= Solution
The wind integrals require a steady <axisymmetric magnetohydrodynamic wind> in <ideal magnetohydrodynamics>, with no viscous azimuthal <torque> or imposed toroidal loop voltage. Axisymmetry by itself is insufficient. Assume smooth connected <poloidal flux surfaces> with nonzero <poloidal magnetic field>; functions of the flux label below are understood on each such branch. In <cylindrical coordinates>, the given <poloidal magnetic flux function> obeys
$$
\mathbf B_p=\frac1s\nabla\chi\times\widehat{\boldsymbol\phi},\qquad
\mathbf B_p\cdot\nabla\chi=0,\qquad
\widehat{\boldsymbol\phi}\times\mathbf B_p=\frac1s\nabla\chi.
$$
For a steady ideal wind, the <electric field> is $\mathbf E=-\mathbf u\times\mathbf B$ and $\nabla\times\mathbf E=0$. Its toroidal component satisfies $\partial_zE_\phi=0$ and $\partial_s(sE_\phi)=0$, so $E_\phi=C/s$. Regularity at the axis, or equivalently the absence of an imposed loop voltage in an annular domain, gives $C=0$. Thus $(\mathbf u_p\times\mathbf B_p)_\phi=0$ and $\mathbf u_p=f\mathbf B_p$. The <continuity equation> and $\nabla\cdot\mathbf B_p=0$ now yield
$$
0=\nabla\cdot(\rho\mathbf u_p)=\mathbf B_p\cdot\nabla(\rho f),\qquad
\rho f=\kappa(\chi).
$$
Consequently $\kappa$ is the <magnetohydrodynamic mass loading>. Including the <toroidal magnetic field>, rewrite the full <velocity> as
$$
\mathbf u=\frac{\kappa}{\rho}\mathbf B+s\omega\widehat{\boldsymbol\phi},\qquad
\omega=\Omega-\frac{\kappa B_\phi}{\rho s}.
$$
Since $\mathbf u\times\mathbf B=\omega\nabla\chi$, steady ideal induction gives $0=\nabla\omega\times\nabla\chi$. Hence $\omega=\omega(\chi)$, establishing
$$
\boxed{\mathbf u=\rho^{-1}\kappa(\chi)\mathbf B+s\omega(\chi)\widehat{\boldsymbol\phi}.}
$$
The invariant $\omega$ is the <field-line angular velocity of an axisymmetric wind>, not generally the fluid's <angular velocity> $\Omega$. When an ideal magnetic surface is anchored to a rigidly rotating star, $\omega$ equals the stellar <angular velocity>. With no poloidal outflow this reduces to <Ferraro's law of isorotation>.
To calculate the magnetic <torque>, use the azimuthal momentum equation. Axisymmetric <pressure> and <Newtonian gravity> have no azimuthal components; the magnetic pressure has none either. The azimuthal acceleration and magnetic tension contain the curvature terms characteristic of <cylindrical coordinates>:
$$
\rho\left(\mathbf u_p\cdot\nabla u_\phi+\frac{u_su_\phi}{s}\right)
=\frac1{\mu_0}\left(\mathbf B_p\cdot\nabla B_\phi+\frac{B_sB_\phi}{s}\right).
$$
Multiplication by $s$ gives the <Maxwell torque conservation in an axisymmetric wind> equation
$$
\rho\mathbf u_p\cdot\nabla(su_\phi)=\frac1{\mu_0}\mathbf B_p\cdot\nabla(sB_\phi).
$$
Using $u_\phi=s\Omega$, $\rho\mathbf u_p=\kappa\mathbf B_p$ and $\mathbf B_p\cdot\nabla\kappa=0$ proves
$$
\mathbf B_p\cdot\nabla\left(\kappa s^2\Omega-\frac{sB_\phi}{\mu_0}\right)=0,\qquad
\boxed{\ell(\chi)=\kappa s^2\Omega-\frac{sB_\phi}{\mu_0}.}
$$
This is the total <angular momentum> transport per unit poloidal <magnetic flux>. The first term is the material contribution and the second is the contribution of the <Maxwell stress tensor>. The specific <magnetohydrodynamic angular-momentum invariant> is $L=\ell/\kappa$ when $\kappa\ne0$; its normalization differs from the paper's $\ell$.
Define the poloidal <Alfvén Mach number> by
$$
M_A^2=\frac{|\mathbf u_p|^2}{|\mathbf B_p|^2/(\mu_0\rho)}=\frac{\mu_0\kappa^2}{\rho}.
$$
Substituting $\Omega=\omega+\kappa B_\phi/(\rho s)$ into $\ell$ gives
$$
\ell=\kappa s^2\omega+\frac{sB_\phi}{\mu_0}(M_A^2-1).
$$
At a smooth crossing of the <Alfvén surface>, $M_A^2=1$ and the <toroidal magnetic field> is finite. The <Alfvén-surface regularity condition for an axisymmetric wind> is therefore
$$
\boxed{\ell=\kappa s_A^2\omega,\qquad L=s_A^2\omega.}
$$
One can also expose the apparent singularity explicitly:
$$
B_\phi=\frac{\mu_0\kappa}{s}\frac{L-s^2\omega}{M_A^2-1},\qquad
\Omega=\frac{M_A^2 L/s^2-\omega}{M_A^2-1}.
$$
Finite passage through the <Alfvén surface> requires the numerator to vanish with the denominator. Equality of the numerator zeros is necessary; a globally smooth wind must also satisfy its other dynamical critical conditions. Here $s_A$ is the cylindrical distance from the rotation axis, not automatically the spherical <Alfvén radius>.
For a tube carrying outward mass flux $d\dot M=\kappa\mathbf B_p\cdot d\mathbf S$, the outward <angular momentum> flux is
$$
d\dot J_{\rm out}=\left(\rho\mathbf u_p s u_\phi-\frac{\mathbf B_p sB_\phi}{\mu_0}\right)\cdot d\mathbf S
=L\,d\dot M=\omega s_A^2\,d\dot M.
$$
Thus the stellar <torque> is $\dot J_\star=-\int\omega s_A^2\,d\dot M$, with the integral over the escaping wind. \b[Magnetic stresses give each mass element an effective angular-momentum lever arm equal to the cylindrical Alfvén crossing radius.] In a strongly sub-Alfvénic region the fluid approximately corotates with the magnetic surface; beyond that region it can carry the accumulated <angular momentum> away. For an outward field and positive $\omega$, a super-Alfvénic expanding tube with $s>s_A$ has $B_\phi<0$, so the magnetic stress transports positive <angular momentum> outwards. If $s_A$ greatly exceeds the stellar surface lever arm, a modest mass-loss rate can cause substantial <wind-driven magnetic braking of a solar-type star>. An unmagnetized outflow would instead carry approximately the material <angular momentum> fixed at its launch radius.
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