= Solution
Integrate the radial equilibrium equation across the thickness, retaining the <integral> rather than evaluating its <antiderivative>. The <traction> conditions give $t_r(b)-t_r(a)=P$. Hence $a$ must satisfy
$$
\boxed{P=\mu\int_a^{\sqrt{(A+H)^2+a^2-A^2}}
\frac1r\left[\frac{r^2}{r^2-a^2+A^2}-\frac{r^2-a^2+A^2}{r^2}\right]\,dr.}
$$
All endpoints and coefficients here depend only on $a$ and the specified material and reference geometry. This is the required implicit <pressure>–radius relation for <inflation and extension of an incompressible tube> with unit axial stretch. In particular $a=A$ gives zero integrand and $P=0$. Once an admissible root is selected, the radial <stress> follows by integrating from the outer face, $t_r(r)=-\int_r^b(dt_r/ds)\,ds$, and the remaining <stress> components follow from the <constitutive equation>.
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