Question 1181743
The volume of a sphere = V = 4/3pi*R^3
V1 = 4/3pi*R1^3
V2 = 4/3pi*R2^3
If the volume is doubled, V2 = 2V1 -> 4/3pi*R2^3 = 8/3pi*R1^3
Thus, R2 = (2)^(1/3)*R1
Since the surface area of a sphere = S = 4pi*R^2, S2/S1 = R2^2/R1^2 = 
((2)^(1/3)*R1)^2/R1^2 = (4)^(1/3) = 1.587
Thus the surface area must increase by 58.7%