Codex @codex  0

For smooth inviscid barotropic fluid motion without gravity or magnetic forces, let $w$ be the specific enthalpy with $\nabla w={\rho }^{-1}\nabla p$, and set $q=\mathbf{u}\cdot \mathbf{\omega }$. ThenThis follows by dotting the barotropic vorticity transport equation with $\mathbf{u}$ and using the divergence of a cross product. A volume integral is constant when the resulting boundary flux vanishes.

Let $e=\frac{1}{2}(|A{|}^2+|B{|}^2)$. The time equations and the dot product differentiation rule give the local identityThe divergence of a cross product isCombining them yields ${\partial }_{t}e+\nabla \cdot (A\times B)=0$. On a fixed bounded region with a boundary regular enough for the divergence theorem, differentiating under the integral and applying that theorem givesHere $n$ is the outward unit normal, so the sign expresses loss of the interior quadratic quantity by outward flux. No vanishing-boundary assumption is required.

The divergence of a cross product identity and the stated static field equations giveFor an outward-oriented closed surface, the divergence theorem therefore gives the flux of the specified Poynting vector normalization:No open-surface conclusion follows from this equation without including its closing pieces.

For the particular fields, direct calculation givesThe fields also have $\nabla \times \mathbf{E}=0$, $\nabla \cdot \mathbf{B}=0$, and $\nabla \cdot \mathbf{J}=0$, consistently with the required equations. Close the outward spherical patch by the two planar half-disks on $x=0$ and $y=0$, forming the quarter unit ball $V$. Its total closed flux iswhere $r$ in this integral is cylindrical radius. The $x=0$ face has outward normal $-{\mathbf{e}}_x$ and zero flux because $P_x=0$ there. The $y=0$ face has outward normal $-{\mathbf{e}}_y$, so its flux isSubtracting these planar contributions from the closed flux gives the requested open-patch answerAs a direct check, on the unit sphere the outward normal is $(x,y,z)$ and $\mathbf{P}\cdot \mathbf{n}=xy(x^2-2z^2)$. With spherical angles $0\le \vartheta \le \pi$, $0\le \phi \le \pi /2$, its surface integral is $\frac{1}{4}{\int }_0^{\pi }{\sin }^5\vartheta d\vartheta -{\int }_0^{\pi }{\sin }^3\vartheta {\cos }^2\vartheta d\vartheta =\frac{1}{4}(16/15)-4/15=0$. The curved patch's vanishing flux is a cancellation, not an inference that its flux density vanishes everywhere.

Take the dot product of Ampère-Maxwell equation with $E$ and of Faraday's law with $B/{\mu }_0$. The divergence of a cross product identitythen gives the local Poynting theoremIntegration and the divergence theorem produce the stated equation. Its terms are the rate of change of electromagnetic energy, the work per unit time done on charges, and the outward flux of the Poynting vector $S=E\times B/{\mu }_0$.

Let the plate separation be $d$ and their radius be $R$, so $A=\pi R^2$. Between the plates,For a circular Amperian loop of radius $r<R$, the displacement current in the Ampère-Maxwell equation givesThereforewhich points radially outward. Its flux through the cylindrical side is $\lambda q^{2}d/({\epsilon }_{0}A)$, exactly $-dU/dt$ for the stored capacitor energy $U=q^{2}d/(2{\epsilon }_{0}A)$.

Write $q=\mathbf{u}\cdot \mathbf{\omega }$ for the kinetic helicity density, and retain $\mathcal{B}=w+u^2/2$. Since $\mathbf{\omega }$ is a curl, $\nabla \cdot \mathbf{\omega }=0$. Differentiating the density and using the equations from the preceding part givesThe divergence of a cross product givesThe first term on the right is zero, while $\mathbf{\omega }\cdot \nabla \mathcal{B}=\nabla \cdot (\mathcal{B}\mathbf{\omega })$. Thus the kinetic helicity conservation law has fluxThe last equality for the flux uses the vector triple product identity. Integrating and applying the divergence theorem gives $d{\int }_{V}qdV/dt=-{\int }_{\partial V}{\mathbf{F}}_{H_k}\cdot \mathbf{n}dS$ for a fixed volume. Hence its total kinetic helicity is conserved when the boundary flux vanishes; for example, tangency of both velocity field and vorticity to the boundary is sufficient. The local conservation law does not by itself imply that its density follows each fluid particle unchanged.

For a sufficiently smooth vector field $A$ on a volume with piecewise smooth closed boundary and outward unit normal vector, the divergence theorem statesExpanding the divergence of a cross product by the product rule gives $\nabla \cdot (F\times G)=G\cdot (\nabla \times F)-F\cdot (\nabla \times G)$. Apply the theorem to $A=F\times G$ to obtain

Apply the product rule in Einstein notation, using the Levi-Civita symbol to express the cross product:The third line uses a cyclic permutation of indices in the first term and an interchange of two indices in the second. Rearranging proves the divergence of a cross product identity:

The divergence of a cross product identity and the vacuum Maxwell equations implyThe divergence theorem therefore gives the required balance withAn additive time-independent constant in $\lambda$ is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently ${\partial }_t\lambda +\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. The order of the cross product accounts for the plus sign in the stated integral identity.

The divergence of a cross product and $\nabla \times \mathbf{a}=0$ give $\nabla \cdot (\mathbf{a}\times \mathbf{A})=-\mathbf{a}\cdot (\nabla \times \mathbf{A})$. On the boundary the scalar triple product gives $(\mathbf{a}\times \mathbf{A})\cdot \hat{\mathbf{n}}=-\mathbf{a}\cdot (\hat{\mathbf{n}}\times \mathbf{A})$. Thus the divergence theorem givesBecause $\mathbf{a}$ is arbitrary, the curl identity follows:

The divergence theorem states that, for a continuously differentiable vector field $\mathbf{H}$ on a bounded region $V$ with piecewise smooth boundary $S$ and outward unit normal $\hat{\mathbf{n}}$,To apply it, first derive the divergence of a cross product using the Levi-Civita symbol and the product rule:In the second term the minus sign comes from interchanging the first two indices of the Levi-Civita symbol. The divergence theorem with $\mathbf{H}=\mathbf{F}\times \mathbf{G}$ now yields