Curl — definition, physical meaning (rotation): Exercises

Reminder of the only formula you need (from the parent): $\nabla \times F = (\partial_{y} F_{3} - \partial_{z} F_{2}, \partial_{z} F_{1} - \partial_{x} F_{3}, \partial_{x} F_{2} - \partial_{y} F_{1})$ Here $\partial_{x}$ means "how fast the thing changes as you nudge $x$ " (a partial derivative), and $(F_{1}, F_{2}, F_{3})$ are the three arrow-lengths of the field in the $x$ , $y$ , $z$ directions.

Level 1 — Recognition

Goal: read the definition and plug in. No cleverness required.

Problem 1.1

Compute the $z$ -component $(\nabla \times F)_{z} = \partial_{x} F_{2} - \partial_{y} F_{1}$ for $F = (3 x, 5 y, 0)$ .

Recall Solution 1.1

Here $F_{1} = 3 x$ and $F_{2} = 5 y$ .

$\partial_{x} F_{2} = \partial_{x} (5 y) = 0$ (there is no $x$ inside $5 y$ , so nudging $x$ changes nothing).
$\partial_{y} F_{1} = \partial_{y} (3 x) = 0$ (no $y$ inside $3 x$ ). $(\nabla \times F)_{z} = 0 - 0 = 0$ A field that only stretches along its own axes never swirls.

Problem 1.2

For $F = (- 2 y, 2 x, 0)$ , find $(\nabla \times F)_{z}$ . Is the field rotating clockwise or counterclockwise?

Recall Solution 1.2

$F_{1} = - 2 y$ , $F_{2} = 2 x$ .

$\partial_{x} F_{2} = \partial_{x} (2 x) = 2$ .
$\partial_{y} F_{1} = \partial_{y} (- 2 y) = - 2$ . $(\nabla \times F)_{z} = 2 - (- 2) = 4$ Positive $\hat{k}$ means counterclockwise (right-hand rule: curl the fingers the way the flow spins, the thumb $+ \hat{k}$ pops out of the page toward you).

Problem 1.3

Which of these has $(\nabla \times F)_{z} = 0$ ? (a) $F = (y, x, 0)$ (b) $F = (- y, x, 0)$

Recall Solution 1.3

(a) $\partial_{x} (x) - \partial_{y} (y) = 1 - 1 = 0$ . Zero curl. (b) $\partial_{x} (x) - \partial_{y} (- y) = 1 - (- 1) = 2$ . Nonzero curl. Answer: (a) has zero $z$ -curl. The tiny sign flip between $y$ and $- y$ is the whole difference between "no spin" and "spins".

Level 2 — Application

Goal: compute the full 3-component curl vector.

Problem 2.1

Compute $\nabla \times F$ for $F = (y, z, x)$ .

Recall Solution 2.1

Use each component $\partial_{next} F_{next} - \partial_{prev} F_{prev}$ cyclically:

$x$ -comp: $\partial_{y} F_{3} - \partial_{z} F_{2} = \partial_{y} (x) - \partial_{z} (z) = 0 - 1 = - 1$ .
$y$ -comp: $\partial_{z} F_{1} - \partial_{x} F_{3} = \partial_{z} (y) - \partial_{x} (x) = 0 - 1 = - 1$ .
$z$ -comp: $\partial_{x} F_{2} - \partial_{y} F_{1} = \partial_{x} (z) - \partial_{y} (y) = 0 - 1 = - 1$ . $\nabla \times F = (- 1, - 1, - 1)$ Every swirl-plane has $- 1$ : the spin axis is the diagonal $(- 1, - 1, - 1)$ .

Problem 2.2

Compute $\nabla \times F$ for $F = (x^{2}, y^{2}, z^{2})$ .

Recall Solution 2.2

Each component of $F$ depends only on its own variable, so every cross-derivative vanishes:

$x$ -comp: $\partial_{y} (z^{2}) - \partial_{z} (y^{2}) = 0 - 0 = 0$ .
$y$ -comp: $\partial_{z} (x^{2}) - \partial_{x} (z^{2}) = 0 - 0 = 0$ .
$z$ -comp: $\partial_{x} (y^{2}) - \partial_{y} (x^{2}) = 0 - 0 = 0$ . $\nabla \times F = (0, 0, 0)$ A "separable" field like this is irrotational.

Problem 2.3

Compute $\nabla \times F$ for $F = (0, 0, x^{2} + y^{2})$ .

Recall Solution 2.3

Only $F_{3} = x^{2} + y^{2}$ is nonzero.

