Chapter 14 - Partial Derivatives - 14.4 Exercises - Page 947: 35

$16 cm^3$

The volume is given by: $V=\pi r^2 h$ The differential form can be evaluated as follows: $dV=\dfrac{\partial V}{\partial r} dr + \dfrac{\partial V}{\partial h} dh$ $dV=[ 2 \pi r h] dr+[ \pi r^2] dh$ Plug in the given data and we have $dV=[ 96 \pi \times 0.04] +(16 \pi) \times (0.08)$ Hence, we have $dV \approx 16 cm^3$