My favourite maths question of all time:
A cylindrical hole 6cm in length is drilled through the centre of a solid sphere, with the axis of the hole passing through the centre of the sphere, what is the remaining volume of the sphere?
To be clear, after drilling the sphere is no longer quite spherical, I want to know the volume of what is left of the sphere, so original sphere minus the stuff drilled out.
Show your working.
If you know the answer already then please don't reply.
The diameter of the hole isn't specified, so I will choose a diameter of >= 6cm. The remaining volume of the sphere will be zero.
Edit: Is this question well-posed mathematical question with a unique solution? Or a typical Arduino forum question with incomplete information and multiple solutions?
It's hard to shift gears.
diameter of the hole is 0..
so the diameter of the sphere is 6, radius 3..
the rest of this eludes me..
the ring that holds the napkin..
therein lies the answer..
gonna be allot of pie..
~q
That's the answer I came up with too. Anything smaller than 6cm would need you to calculate the volume of the hole, taking into account the "domed" ends of the hole.
The first. There is sufficient information in the question to give a single, clear solution.
math can be so tricky..
V = 4/3 Pi r3
r = 3
V= 4/3 Pi (3*3*3)
V= 4/3(3.14)(27)
V= 12.56/3 * 27
V=4.186 * 27
V=113
~q
Why you must be 42! My neighbor is 21 and he's only HALF NUTS!
Yes, I could be wrong, quite possible..
But I still stand by my calculations..
That's what I submit..
~q
I don't recall the diameter of the sphere mentioned or the diameter of the hole.
In a real situation there would need to be integration involved to get the removed volume to subtract from the sphere.
How can you make such a cyclinder when the end surfaces of the cylinder are curvatures and NOT circles?
As for me, the end surfaces of this cylindre are circles
As a mechanical engineer who worked in manufacturing, and also with creating designs for 3d printing, I agree that there's a great number of solutions, if not for:
That's why me must be 42!
Yes!
But, I am referring to the so called cylinder that @PerryBebbington is trying to make out of the vertcal axis of a solid sphere.
It is possible that the solution he means is not a mathematical solution at all
Maybe, if not for:
The removed material will include the volume of a rotated arc.
That will take an integration with limits, same way as calculating the formula for the volume of a sphere only with different limits of integration.
When you know why 4/3 is in the sphere volume formula, you'll get it. It just takes calculus is all.
So it mustn't matter given the information. Not necessarily a trick question. Maybe a trick question, I have no idea but maybe one of those ones where the answer is a constant, like pi.
Since I don't do math for fun because I suck at math and since this made me think of pi and since pi is used with circles and since the puzzle involves a sphere which is like a zillion circles spinning around each other which is really just one continuous circle (think ball of yarn) and since someone else reminded me that the ends of a cylinder are circles... (@b707 in post 13)
my answer is one pi for the sphere, one pi for the one end of the cylinder and one more pi for the other end of the cylinder,
3.14 + 3.14 + 3.14 = 9.42 litres.
Residual volume is
V = (2/3) * Pi * ( hole_length) ** 3
V = (2/3) * 3.14 * (6*6*6) = 452 cm**3
Changed.
Sorry, above is incorrect, I forget to put factor two out of brackets...
V = (4/3) * Pi * ( hole_length/2 ) ** 3
V = (4/3) * 3.14 * (3*3*3) =112.8 cm**3