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

Let x be the radius of the sphere and y be its volume.

$y=\frac{4}{3} \pi x^{3}$

Let $\Delta x$ be the error in the radius.

$x=7$

$\Delta x=0.02$

$\frac{d y}{d x}=4 \pi x^{2}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=7}=196 \pi$

$\therefore \Delta y=d y=\frac{d y}{d x} d x=196 \pi \times 0.02=3.92 \pi$

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