Brain Teasers
Drilling Through A Sphere
I just drilled straight through the center of a solid sphere, resulting in a 6cm long cylindrical hole. What is the volume remaining in the sphere?
Here are some formulas you may find useful:
Volume of sphere = (4/3)Pi(r^3)
Volume of cap = (1/6)Pi(3R^2+h^2)h
Volume of cylinder = Pi(R^2)H
r = radius of sphere
h = height of cap
R = radius of hole (the cap and the cylinder share this radius)
H = height of cylinder (6cm)
The portion of the sphere that was removed can be thought of as a cylinder with a cap at both ends. A cap is flat on one side (touching the cylinder) and spherical on the other side (the original surface of the sphere).
Here are some formulas you may find useful:
Volume of sphere = (4/3)Pi(r^3)
Volume of cap = (1/6)Pi(3R^2+h^2)h
Volume of cylinder = Pi(R^2)H
r = radius of sphere
h = height of cap
R = radius of hole (the cap and the cylinder share this radius)
H = height of cylinder (6cm)
The portion of the sphere that was removed can be thought of as a cylinder with a cap at both ends. A cap is flat on one side (touching the cylinder) and spherical on the other side (the original surface of the sphere).
Hint
Hint for a hard solution:The sphere's diameter (2r) must be the same as the length of the cylinder (H) plus the height of the two caps (2h), so 2r=H+2h.
A right-angled triangle is formed by the radius of the cylinder (R) and half the height of the cylinder (H/2), with the radius of the sphere (r) forming the hypotenuse. Using Pythagoras, R^2+(H/2)^2=r^2.
Hint for an easy solution:
Assume there is a real answer. Therefore, the answer does not depend on the radius of the hole (R). Use whatever value is easiest.
Answer
The volume is exactly 36Pi cm^3 (about 113.1 cm^3).Hard solution:
The sphere's diameter (2r) must be the same as the length of the cylinder (H) plus the height of the two caps (2h), so 2r=H+2h.
Thus, 2h=2r-H and H=6, so 2h=2r-6 or simply h=r-3.
A right-angled triangle is formed by the radius of the cylinder (R) and half the height of the cylinder (H/2), with the radius of the sphere (r) forming the hypotenuse. Using Pythagoras: R^2+(H/2)^2=r^2.
Thus, R^2=r^2-(H/2)^2 and H=6, so R^2=r^2-9.
Volume of cap = (1/6)Pi(3R^2+h^2)h
(replace R^2 and h with values computed above)
= (1/6)Pi(3(r^2-9)+(r-3)^2)(r-3)
= (1/6)Pi(3r^2-27+r^2-6r+9)(r-3)
= (1/6)Pi(4r^2-6r-18)(r-3)
= (1/6)Pi(4r^3-6r^2-18r-12r^2+18r+54)
= (1/6)Pi(4r^3-18r^2+54)
= (2/3)Pi(r^3)-3Pi(r^2)+9Pi
Volume of cylinder = Pi(R^2)H
(replace R^2 with value computed above, and H with given value of 6)
= Pi(r^2-9)6
= 6Pi(r^2)-54Pi
Volume of sphere remaining = sphere - 2 caps - cylinder
= [ (4/3)Pi(r^3) ] - [ 2 ((2/3)Pi(r^3)-3Pi(r^2)+9Pi) ] - [ 6Pi(r^2)-54Pi ]
= [ (4/3)Pi(r^3) ] - [ (4/3)Pi(r^3)-6Pi(r^2)+18Pi ] - [ 6Pi(r^2) - 54Pi ]
= (4/3)Pi(r^3) - (4/3)Pi(r^3) + 6Pi(r^2) - 18Pi - 6Pi(r^2) + 54Pi
= 36Pi
Easy solution:
Assume the hole is infinitely small (i.e. a radius of 0). Then, the volume of the cylinder (and the volume of the caps) is 0. The height of the cap is 0, so the radius of the sphere (r) is 3cm (half of the length of the cylindrical hole). Thus:
Volume of sphere remaining = sphere - 2 caps - cylinder
= (4/3)Pi(r^3) - 0 - 0
= (4/3)Pi(3^3)
= 36Pi
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Comments
ummm, no comment from me either.. lol
Crazy...I'm pretty good at math but I got it wrong...took a while too
You are completly negating that there is a hole in the sphere. The answer be in terms of R (the unknown radius of the hole).
Teenie, the beauty of this problem is that the radius of the hole doesn't matter! The hard answer proves this.
