Worked examples — Curl — definition, physical meaning (rotation)

Before anything, recall the only formula we need, in plain words:

If any symbol here feels new, the parent Curl — definition, physical meaning (rotation) builds each one from the paddlewheel picture. We also lean on Line Integrals and Circulation (what circulation means) and Equality of Mixed Partial Derivatives (Clairaut) (why two identities vanish).

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The scenario matrix

Every curl problem falls into one of these cells. The right column names the example that nails it.

#	Case class	What is tricky about it	Example
A	Positive curl (CCW spin)	sign is $+$	Ex 1
B	Negative curl (CW spin)	sign is $-$	Ex 2
C	Zero curl, straight arrows	expected zero, get zero	Ex 3
D	Zero curl, curved arrows	looks like it must spin — doesn't	Ex 4
E	Nonzero curl, straight arrows (shear)	looks like it can't spin — does	Ex 5
F	Position-dependent curl (varies with point)	answer is a function, not a number	Ex 6
G	Full 3-D curl (tilted axis)	all three components alive	Ex 7
H	Degenerate / constant field	zero because nothing changes	Ex 8
I	Word problem (physical rotation)	translate reality → field	Ex 9
J	Exam twist (prove an identity)	symbolic, no numbers	Ex 10

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Example 1 — Cell A: positive (counterclockwise) curl

Forecast: picture arrows: at $(1,0)$ the field is $(0,1)$ (points up); at $(0,1)$ it is $(-1,0)$ (points left). Whirling counterclockwise. Guess the sign of the $z$-component before reading on.

1. Only the $z$-component can be nonzero (no $z$-dependence, $F_3=0$), so compute ${\partial }_x{F}_2-{\partial }_y{F}_1$. Why this step? $F_1,F_2$ have no $z$ in them and $F_3=0$, so the $x$- and $y$-components of curl are all $\partial (\cdot )0$ or ${\partial }_z(\cdot )=0$. Only the flat-plane swirl survives.
2. ${\partial }_x{F}_2={\partial }_x(x)=1$ and ${\partial }_y{F}_1={\partial }_y(-y)=-1$. Why this step? "How fast does the upward push $x$ grow as I walk right?" ($=1$) and "how fast does the rightward push $-y$ grow as I walk up?" ($=-1$).
3. $(\nabla \times \vec{F}{)}_z=1-(-1)=2$. So $\nabla \times \vec{F}=(0,0,2)$.

Verify: both effects reinforce (moving right speeds up rise, moving up speeds up the leftward push — same rotation), so magnitude $>1$: consistent with $2$. Positive $z$ ⇒ CCW ⇒ matches the forecast. ✓

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Example 2 — Cell B: negative (clockwise) curl

Forecast: this is Example 1 with every arrow reversed, so it should spin clockwise: expect a negative $z$-component.

1. ${\partial }_x{F}_2={\partial }_x(-x)=-1$. Why this step? The downward push $-x$ gets more negative as we step right.
2. ${\partial }_y{F}_1={\partial }_y(y)=1$. Why this step? The rightward push $y$ grows as we step up.
3. $(\nabla \times \vec{F}{)}_z=-1-1=-2$, so $\nabla \times \vec{F}=(0,0,-2)$.

Verify: sign flipped versus Ex 1, magnitude unchanged — exactly what reversing all arrows should do. Negative $\hat{k}$ = clockwise = axle points into the page (right-hand rule). ✓

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Example 3 — Cell C: zero curl, straight parallel arrows

Forecast: arrows all point along $+x$, and their length changes as you move along $x$, not across it. No side-to-side speed difference ⇒ guess zero.

1. $z$-component: ${\partial }_x{F}_2-{\partial }_y{F}_1={\partial }_x(0)-{\partial }_y(x)=0-0=0$. Why this step? $F_1=x$ has no $y$ in it, so ${\partial }_y(x)=0$: the flow does not vary across the paddlewheel (perpendicular to the flow).
2. $x$- and $y$-components involve $F_3=0$ and ${\partial }_z$ of $z$-free fields — all zero.

Verify: the speed changes along the flow (that is divergence: ${\partial }_x(x)=1\ne 0$), not across it, so curl must vanish. This proves curl and divergence are independent, echoing the parent's Divergence — definition and physical meaning contrast. ✓

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Example 4 — Cell D: curved arrows but ZERO curl

Forecast: the arrows literally circle the origin — surely this spins! Most students say curl $=2$ like Example 1. Guess, then watch it collapse.

