Answer
percentage error in volume is 3%
Work Step by Step
We are given that the change in radius is 1
the diameter of the sphere is x=100
then radius $r=\frac{x}{2}$
Volume of sphere: $V=\frac{4}{3}\pi r^3= (\frac{4}{3})\pi (\frac{x^3}{8})= (\frac{1}{3})\pi (\frac{x^3}{2})$
on differentiating the above equation:
$dV=\frac{\pi}{6}(3x^2dx))$
so
%change in volume =change in volume/original volume
$\frac{dV}{V}=\frac{\frac{\pi}{6}(3x^2(dx))}{ (\frac{1}{6})\pi x^3} $
$\frac{dV}{V}=\frac{\frac{\pi}{6}(3(100)^2(1))}{ (\frac{1}{6})\pi (100)^3} $
$\frac{dV}{V}=3$
so the percentage change in volume is 3%
