Chapter 14 - Partial Derivatives - 14.4 Exercises - Page 947: 34

$\dfrac{14 \pi}{5} cm^3$

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, we have $dV=[ 2 \pi \times 2 \times 10] (0.05)+4 (0.2)$ Hence, we have $dV=\dfrac{14 \pi}{5} cm^3$