Question

According to the Washington Post, it is estimated that 9857 cubic meters of oil per day spilled into the Gulf of Mexico on August 2, 2010. Assume that an average of 9857 m3 of oil spilled into the Gulf of Mexico every day and that it formed a hemispherical dome of radius r on the ocean floor.
a)Find the rate of change of the radius with respect to the time when the volume of the oil the spill is 50,000 m3
(b) Since only the oil that is in contact with seawater can mix with seawater, it is important to know how much of the surface area of the oil spill is in contact with seawater. Find the rate of change of the hemisphere’s surface area with respect to time when the volume of the oil spill is 50,000 m3. You should assume that only the top of the hemispherical dome of oil comes in contact with seawater (not the flat bottom, which is in contact with the ocean floor).

Answer 1

a) Given volume = 50,000 m3

The volume of hemisphere is given by

V =   

50,000 =

r^3 = 75000/pi

r = (75,000/pi)^1/3

V =    

= =  

r = (75,000/pi)^1/3

9857 =

b) The surface area of hemisphere =

= 684.6m^2/day

Hence the rate of change of the hemisphere’s surface area with respect to time is 684.6m^2/day.