 Solution

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.