If the radius of a sphere is measured as $9{\rm{ m}}$ with an error of $0.03{\rm{ m}}$, then find the approximate error in calculating its surface area.

The surface area of sphere of radius r is given by :

$S = 4\pi {r^2} \Rightarrow \cfrac{{dS}}{{dr}} = 8\pi r$ [ where S is the surface area of the sphere]

Let $\Delta r$ be the error in measuring radius $\Rightarrow r = 9,{\rm{ }}\Delta r = 0.03$

Hence, $\Delta S \approx \left( {8\pi r} \right)\Delta r = \left\{ {8\pi \left( {9{\rm{ m}}} \right)} \right\}\left( { \pm 0.03{\rm{ m}}} \right) = \pm \left( {2.16\pi } \right){\rm{ }}{{\rm{m}}^2}$
error in calculating the surface area$= \pm {\rm{ }}2.16\pi {\rm{ }}{{\rm{m}}^2}$.