If the radius of a sphere is measured as 4 m with an error of 0.01m, then find the approximate error in calculating its volume.

Concept:

Let the radius of a sphere is r meters,

Volume of the sphere = V = \(\frac{4}{3}{\rm{\pi }}{{\rm{r}}^3}\)

Approximate error in the volume of a sphere is given by:  \({\rm{\Delta V\;}} = {\rm{\;}}\left( {\frac{{{\rm{dV}}}}{{{\rm{dr}}}}} \right){\rm{\Delta r}}\)

Calculation:

Given: Radius of a sphere = 4 m and Δr = 0.01m

Now, the volume of the sphere is given by,

V = \(\frac{4}{3}{\rm{\pi }}{{\rm{r}}^3}\)

Differentiating with respect to r, we get

\( \Rightarrow \frac{{{\rm{dV}}}}{{{\rm{dr}}}} = {\rm{\;}}\frac{{4{\rm{\pi }}}}{3} \times 3{{\rm{r}}^2} = 4{\rm{\pi }}{{\rm{r}}^2}\)

Now,

The approximate error in the volume of sphere = \({\rm{\Delta V\;}} = {\rm{\;}}\left( {\frac{{{\rm{dV}}}}{{{\rm{dr}}}}} \right){\rm{\Delta r}}\)

\(\therefore {\rm{\Delta V\;}} = {\rm{\;}}4{\rm{\pi }}{{\rm{r}}^2} \times {\rm{\Delta r}}\)

⇒ ΔV = 4π × 4² × 0.01 = 0.64π m³

Hence, the approximate error in the volume of a sphere is 0.64π m³