Answer
The percentage change in volume: $2.5\% $
Work Step by Step
We have the volume of a right-circular cone of radius $r$ and height $h$ given by
$V = \frac{\pi }{3}{r^2}h$
Estimate the increase of $V$ if $h$ is increased from $40$ to $41$ cm, that is $\Delta h = 1$:
$\Delta V \approx \frac{{\partial V}}{{\partial h}}\Delta h \approx \frac{\pi }{3}{r^2}\Delta h$
For $r = h = 40$, we get
$\Delta V \simeq \frac{\pi }{3}\cdot{40^2}\cdot1 = \frac{{1600}}{3}\pi $
The percentage change in volume at $r = h = 40$ is
$\frac{{\Delta V}}{V} = \frac{{1600}}{3}\pi \cdot\frac{3}{{\pi \cdot{{40}^2}\cdot40}}\cdot100\% = 2.5\% $
