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For fixed volume define $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ and $J=\nabla \times B/{\mu }_0$. Differentiate and use the ideal magnetohydrodynamic induction equation. The vector identitythen gives the general balanceThe volume term is the work done by the magnetic force on the fluid, with the opposite sign for magnetic energy. Under the force-free assumption of part (b), it vanishes, leaving the requested boundary expression. Without that assumption, or another reason for zero net Lorentz work, a boundary-only balance is not generally valid.

For an axisymmetric boundary its normal has no azimuthal component. The differential rotation is tangential, so $u\cdot dS=0$, and $u\cdot B=R\Omega B_{\varphi }=\Omega f(\psi )$. Therefore ${\dot{E}}_B={\mu }_0^{-1}{\oint }_S\Omega f(\psi )B\cdot dS$. A thin axisymmetric tube between neighboring flux labels carries flux $2\pi d\psi$. Its exit endpoint contributes $+2\pi {\Omega }_{\mathrm{out}}fd\psi$ and its entry endpoint contributes $-2\pi {\Omega }_{\mathrm{in}}fd\psi$. Pair the endpoints, counting each once, to obtainThe endpoint convention fixes the sign. Include each connected tube segment through $V$ once; a field line with no boundary crossing contributes nothing. This is magnetic-energy injection by differential boundary rotation. A common angular velocity at both ends produces no such injection.