A flat-bottomed circular bowl of parabolic cross-section has an interior base diameter of 16cm, an interior height of 10cm and its interior diameter across the top is 24cm. What is its capacity (volume) in litres?
This bowl is a section of a solid parabola.
Consider the volume of revolution of a parabola, y = a.x^2, about its vertical axis (y-axis), taken between the limits y1 and y2.
We are given that the height of the bowl is h = y2 – y1 = 10 cm.
Since the internal diameter, across the bottom of the bowl, is 16 cm, then the x-coordinate of this lower part of the bowl is x1 = 8 cm and its y-coordinate is y1.
Since the internal diameter, across the top of the bowl, is 24 cm, then the x-coordinate of this upper part of the bowl is x2 = 12 cm and its y-coordinate is y2.
We thus have two points on the bowl, P1(x1, y1) = (8, y1) and P2(x2, y2) = (12, y2) which are related by the equation, y=a.x^2.
Using this equation with P1 and P2,
y1 = a.x1^2
y2 = a.x2^2
Substituting values,
y1 = a.64
y2 = a.144
Since the height of the bowl is given as 10 cm, then we have y2 – y1 = 10.
Subtracting the previous two equations,
y2 – y1 = 10 = 144a – 64a = 80a
10 = 80a
a = 1/8
Then, y1 = 8 cm, y2 = 18 cm.
Therefore the defining equation of the parabola is: y = (1/8)x^2.
We can now re-describe the bowl as the volume of revolution of a parabola, y = (1/8)x^2, about its vertical axis (y-axis), taken between the limits y1 = 8 and y2 = 18.
This volume is given by the formula, V = int [ y = y1 .. y2] (1/a).π.y dy.
V = int [ y = 8 .. 18] (8).π.y dy.
Integrating,
V = [4.π.y^2] [18 .. 8]
V = 4.π.{324 – 64} = 4.π.{260}
V = 1040.π. cu cm.
V = 3.267 litres