Amelting spherical block of ice has its surface area changing at the rate of 24 inch2 /sec at the time the radius is 3 inches. how fast is the volume of the ice changing at that moment

The surface area of the sphere is:
 A = 4 * pi * r ^ 2
 Deriving we have:
 A '= 8 * pi * r * r'
 We clear the speed:
 r '= A' / (8 * pi * r)
 Substituting values:
 r '= 24 / (8 * pi * 3)
 r '= 1 / pi inch / sec
 Then, the volume of the sphere is:
 V = (4/3) * (pi) * (r ^ 3)
 Deriving we have:
 V '= (3) * (4/3) * (pi) * (r ^ 2) * (r')
 Substituting values:
 V '= (3) * (4/3) * (pi) * (3 ^ 2) * (1 / pi)
 Rewriting:
 V '= (4) * (9)
 V '= 36 inch3 / sec
 
 the volume of the ice is changing at 36 inch3 / sec at that moment