= Solution
For fixed volume define $E_B=\int_VB^2/(2\mu_0)\,dV$ and $J=\nabla\times B/\mu_0$. Differentiate and use the <ideal magnetohydrodynamic induction equation>. The vector identity
$$
B\cdot\nabla\times(u\times B)
=\nabla\cdot[(u\times B)\times B]+(u\times B)\cdot\nabla\times B
$$
then gives the general balance
$$
\frac{dE_B}{dt}=\frac1{\mu_0}\oint_S[(u\cdot B)B-B^2u]\cdot dS
-\int_Vu\cdot(J\times B)\,dV.
$$
The 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 $u\cdot B=R\Omega B_\phi=\Omega f(\psi)$. 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 $+2\pi\Omega_{\rm out}f\,d\psi$ and its entry endpoint contributes $-2\pi\Omega_{\rm in}f\,d\psi$. Pair the endpoints, counting each once, to obtain
$$
\boxed{\frac{dE_B}{dt}=\frac{2\pi}{\mu_0}\int f(\psi)\Delta\Omega(\psi)\,d\psi,
\quad\Delta\Omega=\Omega_{\rm out}-\Omega_{\rm in}.}
$$
The 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.
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