Air is escaping from a spherical balloon at the rate of 4cc/min. how fast is the surface area of the balloon shrinking when the radius is 24 cm? (a sphere of radius r has volume 4 3 πr3 and surface area 4πr2 .)

The volume of the ball is given by:
 V = (4/3) πr ^ 3
 Deriving we have:
 V '= (3) (4/3) (π) (r ^ 2) (r')
 Clearing r 'we have:
 r '= V' / ((3) (4/3) (π) (r ^ 2))
 Substituting values:
 r '= 4 / ((3) (4/3) (π) ((24) ^ 2))
 r '= 0.000552621 c / min
 Then, the surface area is:
 A = 4πr2
 Deriving we have:
 A '= 8πrr'
 Substituting values:
 A '= 8π (24) (0.000552621)
 A '= 0.33 cm ^ 2 / min
 
 The surface area of the balloon is shrinking at:
 A '= 0.33 cm ^ 2 / min