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The electric-dipole approximation for radiation requires the source size to be much smaller than the radiation wavelength:The displayed $1/r$ fields also describe the radiation zone, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.

Write $t_r=t-r/c$. The given vector potential and the time derivative of the electric dipole moment giveWhen taking the curl, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\hat{\mathbf{x}}/c$. Hence

The assumed electric field $\mathbf{E}=-c\hat{\mathbf{x}}\times \mathbf{B}$ and the transversality $\hat{\mathbf{x}}\cdot \mathbf{B}=0$ make the Poynting vectorOn a sphere of radius $R$, the power through surface element $R^{2}d\Omega$ is thereforeChoose the polar axis along $\ddot{\mathbf{p}}$ and use the angular distribution of electric dipole radiation, for which $|\hat{\mathbf{x}}\times \ddot{\mathbf{p}}{|}^2=|\ddot{\mathbf{p}}{|}^2{\sin }^2\theta$. Since $\int {\sin }^2\theta d\Omega =8\pi /3$,

For the charged particle in simple harmonic motion,ThusHere the angular frequency obeys $\omega =2\pi /T$, and the average uses $⟨{\sin }^2⟩=1/2$. The dipole condition is

Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy isso the radiation damping of a harmonically oscillating charge givesThis is exponential decay with rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$. Its half-life isFinally, the largest particle speed is $v_{\max }=A\omega$. Therefore the dipole condition $A\omega /c\ll 1$ itself implies $v_{\max }\ll c$, precisely the nonrelativistic limit required by the mechanical energy formula.

Writing $a_k{x}_k=\mathbf{a}\cdot \mathbf{x}$ and differentiating givesIts symmetric second-rank tensor part and antisymmetric second-rank tensor part areTaking the trace makes the two symmetric terms cancel, so the divergence isContracting the antisymmetric part with the Levi-Civita symbol gives the curl

The surface is an annular band of the downward-opening paraboloid, bounded by the circles $r=1,z=4$ and $r=2,z=1$. Choose the upward orientation.

The curl of $\mathbf{A}=(3y,-xz,y{z}^2)$ isFor the graph $z=5-x^2-y^2$, the upward vector surface integral element isOver the annulus $1<r<2$, the odd term $2x{z}^2$ integrates to zero. Therefore

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$, the vector field has cylindrical componentsThe divergence in cylindrical coordinates is thereforeDirect use of the curl in cylindrical coordinates givesandThus, definingwe obtain

On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

Each field has only an azimuthal component and is independent of $r$ and $\varphi$. The curl in spherical coordinates therefore reduces toand its $\varphi$ component vanishes. The half-angle identitiesshow in both cases that

Apply part (a) with $g(r)=r^{-3}$. Thenaway from the origin, explicitly confirming that each resulting curl has zero divergence.

Using the polar-coordinate curl with $u_{\theta }=r\Omega (r)$ gives

Take the curl of the integral from part (a). Since $J(r^{\prime })$ is independent of $r$,This is the Biot-Savart law.

Linearizing the cosmological Euler equation givesTaking the curl removes the gradient force, so the vorticity $\mathbf{w}=\nabla \times \mathbf{v}$ obeysSince $\mathcal{H}=a^{\prime }/a$, its solution isThus expansion dilutes linear vorticity; in an Einstein-de Sitter universe, $a\propto {\tau }^2$ and $\mathbf{w}\propto {\tau }^{-2}$.

The steady incompressible Stokes flow equations areTaking the curl eliminates the pressure gradient and commutes with the Laplacian, so for the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$,This is the vorticity part of Harmonic pressure and vorticity in Stokes flow.

For a two-dimensional incompressible flow, choose the stream function conventionThenApplying the harmonic-vorticity equation givesthe Biharmonic stream function for planar Stokes flow equation.

In the Cartesian basis, writing $F=(F_1,F_2,F_3)$,defines the divergence, while the curl isEquivalently,

For a smooth divergence-free field on ${\mathbb{R}}^3$, the differential form $F={\iota }_{\mathbf{B}}(d{x}^1\wedge d{x}^2\wedge d{x}^3)$ is closed. The radial homotopy formula gives a global magnetic vector potential $\mathbf{A}(\mathbf{x})={\int }_0^{1}s\mathbf{B}(s\mathbf{x})\times \mathbf{x}ds$ with curl $\mathbf{B}$. For constant $\mathbf{B}$, this reduces to $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

When the only nonpotential force points radially, its curl has no radial component. The radial absolute vorticity therefore obeys a local flux conservation law, although vortex stretching can change it along a fluid trajectory.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.

The vertical curl of surface wind stress is $({\nabla }_h\times \mathbf{\tau }{)}_z={\tau }_{y,x}-{\tau }_{x,y}$. In a layer of density ${\rho }_0$ and depth $h$, its normalized potential vorticity source is $W=({\nabla }_h\times \mathbf{\tau }{)}_z/({\rho }_{0}h)$. It drives Sverdrup balance in the weakly nonlinear steady ocean interior.