If the volume of a sphere is increased by 25%, what is the percentage increase in the surface area, correct to the nearest whole?

1 Answer
Jul 29, 2017

Percentage incease in surface area is 16.04%16.04%

Explanation:

If radius of sphere is rr, volume VV is given by 4/3pir^343πr3 and surface area SS is 4pir^24πr2

Hence Vpropr^3Vr3 and Spropr^2Sr2 i.e. rpropS^(1/2)rS12

and hence VpropS^(3/2)VS32 or SpropV^(2/3)SV23

as volume rises by 25%25%, V->1.25VV1.25V

hence S->Sxx(1.25)^(2/3)=1.1604SSS×(1.25)23=1.1604S

i.e. percentage incease in surface area is 16.04%16.04%