## College Physics (4th Edition)

The armadillo must swallow a volume of $1.17\times 10^{-3}~m^3$ of air.
We can find the original volume $V$ of the armadillo: $V = \frac{m}{\rho} = \frac{7.0~kg}{1200~kg/m^3} = 5.83\times 10^{-3}~m^3$ We can assume that the additional volume of air has a negligible mass compared to the armadillo's original mass. We can find the volume $V'$ which will give the armadillo a density of $1000~kg/m^3$: $V' = \frac{m}{\rho} = \frac{7.0~kg}{1000~kg/m^3} = 7.0\times 10^{-3}~m^3$ The required volume of the air is equal to the difference of $V'$ and $V$: $V'-V = (7.0\times 10^{-3}~m^3)-(5.83\times 10^{-3}~m^3)$ $V'-V = 1.17\times 10^{-3}~m^3$ The armadillo must swallow a volume of $1.17\times 10^{-3}~m^3$ of air.