Question: use python to complete the following: a. Open the starter code in the file: soda_can.py # A class to model soda cans # Implement a

use python to complete the following:

a. Open the starter code in the file: soda_can.py

# A class to model soda cans # Implement a class SodaCan with methods getSurfaceArea() and getVolume(). # The height and radius of a soda can object should be specified upon instantiation.  import math class SodaCan(object): def __init__(self, height, radius): self.height = height self.radius = radius self.contents = None def getHeight(self): return self.height def getRadius(self): return self.radius def setHeight(self, height): self.height = height def setRadius(self, radius): self.radius = radius def getVolume(self): V = math.pi*(r**2)*h return 0 # fix this stub   def getSurfaceArea(self): return 0 # fix this stub  if __name__ == "__main__": # b. Construct a soda can with height h = 5 inches and radius r = 2.5 inches.  # Then print out its volume V and surface area A. Use your methods getVolume and getSurfaceArea.  can = SodaCan(5, 2.5) V = can.getVolume() S = can.getSurfaceArea() print("The volume of the can is V = %0.3f" % V) print("The surface area of the can is S = %0.3f" % S) # construct a second soda can with radius 2 inches and height 8 inches.  # Which holds a greater volume?  # Use an if statement to decide.

Complete the methods shown below to return the volume and surface area of the soda can. The following formulas should be familiar.

= ^2h

= 2 ^2 + 2h

Since the math module has been imported at the top, you can use math.pi any time you need to use .

def getVolume(self): return 0 # fix this stub

def getSurfaceArea(self):

return 0 # fix this stub

b. In the main section, construct a cylindrical soda can with radius 2.5 inches and height 5 inches. Then print out its volume V and surface area A. Use your methods getVolume and getSurfaceArea.

c. Now construct a second soda can with radius 2 inches and height 8 inches. Which holds a greater volume? Use an if statement to decide.

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