The function 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 where
is the radius. Therefore, the volume of a hemisphere is
.
In this example, we have so that the volume of the hemisphere (a solid of revolution) in this case is
.
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
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 where
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 , lower radius
and height
has volume
In the example here, is the radius we wish to find (half of the diameter we wish to find), and
is the radius of the original hemisphere before drilling, ie
.
The value of is the value of
corresponding to
on the circle radius 3 centered on the origin, and so satisfies
so that
in terms of
is given by
Plugging this in to the formula for the spherical segment object we have
We know that V is 2/3 of the volume of the original hemisphere, giving
Solving this, it follows that the radius of the drilled section is approximately
and the diameter is approx
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