Volume of a Sphere

[xilman]

Originally Posted by Patrick123

It is fascinating, this is how I originally worked it out.

This is my solution.

If the question has a unique answer, and it phrased as if it must, the solution must be independent of the diameter of the sphere or the radius of the hole. In particular, the special case of a hole of zero radius and hence zero volume must yield the unique answer. This hole must clearly be drillled through a 6" diameter sphere.

Paul

[mfgoode]

Originally Posted by xilman

This is my solution.

If the question has a unique answer, and it phrased as if it must, the solution must be independent of the diameter of the sphere or the radius of the hole. In particular, the special case of a hole of zero radius and hence zero volume must yield the unique answer. This hole must clearly be drillled through a 6" diameter sphere.

Paul

Kindly re-read my post no.10.
Any sphere of diameter 6 inches and Above will yield a residue of 36 pi for a 6 inch long hole, including our earth! Fascinating isn't it?
Mally

[Fusion_power]

Draw a square. It can be any size. Now draw a circle so that it fills the square. What is the relationship of the area of the circle to the area of the square?

Now draw the same square and put 4 identical circles inside it so they fill the square. What is the relationship of the area of the circle to the area of the square?

Now draw the same square with 9 circles and figure the areas.

What would it be with 16 circles?

Hint, an easy approach to this is to use a square with sides 6 long. Use inches, cm, etc, whatever makes you happy.

Fusion

[drew]

Originally Posted by Fusion_power

Draw a square. It can be any size. Now draw a circle so that it fills the square. What is the relationship of the area of the circle to the area of the square?

Now draw the same square and put 4 identical circles inside it so they fill the square. What is the relationship of the area of the circle to the area of the square?

Now draw the same square with 9 circles and figure the areas.

What would it be with 16 circles?

Hint, an easy approach to this is to use a square with sides 6 long. Use inches, cm, etc, whatever makes you happy.

Fusion

Is this another puzzle?

The way you described, the picture can still be reduced to smaller squares, so the ratios of areas are the same due to similarity. The only way to improve this is to change the packing. Hexagonal packing will be an improvement over the square packing you described.

You can do even better if you allow circles of various sizes.

Drew

[Fusion_power]

Its not a packing puzzle, its a relationship demonstration. You will see the relationship if you solve the elementary math involved. Leave the result in the form X(pi). You should also see the relationship to the above about a hole drilled into a sphere.

Fusion

[drew]

Originally Posted by Fusion_power

Its not a packing puzzle, its a relationship demonstration. You will see the relationship if you solve the elementary math involved. Leave the result in the form X(pi). You should also see the relationship to the above about a hole drilled into a sphere.

Fusion

The ratio of circle area to square area is pi/4 for all cases. (pi*R²/(2R)²)

Like I said. The ratio will always be the same due to similarity. It can always be divided into a number of circles inscribed in squares.

Drew

[Numbers]

Fusion,
When you say the circle "fills" the square, do you mean the sides of the square are tangent to the circle, or do you mean the circle touches the square at its corners?

Thanks,

[mfgoode]

Originally Posted by Numbers

Fusion,
When you say the circle "fills" the square, do you mean the sides of the square are tangent to the circle, or do you mean the circle touches the square at its corners?

Thanks,

Ah numbers, you mean the circle is inscribed in the square or it circumscribes the square. With the right math terminology you cant go wrong!
Mally

[Fusion_power]

Inscribed in the square meets the conditions stated. However, a modified set of conditions would give a similar result for a circumscribed circle(s).

Fusion