The function y = sqrt(9 -x^2) defines a circle of radius 3 centered on the origin.
The region bounded above by the function and below by the y-axis is a half circle.
Rotating this about the y-axis gives a half sphere (hemisphere) with radius 3.
The volume of a sphere is 4/3 pi r^3 where r is the radius. Therefore, the volume of a hemisphere is 2/3 pi r^3 .
In this example, we have r=3 so that the volume of the hemisphere (a solid of revolution) in this case is 18pi .
If a hole, centered on the axis of revolution (the y-axis here) is drilled through the solid so that 1/3 of the volume is removed then the volume of the remaining volume is 12pi
The hole that is drilled out is also a solid of revolution about the y-axis, where the relevant interval on the x-axis is 0<=x<=a where a is the radius of the hole. Also, what is left of the hemisphere is a solid of revolution, called a spherical segment.
A spherical segment with upper radius a , lower radius b and height h has volume
V = 1/6pi h (3a^2 + 3b^2 + h^2)
In the example here, a is the radius we wish to find (half of the diameter we wish to find), and b is the radius of the original hemisphere before drilling, ie r=3 .
The value of h is the value of y corresponding to x=a on the circle radius 3 centered on the origin, and so satisfies
a^2 + h^2 = 9 so that h^2 in terms of a^2 is given by
h^2 = 9 - a^2
Plugging this in to the formula for the spherical segment object we have
V = 1/6 pi sqrt(9-a^2)(3a^2 + 3r^2 + 9 -a^2) = 1/6pi sqrt(9-a^2)(2a^2 + 36)
We know that V is 2/3 of the volume of the original hemisphere, giving
V = 12pi = 1/6 pi sqrt(9-a^2)(2a^2 + 36)
(9-a^2)(2a^2+36)^2 = 36(144)
Solving this, it follows that the radius of the drilled section a is approximately
a = 2.62900'
and the diameter is approx
2a = 5.258'
http://mathworld.wolfram.com/SolidofRevolution.html
http://mathworld.wolfram.com/SphericalSegment.html
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