The diameter of a sphere is measured to be $40 cm$. If an error of $0.02 cm$ is made in it, then find approximate errors in volume and surface area of the sphere.

diameter $d=40, r=\frac{d}{2}, \delta d=0.02$

Let $V$ be the volume of sphere

$\therefore V=\frac{4}{3}\pi r^3=\frac{4\pi }{3}{\left(\frac{d}{2}\right)}^3=\frac{4\pi }{3}\frac{d^3}{8}=\frac{\pi d^3}{6}$

Error in volume $\mathrm{\Delta }v=\frac{\pi }{6}\left(3d^2\right)\mathrm{\Delta }d$

$\Rightarrow \mathrm{\Delta }v=\frac{\pi }{2}\left(40{)}^2\right(0.02)$

$=\pi \left(1600\right)\left(0.01\right)=1.6\pi$

Surface area $S=4\pi r^2\Rightarrow S=4\pi {\left[\frac{d}{2}\right]}^2$

$=4\pi \frac{d^2}{4}\Rightarrow S=\pi d^2$

Error in surface area 

$\Rightarrow \mathrm{\Delta }S=\pi \left(2d\right)\left(\mathrm{\Delta }d\right)=2\pi \left(40\right)\left(0.02\right)=1.6\pi$