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Volume preservation by the elastic accommodation in morphoelasticity. If $\mathbf{F}=\mathbf{AG}$, elastic incompressibility imposes $\det \mathbf{A}=1$, and hence $\det \mathbf{F}=\det \mathbf{G}$. Total volume can increase through growth while the elastic response remains incompressible. A hydrostatic pressure acts as a Lagrange multiplier enforcing this constraint in the Cauchy stress tensor.

The factor describing local stress-free growth in the morphoelasticity convention $\mathbf{F}=\mathbf{AG}$. Its determinant is the local grown-volume ratio. For $\mathbf{G}=\mathrm{diag}(1,1,\gamma )$, growth changes only the preferred third-direction length, by factor $\gamma$.

In morphoelasticity, the deformation gradient factors as $\mathbf{F}=\mathbf{AG}$, where $\mathbf{G}$ accounts for local stress-free growth and $\mathbf{A}$ accounts for the subsequent elastic accommodation. Elastic incompressibility means that the latter preserves volume:It does not forbid volume increase by growth: here $\det \mathbf{F}=\det \mathbf{G}=\gamma$.

For the homogeneous diagonal deformation, the two unloaded transverse directions are equivalent. Taking the transversely symmetric branch implicit in the paper's hint gives ${\alpha }_1={\alpha }_2>0$. Set ${\alpha }_3=\alpha >0$. The determinant constraint is thensoEquality of the transverse stretches uses that symmetry assumption; elastic incompressibility alone would only give ${\alpha }_1{\alpha }_2{\alpha }_3=1$. The total deformation gradient and the actual axial length arebecause the initial cuboid is a unit cube.