Calculus about volume of sphere rate of change can someone help me?

Question

A ball made of ice is melting at a rate of 2ft^3 per hour. if it remains spherical, at what rate is the radius changing when the radius of the ball is 20 inches?

Answer 1

The ball made of ice is melting at at rate      2 ft^3

                                                  So             dV/dt =  -2

Volume of the sphere is      V = 4/3× ∏ × r^3

                                        dV/dt = 4/3 × ∏ ×3 × r^2 × dr/dt

                               Since    r = 20 inches

                                              = 20/12 = 5/3 ft

                   Now              -2 = 4/3 × ∏ ×3 × (5/3)^2 × dr/dt

                                         -2 = 4∏ × 25 /9 × dr/dt

                                          (-2/4∏)× 9/25 = dr/dt

                                           dr/dt = - 9/50×∏

So, the change in radius is     -9/50×∏    feet per hour