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Apply equal compressive pressure $p$ in all directions; here $\sigma$ denotes Poisson's ratio. The three principal normal stresses are $-p$. For an isotropic linearly elastic material, Hooke's law givesand similarly for $y,z$. At small strain, the fractional volume change is the sum of these normal strains:The bulk modulus is $B=-p/(\Delta V/V)$, giving the isotropic elastic modulus relation,The result assumes linear elasticity and fixed isotropic moduli. In the incompressible limit $\sigma \to 1/2$ with finite Young's modulus, the bulk modulus tends to infinity.

For linear elasticity of an isotropic material, the isotropic elastic modulus relation givesThe printed Young's modulus is $E=10^{-3}\mathrm{GPa}=10^6\mathrm{Pa}$, and the printed Poisson's ratio is $\nu =1.47$. Formal substitution givesWith compressive pressure positive, the linear volume response is $\Delta V/V=-p/B$. Thus a thirty per cent reduction formally requiresHere $1\mathrm{atm}=101325\mathrm{Pa}$. The negative pressure is a formal tensile load, not the requested positive isotropic compression.

The original PDF really prints $1.47$. It is incompatible with a stable ordinary isotropic elastic solid: positive bulk modulus and shear modulus require $-1<\nu <1/2$. Consequently there is no physically admissible positive-pressure answer with the stated constants. If $0.47$ was intended, the corresponding linear estimates would instead beThat is an explicit alternative, not a correction established by the source. A thirty per cent volume strain is also outside a controlled small-strain approximation; accurate rubber behaviour at that deformation needs a finite-strain constitutive law. The initial volume cancels because the specified change is a fractional one.

Under a uniform compressive pressure $p$, each normal stress is $-p$. The isotropic Hooke's law gives each longitudinal strain asThus $\Delta V/V=-3p(1-2\sigma )/E$. By the definition of bulk modulus, $B=-p/(\Delta V/V)$, so the isotropic elastic modulus relation isPositive finite Young's modulus and bulk modulus requireThe incompressible limit is $\sigma \to 1/2$ with $B\to \infty$ at finite $E$. If positivity of the shear modulus $G_s=E/[2(1+\sigma )]$ is also imposed, the full stable isotropic range is $-1<\sigma <1/2$. Negative Poisson's ratio is not ruled out; the displayed bulk-modulus relation alone supplies the upper bound.