show tht relative error in computing volume of sphere due to error in measuring the radius is approx 3 times relative error in the radius?

Let dr be the error in radius
Let dV be the error in volume
Volume = (4/3) pi r^3 , r being the radius
dV = (4/3) pi (3r^2) dr
dV = 4 pi r^2 dr

Let R(r) be the relative error associated with the radius r
Let R(V) be the relative error associated with the volume V

R(r) = | (r- r') / r | = | dr /r|
R(V) = |dV/ V|
|dV/V| = |4 pi r^2 dr / ((4/3) pi r^3 )| = 3 dr /r = 3 |dr /r|

R(V) = 3 R(r)