= Solution
Let $b_0$ be the outer radius when $a=a_0$. The area relation from part (i) gives
$$
b^2=b_0^2+a^2-a_0^2
=b_0^2+2a_0^2\epsilon+O(\epsilon^2),
$$
so
$$
\boxed{b=b_0\left(1+\frac{a_0^2}{b_0^2}\epsilon\right)+O(\epsilon^2)}.
$$
Moreover,
$$
A=a\dot a=a_0^2\dot\epsilon+O(\epsilon^2),
\qquad
\dot A=a_0^2\ddot\epsilon+O(\epsilon^2).
$$
The term $A^2$ in the pressure is second order, while the logarithm multiplying $\dot A$ may be evaluated at equilibrium. The <linearization> is therefore
$$
\boxed{p(a,t)-p_\infty
=\rho a_0^2\log\frac{b_0}{a_0}\,\ddot\epsilon+O(\epsilon^2)}.
$$
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