Curl and Stokes' Theorem

The Curl Operator

For a vector field $\mathbf{F} = (F_x, F_y, F_z)$, the curl is:

$\nabla \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}$

Explicitly:

$\nabla \times \mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)$

The result is a vector field.

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Physical Interpretation

$\nabla \times \mathbf{F}$ measures rotation: the tendency to cause circulation.

• $\nabla \times \mathbf{F} \neq \mathbf{0}$: Field has rotational component
• $\nabla \times \mathbf{F} = \mathbf{0}$: Irrotational

Example: Paddle wheel in a fluid placed at point $\mathbf{r}$:

• Angular velocity proportional to $|\nabla \times \mathbf{v}|$
• Axis of rotation aligned with $\nabla \times \mathbf{v}$

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Example

For $\mathbf{F} = (-y, x, 0)$:

$\nabla \times \mathbf{F} = \left(0 - 0, 0 - 0, 1 - (-1)\right) = (0, 0, 2)$

This field has uniform rotation about $z$-axis.

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Stokes’ Theorem

For an open surface $S$ with boundary curve $C$: $\int_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}$

The flux of curl through a surface equals the circulation around its boundary.

Orientation: Right-hand rule — fingers curl along $C$, thumb points along $\mathbf{n}$.

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Proof Sketch (Rectangle)

Consider rectangle $S$: $0 \leq x \leq a$, $0 \leq y \leq b$ in $z = 0$ plane.

Boundary consists of four edges:

• $C_1$: $x$ from 0 to $a$ at $y = 0$
• $C_2$: $y$ from 0 to $b$ at $x = a$
• $C_3$: $x$ from $a$ to 0 at $y = b$ (reversed)
• $C_4$: $y$ from $b$ to 0 at $x = 0$ (reversed)

Circulation:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_{C_1} F_x\,dx + \int_{C_2} F_y\,dy + \int_{C_3} F_x\,dx + \int_{C_4} F_y\,dy$

$= \int_0^a [F_x(x, 0) - F_x(x, b)]\,dx + \int_0^b [F_y(a, y) - F_y(0, y)]\,dy$

$= \int_0^a \int_0^b \frac{\partial F_y}{\partial x}\,dy\,dx - \int_0^a \int_0^b \frac{\partial F_x}{\partial y}\,dy\,dx$

$= \int_S \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)dS = \int_S (\nabla \times \mathbf{F})_z\,dS$

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Path Independence

Stokes’ theorem implies: for $\nabla \times \mathbf{F} = \mathbf{0}$, $\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ for any $C$.

Consequence: The line integral from $A$ to $B$ is path-independent.

Equivalent to $\mathbf{F}$ being conservative.

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Relations Between Operators

Curl of Gradient = 0

$\nabla \times (\nabla \phi) = \mathbf{0}$

Proof: The $x$-component is $\frac{\partial^2\phi}{\partial y\partial z} - \frac{\partial^2\phi}{\partial z\partial y} = 0$ by equality of mixed partials.

Divergence of Curl = 0

$\nabla \cdot (\nabla \times \mathbf{F}) = 0$

Proof: Each term in the divergence pairs with another that cancels it.

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Summary of Relationships

Operator combination	Result	Reason
$\nabla \times \nabla \phi$	$\mathbf{0}$	Mixed partials equal
$\nabla \cdot (\nabla \times \mathbf{F})$	0	Terms pair and cancel
$\nabla \cdot \nabla \phi$	$\nabla^2\phi$	Laplacian (non-zero in general)

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Irrotational Fields

$\nabla \times \mathbf{F} = \mathbf{0}$ everywhere $\Leftrightarrow$ $\mathbf{F}$ is conservative $\Leftrightarrow$ $\mathbf{F} = \nabla \phi$

Example: Gravitational field $\mathbf{g} = -\nabla \phi_G$ is irrotational.

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Why Stokes’ Theorem Matters

Stokes’ theorem converts between:

• Surface integrals (flux of curl)
• Line integrals (circulation)

This connects:

• Circulation to rotation in fluid mechanics
• Electromagnetic induction (Faraday’s law)
• Path independence of conservative fields