The hole in the sphere?

Question

A 6cm long hole is drilled through a solid sphere. What is the volume of material left?

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Answer 1

Good question.

Standard Organisations has defined common drill bit sizes which ranges from 0.02 cm to 2.5 cm in diameter.

Now,

Volume = ( Base Area of the drill bit = π * r^2 ) * height

Min Volume drilled out = [π * (0.01)^2 *  6] cu. cm

= 6π/10000 cu. cm = 0.0018849 cu. cm

Max Volume drilled out = [π * (1.25)^2 * 6] cu. cm

= 9.375π cu. cm = 29.4524311 cu.cm

Answer 2

Well it can be shown that the initial volume of the sphere doesnt matter.

The leftover volume of a sphere with a 6cm long hole drilled through it is the same as the volume left behind of a sphere of 1/2 the diameter with a 0cm long hold drilled through it

which is 4/3*(pi*(3)^3)) = 36pi  

Answer 3

You've removed an important piece of information: the hole is drilled straight through the center of the sphere and the 6cm is measured as the length of the hole, not the length of the part removed.

Given these, the answer is (although not obviously uniquely defined)

4/3 pi (6/2)^3 = 36pi.

Answer 4

The diameter of the 6cm long hole would be a big help

Answer 5

I don't think that we do.  We don't know the diameter of the hole that was drilled.

We know since it's 6cm long, that the radius is 3cm.

So we can find the volume of the orignal sphere, but depending on what the radius of the hole is, it would change the value of what's left.

Answer 6

No, we need one more piece of info.

What is the diameter of the hole being drilled?

The diameter of the sphere is 6.

Volume of Sphere = 4/3 pi (6/2) ^3

=36 pi cm^3

and the removed material = 6 * pi (radius of hole in centimeters)^ 2

Answer 7

4/3 π (6/2)³ = 36π

***

All right, then.

Ho = 6/2 = 3

Cross-section of a sphere of radius R with a hole of raduis r at height h is  

S(h) = πρ²(h) - πr² = π(ρ²(h) - r²) = π((R²-h²) - r²)

S(h) = π[ (R²-r²) - h²]

The volume is given by

dV = S(h) dh

V = ∫S(h) dh =  ÃƒÂ¢Ã‚ˆÂ«ÃÂ€[(R²-r²) - h²] dh

The integrand π[(R²-r²) - h²] is quadratric function  of h, for such integrals simpson quadrature is exact:

V = 1/6 * 2Ho [ S(h= -Ho) + 4S(h=0) + S(h=+Ho) ]

Both top and bottom cross-sections have areas equal to zero:

S(h= -Ho) = S(h=+Ho) = 0

V = 1/6 * 2Ho [ 4S(h=0)]

V = 4/3 Ho S(h=0) = 4/3 Ho π (R² - r²)

V = 4/3 π Ho Ho² = 4/3 π Ho³

Answer 8

I suspect foul play.

Answer 9

This is a well known problem. Since you are not told the diameter of the hole or sphere but are asked for a numerical answer it implies that the diameter of the hole can be anything. In which case you choose to make it infinitely small i.e. zero. This leaves the sphere as radius 3cm. That is why the answer two above mine is correct.

This problem can be done by assuming that R, r are the radii of the sphere and hole. About six pages of algebra later you find that both of them fall out and leave the numerical answer given.

Answer 10

If the hole had a diameter approaching 0, then the diameter of the sphere would approach 6 cm. In this case, the volume of the remaining material would approach 4pi3^3/3 = 36pi cm^3.

If the diameter of the hole is x, then the diameter of the sphere increases so that the remainig material is still 36pi cm^3  

Answer 11

The volume of material left is exactly the same as what you started with, it's just not all in the sphere. A numerical answer is impossible since the size of the sphere is unknown - the hole isn't necessarily drilled through the centre of the sphere hence we don't know the sphere's radius.