never negative, coasting to a halt at the origin and picking up speed again on the far side. The same pattern runs along the other two axes. None of that motion is rotation, which is why the curl misses it entirely.

\nabla \cdot \mathbf{F} = 2x + 2y + 2z, \qquad \mathbf{F} = \nabla \left( \frac{x^3 + y^3 + z^3}{3} \right)

The divergence is nonzero except on the plane $x + y + z = 0$ , so the field has sources and sinks nearly everywhere. The gradient statement is the other half of the story: on a simply connected domain, zero curl is exactly the condition for a potential, and $\frac{x^3+y^3+z^3}{3}$ is one.

The mistake: computing the divergence and calling it the curl

The two operators are built from the same nine partials of the field but use disjoint halves of them. Divergence takes the three matched ones, $\frac{\partial P}{\partial x}$ , $\frac{\partial Q}{\partial y}$ and $\frac{\partial R}{\partial z}$ . Curl takes the other six.

On this field the matched three are the only nonzero partials, so a student who confuses the operators gets $2x + 2y + 2z$ and thinks the curl is nonzero. The right answer is that curl and divergence disagree here as strongly as they can.

Divergence returns a scalar, curl returns a vector; a single expression as your curl answer is already wrong.
Divergence uses each component with its own variable; curl never does.
A quick shape check: if $\nabla \times \mathbf{F}$ came out as one number rather than three, go back to the determinant.

The build differentiates P, Q and R numerically at seeded points and fails if any curl component stops coming out zero.

Frequently asked questions

If the curl is zero, is the field conservative?

Yes here. The field is smooth on all of $\mathbb{R}^3$ , which is simply connected, so zero curl guarantees a potential, and $f = \frac{x^3 + y^3 + z^3}{3}$ works. Check by differentiating: $\frac{\partial f}{\partial x} = x^2$ , and the other two match the same way.

Can a field have zero curl and zero divergence at once?

Yes. $\langle yz,\ xz,\ xy \rangle$ is one: it is the gradient of $xyz$ , so its curl vanishes, and no component contains its own variable, so its divergence vanishes too. Such fields are the gradients of harmonic functions, and they are the quiet case rather than the typical one.