= Solution
{c}
First let $A\downarrow0$ at fixed positive cavity radius $a$ and fixed reference outer radius $B$. Then
$$
\lambda_a=\frac aA\longrightarrow\infty,\qquad
\lambda_b\longrightarrow\left(1+\frac{a^3}{B^3}\right)^{1/3},
$$
and the <pressure> sustaining a finite cavity is
$$
P(a)=-\int_{(1+a^3/B^3)^{1/3}}^\infty
\frac{h'(\lambda)}{\lambda^3-1}\,d\lambda.
$$
\b[The onset of cavitation is the additional limit $a\downarrow0$.] The outer stretch then tends to one, and the signed critical <pressure> is
$$
\boxed{P_{\mathrm{crit}}=-\int_1^\infty
\frac{h'(\lambda)}{\lambda^3-1}\,d\lambda.}
$$
When this <improper integral> is positive and finite before the minus sign, the tensile loading magnitude is $-P_{\mathrm{crit}}>0$. Thus the cavity opens under tension, even though the load variable is named <pressure>. The critical value is independent of the original sphere radius $B$.
For $A=0$ and $a>0$, the <deformation map> is smooth away from $R=0$ but singular at the original centre: $r=(R^3+a^3)^{1/3}$ approaches $a$, rather than zero, as $R\downarrow0$. This is the <elastic cavitation> mechanism; it cannot be represented by a smooth <deformation map> through the centre.
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