= Solution
{c}
Take $s=\sigma$ up to an irrelevant additive constant, $r=a$ constant, $z=z_0-\sigma$, and $\theta=\pi/2$. These satisfy the geometric relations and give the <principal stretches>
$$
\lambda_s=1,\qquad\lambda_\varphi=\frac a{a_0}.
$$
The <constitutive equations> therefore produce constant <membrane tensions>
$$
t_s=f_s(1,a/a_0),\qquad t_\varphi=f_\varphi(1,a/a_0).
$$
The tangential <mechanical equilibrium> equations hold automatically, while the normal equation reduces to
$$
\boxed{Pa=f_\varphi\left(1,\frac a{a_0}\right).}
$$
Every positive root of this algebraic equation provides a cylindrical solution with no axial <stretch ratio> change. There is no further constraint on $t_s$ in the interior equations when end <boundary conditions> are discarded. In particular, a closed-end pressure balance would impose additional axial loading data that are absent here. Existence of a positive root for an arbitrary specified $P$ depends on the given <constitutive equation>.
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