If the radius of a sphere is measured as 10cm with an error of 0.01 cm, then find the approximate error in calculating its volume.

$\text{ measuring the radius, then }\mathrm{r}=10\mathrm{cm}\text{ and }\mathrm{\Delta }r=0.01\mathrm{cm}\text{. }$

    Now the volume V of the sphere is given by

$\begin{array}{l}V=\frac{4}{3}\pi r^3\\ \frac{dV}{dr}=4\pi r^2\\ dV=\left(\frac{dV}{dr}\right)\mathrm{\Delta }r\\ =\left(4\pi r^2\right)\mathrm{\Delta }r\\ =4\pi \left(10{)}^2\right(0.01){\mathrm{cm}}^3\\ =4\pi {\mathrm{cm}}^3\end{array}$

Thus, the approximate error in calculating the volume is $4\pi c{m}^3$