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
Shwetank Mauria
Jul 29, 2017

Percentage incease in surface area is #16.04%#

Explanation:

If radius of sphere is #r#, volume #V# is given by #4/3pir^3# and surface area #S# is #4pir^2#

Hence #Vpropr^3# and #Spropr^2# i.e. #rpropS^(1/2)#

and hence #VpropS^(3/2)# or #SpropV^(2/3)#

as volume rises by #25%#, #V->1.25V#

hence #S->Sxx(1.25)^(2/3)=1.1604S#

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

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