$x$ -comp: $\partial_{y} (x^{2} + y^{2}) - \partial_{z} (0) = 2 y - 0 = 2 y$ .
$y$ -comp: $\partial_{z} (0) - \partial_{x} (x^{2} + y^{2}) = 0 - 2 x = - 2 x$ .
$z$ -comp: $\partial_{x} (0) - \partial_{y} (0) = 0$ . $\nabla \times F = (2 y, - 2 x, 0)$ The upward push $F_{3}$ grows as you leave the $z$ -axis, so a paddlewheel tipped sideways gets torqued — the curl lives in the horizontal plane.

Level 3 — Analysis

Goal: connect the number to the physical picture and to a paddlewheel.

Problem 3.1

A paddlewheel is dropped into the 2D field $F = (0, x, 0)$ (arrows all point vertically, longer as $x$ grows — see figure). Predict the spin direction, then confirm with curl.

Figure — Curl — definition, physical meaning (rotation)

Recall Solution 3.1

Physical prediction: to the right ( $x$ large) the upward arrows are longer/faster; to the left they are shorter. So the wheel's right side is pushed up harder than its left → it rotates counterclockwise ( $+ \hat{k}$ ). Curl check: $(\nabla \times F)_{z} = \partial_{x} (x) - \partial_{y} (0) = 1 - 0 = 1$ . Positive, matching the counterclockwise prediction. Note the arrows are perfectly straight — yet the wheel spins. This is shear, not curved flow.

Problem 3.2

For which constant $c$ is $F = (cy, x, 0)$ irrotational (curl $= 0$ )?

Recall Solution 3.2

$(\nabla \times F)_{z} = \partial_{x} (x) - \partial_{y} (cy) = 1 - c$ . Set to zero: $1 - c = 0 \Rightarrow c = 1$ . With $c = 1$ we get $F = (y, x, 0)$ , which is the gradient of $ϕ = x y$ — and gradients are always irrotational (parent identity $\nabla \times \nabla ϕ = 0$ ; see Gradient and Conservative Fields).

Problem 3.3

The field $F = (- y, x, 0)$ has curl $(0, 0, 2)$ . Using "magnitude of curl $= 2 \times$ angular speed", find the angular speed $ω$ of a tiny floating chip at the origin.

Recall Solution 3.3

$∣\nabla \times F ∣ = 2$ , and $∣ curl ∣ = 2 ω$ , so $ω = \frac{2}{2} = 1 radian per unit time.$ The field genuinely rotates like a rigid turntable at $1$ rad/s; curl over-counts angular speed by exactly a factor $2$ (each of the two cross-terms contributes one $ω$ ).

Level 4 — Synthesis

Goal: combine curl with divergence, gradients and Stokes/Green.

Problem 4.1

Take $F = (x, y, 0)$ . Compute both its divergence $\nabla \cdot F = \partial_{x} F_{1} + \partial_{y} F_{2} + \partial_{z} F_{3}$ and its curl. What does the pair of answers prove?

Recall Solution 4.1

Divergence: $\partial_{x} (x) + \partial_{y} (y) + \partial_{z} (0) = 1 + 1 + 0 = 2$ . Curl: every cross-derivative is $0$ , so $\nabla \times F = (0, 0, 0)$ . A field can have strong divergence yet zero curl. This proves divergence and curl answer independent questions — see Divergence — definition and physical meaning. This one only spreads out (a source), never swirls.

Problem 4.2

Let $ϕ = x^{2} y + z$ . Compute $F = \nabla ϕ = (\partial_{x} ϕ, \partial_{y} ϕ, \partial_{z} ϕ)$ and then $\nabla \times F$ . Explain the result via mixed partials.

Recall Solution 4.2

$\nabla ϕ = (2 x y, x^{2}, 1)$ . Curl:

$x$ -comp: $\partial_{y} (1) - \partial_{z} (x^{2}) = 0 - 0 = 0$ .
$y$ -comp: $\partial_{z} (2 x y) - \partial_{x} (1) = 0 - 0 = 0$ .
$z$ -comp: $\partial_{x} (x^{2}) - \partial_{y} (2 x y) = 2 x - 2 x = 0$ . $\nabla \times \nabla ϕ = (0, 0, 0)$ The $z$ -component is exactly $\partial_{x} \partial_{y} ϕ - \partial_{y} \partial_{x} ϕ$ , which is $0$ by Clairaut's theorem. Every gradient field is irrotational.

