A solid is generated by revolving the region bounded by `y=sqrt(9-x^2)` and `y = 0` about the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-third of the volume is removed. Find the diameter of the hole.

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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...

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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'

Approved by eNotes Editorial Team