Change in area $=\frac{\Delta\left(4 \pi r ^2\right)}{\left(4 \pi r ^2\right)})=\frac{\Delta r ^2}{ r ^2}=\frac{2 \Delta r }{ r }=2 X$

Change in Volume $=\frac{\Delta\left(4 / 3 \pi r^3\right)}{\left(4 / 3 \pi r^3\right)}=\frac{\Delta r^3}{r^3}=\frac{3 \Delta r}{r}=3 X$

Thus change in volume is maximum.