Question

Given a sphere of diameter, d. What is the percentage increase in its volume when surface area increases by 21%?

Answer 1

LEt me start over

Answer 2

Let d = diameter, r = radius, V for volume, S for surface area
Since d = 2*r , so r = d/2
V = 4/3 Pi * r^3 = 4/3 * Pi * (d/2)^3 = Pi/ 6 * D^3
S = 4 * Pi * r^2 = 4 * pi ( d/2)^2 = pi * d^2
ok so far?

Answer 3

the initial surface area and initial diameter we will call S1, D1. After the increase the new surface area and diameter are S2 , D2.
S1 = Pi * D1^2
solve for D1
D1 = sqrt( S1 / pi )
S2 = Pi * D2^2
but we are given that S2 = 1.21*S1
therefore by substitution
S2= 1.21 * Pi * D1^2
and
1.21 * Pi * D1^2=Pi * D2^2
solve for D2
D2 = sqrt( 1.21 * D1^2)

Now V2 = pi/6 * D2^3.
V2 = pi/6 ( sqrt(1.21*D1^2))^3
V2= pi/6 * (1.21 * D1^2)^(3/2)
V2 = pi/6 * 1.21^(3/2) * D1 ^3 = (pi/6 * D1^3) * 1.21^3/2 = V1 * 1.21 ^(3/2)