I showed that R^2=r^2-9. The radius of the hole depends on the radius of the sphere - but they cancel out in the final answer - this works for many sizes of spheres and holes, as long as the hole is 6cm long as given in the problem.
If R=4, r=5. If R=1, r=square root of 10. If R=0, r=3 (it isn't much of a hole, but it's like a limit in calculus).
I showed that R^2=r^2-9. The radius of the hole depends on the radius of the sphere - but they cancel out in the final answer - this works for many sizes of spheres and holes, as long as the hole is 6cm long as given in the problem.
If R=4, r=5. If R=1, r=square root of 10. If R=0, r=3 (it isn't much of a hole, but it's like a limit in calculus).
Not my cup of tea, sorry.
This is a standard problem in Calculus solved by using integration to calculate the volume of the remaining solid formed by rotating the remaining part of the circumference of the sphere about its axis. I like the twist where the radius of the sphere and the hole are not given but are co dependent in order to produce the required 6 unit length hole.
I haven't think a lot about it, but my first impression is that the "easy" way to solve the problem is not easy at all. The problem has 4 variables (r, H, R and h) and only 2 equations (explained in the "hard" solution), so the "remaining volume V" in principle depends on 2 variables. Surprisingly, one of them cancels out and the final expression is
V=(pi/6)*H^3
so the calculation works for any R (particularly for R=0). BUT THIS MUST BE PROVEN FIRST!
V=(pi/6)*H^3
so the calculation works for any R (particularly for R=0). BUT THIS MUST BE PROVEN FIRST!
Tricky, maybe not too bad, but most importantly, it is not specified that, say the cylindrical whole STOPS at the center. I assumed it went all the way through, because you said it went "through the center" not even "through TO the center" and so thats a potentially confusing part. otherwise, i didn't do the math, but i doubt the answer would be the same. That's why the cap-idea made not too much sense, as i thought it was rounded on both ends.
lessthanjake, the cylindrical hole doesn't stop at the centre, your original understanding was correct - it goes all the way through.
Assuming the problem has a single solution, it can be inferred that the as the "bore" of the drill increases, the radius of the original sphere must increase (to keep the 6cm length constant), and the resulting increase in original volume must equal the increase in drilled out volume.
So the solution with an infinitesimal "bore" must work.
But all this hinges on the question being properly written, which is not always the case on this site.
So the solution with an infinitesimal "bore" must work.
But all this hinges on the question being properly written, which is not always the case on this site.
May 08, 2011
There is an easier way to do the hard solution doesn't require formulae for the end caps so you can see for yourself more easily why it is true.
Uses Archimedes theorem, that you have two different 3D shapes, and if all the horizontal slices through the shapes have the same area then the volumes are the same. That's kind of intuitive (though needs calculus to prove)
Let 2a be the length of the hole. Then let one of the volumes be the sphere with radius a, so with volume 4/3*pi*a^3
Let the other volume be the sphere with radius r > a with hole of length 2a through it
Cut through both volumes at a height h above centre of the sphere. Then at this height, the sphere radius a has cross section area pi * (a^2 - h^2) by pythagorous's theorem (radius of the cross sectional disk is sqrt(a^2 - h^2) )
At the same height h, the sphere radius r with hole length 2a has an annular cross section of area pi * (r^2 - h^2) - pi * (r^2 - a^2) and the pi * r^2 terms cancel to give pi * (a^2 - h^2) as before.
For a more detailed description of this solution see Dr. Math discussion here:
http://tinyurl.com/4y5au8w
Uses Archimedes theorem, that you have two different 3D shapes, and if all the horizontal slices through the shapes have the same area then the volumes are the same. That's kind of intuitive (though needs calculus to prove)
Let 2a be the length of the hole. Then let one of the volumes be the sphere with radius a, so with volume 4/3*pi*a^3
Let the other volume be the sphere with radius r > a with hole of length 2a through it
Cut through both volumes at a height h above centre of the sphere. Then at this height, the sphere radius a has cross section area pi * (a^2 - h^2) by pythagorous's theorem (radius of the cross sectional disk is sqrt(a^2 - h^2) )
At the same height h, the sphere radius r with hole length 2a has an annular cross section of area pi * (r^2 - h^2) - pi * (r^2 - a^2) and the pi * r^2 terms cancel to give pi * (a^2 - h^2) as before.
For a more detailed description of this solution see Dr. Math discussion here:
http://tinyurl.com/4y5au8w
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