If the radius of a sphere is measured as $7$m with an error of $0.02$m, then find the approximate error in calculating its volume.

The volume of the sphere of radius r is given by :

$V = \cfrac{4}{3}\pi {r^3} \Rightarrow \cfrac{{dV}}{{dr}} = \left( {\cfrac{4}{3}\pi } \right)\left( {3{r^2}} \right) = 4\pi {r^2}$ [ where V is the volume of the sphere]

Let $\Delta r$ be error in measuring radius
$\Rightarrow r = 7m$ and $\Delta r = 0.02\,m$.

Hence, $\Delta V \approx \left( {4\pi {r^2}} \right)\Delta r = \left( {4\pi \left( {{7^2}} \right)} \right)\left( { \pm 0.02} \right){m^3} = \pm \,3.92\pi \,{m^3}$

Therefore, error in calculating the volume $= \pm 3.92\pi \,{m^3}$