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 ` ` <br> `
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'
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