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 correct answer is $3.92 \pi m^3$

Volume of a sphere = $V$ = $\dfrac{4}{3} \pi r^3$

⇒ $\dfrac{dV}{dr}$ = $4 \pi r^2$ and $\triangle{r} = 0.02 m$

∴ $\triangle{V} \approx (4 \pi r^2) \triangle{r}$

= $(4 \pi r^2) \left( 0.02 \right)$

= $0.08 \pi r^2m^3$

= $0.08 \pi (7)^2m^3$

= $3.92 \pi m^3$