If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.

Explanation

Given percentage error is `0.1%r`
`0.1%r=0.01r`
we know that,
`V=4/3 pir^3`
`(dV)/(dr)=4/3xx3xxpir^2`
`Delta V = 4xxpixxr^2 dr`
`=4xxpixxr^2xx0.01r`
`=0.004pir^3`
percentage error in `V=(0.004pir^3)/(4/3 pir^3)xx100`
`=3%`