#Example 10.1
#page no. 252
# Given that
#three metal piece being cast have the same volume but different shapes
#shapes are sphere,cube,cylinder(height=diameter)
print("\n #solidification time for various shapes# \n")
#solidification time is inversely proportional to the square of surface area
#for sphere
r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3
A=4*3.14*((r)**2)
time1=1./(A)**2.
print'%s %.6f %s' %("\n the solidification time for the sphere is ",time1,"C")
#for cube
a=1#edge of the cube
A=6*a**2
time2=1./(A)**2
print'%s %.6f %s' %("\n the solidification time for the cube is ",time2,"C")
#for cylinder
#given height =diameter
#radius=2*height
r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h
A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)
time3=1./(A)**2.
print'%s %.6f %s' %("\n the solidification time for the sphere is ",time3,"C")