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

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then,

$r=7 \mathrm{~m}$ and $\Delta r=0.02 \mathrm{~m}$

Now, the volume V of the sphere is given by,

$V=\frac{4}{3} \pi r^{3}$

$\begin{aligned} \therefore \frac{d V}{d r} &=4 \pi r^{2} \\ \therefore d V &=\left(\frac{d V}{d r}\right) \Delta r \\ &=\left(4 \pi r^{2}\right) \Delta r \\ &=4 \pi(7)^{2}(0.02) \mathrm{m}^{3}=3.92 \pi \mathrm{m}^{3} \end{aligned}$

Hence, the approximate error in calculating the volume is $3.92 \pi \mathrm{m}^{3}$.