1. Write $r^2=x^2+y^2$. Then $F_2=x/r^2$, $F_1=-y/r^2$. Why this step? Naming $r^2$ keeps the quotient rule readable.
2. ${\partial }_x{F}_2={\partial }_x\left(\frac{x}{x^2+y^2}\right)=\frac{(x^2+y^2)-x(2x)}{(x^2+y^2{)}^2}=\frac{y^2-x^2}{(x^2+y^2{)}^2}$. Why this step? Quotient rule: (bottom·d(top) − top·d(bottom))/bottom². Here ${\partial }_x{r}^2=2x$.
3. ${\partial }_y{F}_1={\partial }_y\left(\frac{-y}{x^2+y^2}\right)=\frac{-(x^2+y^2)+y(2y)}{(x^2+y^2{)}^2}=\frac{y^2-x^2}{(x^2+y^2{)}^2}$. Why this step? Same rule; ${\partial }_y(-y)=-1$ on top, ${\partial }_y{r}^2=2y$.
4. $(\nabla \times \vec{F}{)}_z=\frac{y^2-x^2}{(x^2+y^2{)}^2}-\frac{y^2-x^2}{(x^2+y^2{)}^2}=0$.

Verify: the two identical fractions cancel exactly ⇒ curl $=0$ everywhere except the origin, where the field blows up. The whirling look comes from arrows getting slower as $r$ grows in just the right way to cancel the turning. (This is the field behind why circulation can be nonzero yet curl zero — a subtlety for Stokes' Theorem and Gradient and Conservative Fields.) ✓

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Example 5 — Cell E: straight arrows, NONZERO curl (shear)

Forecast: every arrow points horizontally (no curve at all). Instinct screams "zero". But feel the paddlewheel: below the axle the flow is fast one way, above it fast the other way…

1. $(\nabla \times \vec{F}{)}_z={\partial }_x(0)-{\partial }_y(-y)=0-(-1)=1$. Why this step? The horizontal push $-y$ is strongly rightward below the $x$-axis and leftward above it — a shear. ${\partial }_y(-y)=-1$ captures that top/bottom speed difference.

Verify: stand a paddlewheel on the $x$-axis: the bottom water shoves it one way, the top the other — it turns. Nonzero curl with dead-straight arrows, confirming curl reads shear across the flow, not visual curvature. ✓

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Example 6 — Cell F: curl that varies with position

Forecast: these products of $x$ and $y$ mean the swirl changes point to point — the answer will be a formula, not a single number. Where might it be zero?

1. ${\partial }_x{F}_2={\partial }_x(y^{2}x)=y^2$. Why this step? Treat $y$ as a constant while differentiating in $x$; $y^{2}x\to y^2$.
2. ${\partial }_y{F}_1={\partial }_y(x^{2}y)=x^2$. Why this step? Treat $x$ as constant; $x^{2}y\to x^2$.
3. $(\nabla \times \vec{F}{)}_z=y^2-x^2$. So $\nabla \times \vec{F}=(0,0,y^2-x^2)$.
4. This is zero exactly on the lines $y=\pm x$. Why this step? $y^2-x^2=(y-x)(y+x)=0$ when $y=x$ or $y=-x$.

Verify: pick $(1,0)$: curl $=0-1=-1$ (CW there). Pick $(0,1)$: curl $=1-0=+1$ (CCW there). The field spins opposite ways in different regions, separated by the diagonals $y=\pm x$ — exactly where curl $=0$. ✓

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Example 7 — Cell G: a genuinely 3-D curl (tilted axis)

Forecast: all three variables appear once, so probably all three curl components come out equal — a diagonal axis.

1. $x$-comp $={\partial }_y{F}_3-{\partial }_z{F}_2={\partial }_y(y)-{\partial }_z(x)=1-0=1$.
2. $y$-comp $={\partial }_z{F}_1-{\partial }_x{F}_3={\partial }_z(z)-{\partial }_x(y)=1-0=1$.
3. $z$-comp $={\partial }_x{F}_2-{\partial }_y{F}_1={\partial }_x(x)-{\partial }_y(z)=1-0=1$. Why these steps? Each row uses the cyclic pattern $x\to y\to z\to x$: derivative of the next field minus derivative of the previous field.
4. So $\nabla \times \vec{F}=(1,1,1)$; the spin axis is the unit vector $\frac{1}{\sqrt{3}}(1,1,1)$.

