= Solution
{c}
The <radial unit vector in spherical coordinates> is $\mathbf e_R=\mathbf X/R$, so $\mathbf X\otimes\mathbf X/R^2=\mathbf e_R\otimes\mathbf e_R$. The <identity matrix> is the sum of the three basis <outer products>. Substitution in the given <deformation gradient> gives
$$
\mathbf F=f'(R)\mathbf e_R\otimes\mathbf e_R
+\frac{f(R)}R\left(\mathbf e_\Theta\otimes\mathbf e_\Theta
+\mathbf e_\Phi\otimes\mathbf e_\Phi\right)
=\operatorname{diag}\left(f'(R),\frac{f(R)}R,\frac{f(R)}R\right).
$$
The reference and current radial directions coincide for this <spherically symmetric deformation>. The positive <principal stretches> are therefore $\lambda_r=f'(R)$ and $\lambda_\theta=\lambda_\phi=f(R)/R$. <Incompressibility> imposes
$$
f'(R)\frac{f(R)^2}{R^2}=1,
\qquad \frac{d}{dR}f(R)^3=3R^2.
$$
Integrating and using $f(A)=a$ gives
$$
\boxed{r=f(R)=\left(R^3+a^3-A^3\right)^{1/3},\qquad
b^3=B^3+a^3-A^3.}
$$
In particular $b^3-a^3=B^3-A^3$, expressing conservation of shell <volume>. Writing $\lambda=r/R$ also gives $\lambda_r=\lambda^{-2}$ and $\lambda_\theta=\lambda_\phi=\lambda$.
Back to article page