Problem 4.3

Use Green's Theorem to find the circulation $\oint_{C} F \cdot d r$ of $F = (- y, x, 0)$ around the unit circle $C$ (radius $1$ , counterclockwise). Green: $\oint_{C} F \cdot d r = \iint_{D} (\partial_{x} F_{2} - \partial_{y} F_{1}) d A$ .

Recall Solution 4.3

The integrand is the $z$ -curl: $\partial_{x} (x) - \partial_{y} (- y) = 1 - (- 1) = 2$ (constant). $\oint_{C} F \cdot d r = \iint_{D} 2 d A = 2 \cdot (area of unit disk) = 2 π .$ $2 π$ Because curl is constant, "circulation = curl $\times$ area" — the tiny-loop definition scaled up. This is Green's Theorem, the 2D face of Stokes' Theorem.

Level 5 — Mastery

Goal: prove general statements and handle degenerate cases.

Problem 5.1

Prove that for any smooth $F$ , $\nabla \cdot (\nabla \times F) = 0$ .

Recall Solution 5.1

Write $\nabla \times F = (\partial_{y} F_{3} - \partial_{z} F_{2}, \partial_{z} F_{1} - \partial_{x} F_{3}, \partial_{x} F_{2} - \partial_{y} F_{1})$ . Take its divergence (sum of $\partial_{x}$ , $\partial_{y}$ , $\partial_{z}$ of the three components): $\partial_{x} (\partial_{y} F_{3} - \partial_{z} F_{2}) + \partial_{y} (\partial_{z} F_{1} - \partial_{x} F_{3}) + \partial_{z} (\partial_{x} F_{2} - \partial_{y} F_{1}) .$ Group the six terms into pairs: $(\partial_{x} \partial_{y} F_{3} - \partial_{y} \partial_{x} F_{3}) + (\partial_{z} \partial_{x} F_{2} - \partial_{x} \partial_{z} F_{2}) + (\partial_{y} \partial_{z} F_{1} - \partial_{z} \partial_{y} F_{1}) .$ Each parenthesis is zero by Clairaut's theorem (order of mixed partials doesn't matter). So the total is $0$ . "A pure swirl has no source."

Problem 5.2

Find every constant $a$ so that $F = (a y + z, x + 2 z, x + 3 y)$ is irrotational, or show none/all work.

Recall Solution 5.2

Compute the three curl components:

$x$ -comp: $\partial_{y} (x + 3 y) - \partial_{z} (x + 2 z) = 3 - 2 = 1$ .
$y$ -comp: $\partial_{z} (a y + z) - \partial_{x} (x + 3 y) = 1 - 1 = 0$ .
$z$ -comp: $\partial_{x} (x + 2 z) - \partial_{y} (a y + z) = 1 - a$ . The $x$ -component is $1 \neq = 0$ regardless of $a$ . So the curl can never be zero. $No value of a makes it irrotational.$ (Even if we chose $a = 1$ to kill the $z$ -component, the stubborn $x$ -component survives.) A single nonzero component is enough to break irrotationality.

Problem 5.3

Degenerate cases. For each, state the curl and interpret. (a) The zero field $F = (0, 0, 0)$ . (b) A uniform field $F = (c_{1}, c_{2}, c_{3})$ with constants $c_{i}$ . (c) A field with a single nonzero constant component $F = (0, 0, K)$ .

Recall Solution 5.3

(a) Every derivative of $0$ is $0$ : $\nabla \times F = (0, 0, 0)$ . No flow, no spin — the trivial baseline. (b) Derivatives of constants are $0$ , so all six terms vanish: $\nabla \times F = (0, 0, 0)$ . A uniform "wind blowing everywhere the same way" pushes a paddlewheel bodily but never spins it. (c) Same reasoning — $K$ is constant, all derivatives $0$ : $\nabla \times F = (0, 0, 0)$ . Constant fields are always irrotational; curl only sees change, and constants never change.

Recall Final self-check (cover the answers)

Curl of $(- 2 y, 2 x, 0)$ ? ::: $(0, 0, 4)$ Curl of $(y, z, x)$ ? ::: $(- 1, - 1, - 1)$ Circulation of $(- y, x, 0)$ around the unit circle? ::: $2 π$ Does any $a$ make $(a y + z, x + 2 z, x + 3 y)$ irrotational? ::: No — the $x$ -component of curl is $1$ for every $a$ . Curl of any constant field? ::: $(0, 0, 0)$

Connections

Curl — definition, physical meaning (rotation) (parent)
Divergence — definition and physical meaning
Green's Theorem · Stokes' Theorem · Line Integrals and Circulation
Gradient and Conservative Fields · Equality of Mixed Partial Derivatives (Clairaut)