Verify: magnitude $|(1,1,1)|=\sqrt{3}$. Each coordinate plane carries unit swirl, so the total rotation axis tilts equally toward all three axes — the body diagonal of a cube. ✓

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Example 8 — Cell H: degenerate constant field

Forecast: nothing changes from point to point, so nothing can twist a paddlewheel. Guess $\vec{0}$.

1. Every partial derivative of a constant is $0$, e.g. ${\partial }_x(-2)=0$, ${\partial }_y(7)=0$, etc. Why this step? A derivative measures change; a constant never changes.
2. Every one of the six terms is $0$, so $\nabla \times \vec{F}=(0,0,0)$.

Verify: a uniform flow (river moving as one rigid sheet) drags a wheel along without turning it. Zero curl matches the physics. Note this field is also a gradient ($\nabla (5x-2y+7z)$), so by the parent's identity it must be irrotational — a cross-check via Gradient and Conservative Fields. ✓

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Example 9 — Cell I: word problem (a spinning turntable)

Forecast: rigid rotation is the purest spin there is, so curl should point straight up ($+\hat{k}$) and be proportional to $\omega$. Guess the constant.

1. Substitute $\omega =3$: $\vec{v}=(-3y,3x,0)$. Why this step? Turn the physics ("spins at 3 rad/s") into a concrete field.
2. $(\nabla \times \vec{v}{)}_z={\partial }_x(3x)-{\partial }_y(-3y)=3-(-3)=6$. Why this step? Same shear calculation as Example 1, now scaled by $\omega$.
3. So $\nabla \times \vec{v}=(0,0,6)=2\omega \hat{k}$. Why this step? $6=2\times 3$: curl equals twice the angular speed.

Verify (units + meaning): $\vec{v}$ has units m/s, its spatial derivative m/s per m $=$ 1/s, matching $\omega$'s rad/s. The famous rule "curl of a rigid rotation $=2\vec{\omega }$" gives $2(3)=6$. ✓

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Example 10 — Cell J: exam twist (prove an identity, no numbers)

Forecast: a gradient is "pure downhill flow" and downhill flow can't swirl, so expect exact zero — the reason will be that mixed partials commute.

1. Write the gradient: $\nabla \varphi =({\partial }_x\varphi ,{\partial }_y\varphi ,{\partial }_z\varphi )$, so here $F_1={\partial }_x\varphi$, $F_2={\partial }_y\varphi$, $F_3={\partial }_z\varphi$. Why this step? Name the components so the curl formula can act on them.
2. $z$-component of the curl: ${\partial }_x{F}_2-{\partial }_y{F}_1={\partial }_x({\partial }_y\varphi )-{\partial }_y({\partial }_x\varphi )$. Why this step? Plug $F_2,F_1$ into the $z$-formula from the top of the page.
3. By Equality of Mixed Partial Derivatives (Clairaut), ${\partial }_x{\partial }_y\varphi ={\partial }_y{\partial }_x\varphi$, so this difference is $0$. Why this step? Clairaut says the order of partial differentiation does not matter for smooth $\varphi$.
4. The $x$- and $y$-components vanish identically by the same swap (${\partial }_y{\partial }_z\varphi ={\partial }_z{\partial }_y\varphi$, etc.). Hence $\nabla \times (\nabla \varphi )=\vec{0}$.

Verify: test on $\varphi =x^{2}y$. Then $\nabla \varphi =(2xy,x^2,0)$ and $z$-comp of curl $={\partial }_x(x^2)-{\partial }_y(2xy)=2x-2x=0$. The general proof holds on this concrete case. ✓

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Recall One-line recap of every cell

Sign of $z$-curl tells CCW ($+$) vs CW ($-$); curl reads shear across the flow, not curved arrows (Ex 4 vs Ex 5); it can be a whole function (Ex 6); rigid rotation gives $2\vec{\omega }$ (Ex 9); a gradient is always irrotational (Ex 10).

Active Recall

Connections

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