Multivariable calculus

Curl of the Gradient Field (2xy, x^2 + z^2, 2yz)

The curl of the field with components 2xy, x squared plus z squared, and 2yz is the zero vector. Each pair of cross partials cancels: the x-partial of the second component and the y-partial of the first are both 2x, and the same happens twice more. The field is the gradient of x squared times y plus y times z squared.

P(x,y,z) = 2xy

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

R(x,y,z) = 2yz

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

Check the three cross derivative pairs

A curl component is a difference of two partials, so it vanishes when the two agree. Take the pairs in order and write both halves down, even the ones that are obviously zero.

\frac{\partial R}{\partial y} = 2z, \qquad \frac{\partial Q}{\partial z} = 2z \quad \Rightarrow \quad 2z - 2z = 0

\frac{\partial P}{\partial z} = 0, \qquad \frac{\partial R}{\partial x} = 0 \quad \Rightarrow \quad 0 - 0 = 0

\frac{\partial Q}{\partial x} = 2x, \qquad \frac{\partial P}{\partial y} = 2x \quad \Rightarrow \quad 2x - 2x = 0

Three zeros, so $\nabla \times \mathbf{F} = \mathbf{0}$ at every point of space. The cancellation is not luck: it is the signature of a field that came from a potential.

Recover the potential

This field is defined and smooth on all of $\mathbb{R}^3$ , which is simply connected, so zero curl guarantees a scalar $f$ with $\nabla f = \mathbf{F}$ . Build it one variable at a time, starting from $P$ .

f = \int 2xy \,dx = x^2 y + g(y,z)

The constant of integration is a whole function of the other two variables, since anything without an $x$ differentiates to zero in $x$ . Pin it down by matching $Q$ .

\frac{\partial f}{\partial y} = x^2 + \frac{\partial g}{\partial y} = x^2 + z^2 \quad \Rightarrow \quad g = yz^2 + h(z)

Now match $R$ : $\frac{\partial f}{\partial z} = 2yz + h'(z)$ must equal $2yz$ , so $h$ is constant. The potential is $f = x^2 y + y z^2$ , and differentiating it back reproduces all three components.

\nabla \left( x^2 y + y z^2 \right) = \langle 2xy,\ x^2 + z^2,\ 2yz \rangle

The mistake: stopping after one component

In two variables the whole conservative test is $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ , and that habit carries over badly. In three variables it is only the third component of the curl, and a field can pass it while the other two components are wide awake.

Take $\langle 2xy,\ x^2,\ y \rangle$ . The third component of its curl is $2x - 2x = 0$ , exactly as here, but the first is $\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 1 - 0 = 1$ , so that field has no potential. One component proves nothing.

The second slip is skipping the domain. Zero curl gives a potential on a simply connected region, and nowhere else automatically. Polynomials on all of $\mathbb{R}^3$ are the easy case; a field with a line or a plane removed from its domain needs a second thought.

All three components are recomputed on every build by numerically differentiating P, Q and R, so a zero claimed here is a zero the machine has seen.

Frequently asked questions

Does zero curl always mean a potential exists?

On a simply connected domain, yes. This field is smooth on all of $\mathbb{R}^3$ , so zero curl is enough and $f = x^2 y + y z^2$ is the potential. On a domain with a line removed, such as the space around the $z$ -axis, zero curl does not settle the question and you have to test a loop directly.

Should I check the curl before hunting for a potential?

Yes, it is much faster. Three subtractions either rule the field out or clear it, whereas the potential hunt costs three integrations and a matching step. If any curl component is nonzero the hunt is guaranteed to fail partway through, usually at the stage where an impossible condition on $h(z)$ appears.