Hole in a sphere problem
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6 members have voted
December 2nd, 2017 at 8:54:00 PM
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A hole is drilled straight through the center of a sphere. The hole measures 6" deep. At what radius of the sphere is the volume of the remaining portion of the sphere, after the drilling, maximized, and what is that area?
As usual, free beer to the first acceptable answer and solution.
Note: Section in bold added for clarity.
As usual, free beer to the first acceptable answer and solution.
Note: Section in bold added for clarity.
Last edited by: Wizard on Dec 3, 2017
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 2nd, 2017 at 11:42:14 PM
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What's the radius of the hole? Since that wasn't given, I'd think this could be answered without knowing that, which now confuses mee. :( Seems like the sphere's area would be maximized when the radius of the sphere is 2.999999
December 3rd, 2017 at 1:24:58 AM
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Quote: WizardA hole is drilled straight through the center of a sphere. The hole measures 6" deep. At what radius of the sphere is the area of the remaining portion, after the drilling, maximized, and what is that area?
As usual, free beer to the first acceptable answer and solution.
3 inches, since the hole is 6 inches deep and goes through the center of the sphere? In fact, 3 inches can be the only answer, since the diameter of the sphere is 6 inches.
Unless I'm missing something?
Unless I'm missing something?
Simplicity is the ultimate sophistication - Leonardo da Vinci
December 3rd, 2017 at 3:05:02 AM
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Quote:The key here is to realise that the remaining volume of a sphere with a 6" long hole drilled in it is the same as the remaining volume of a 6" diameter sphere with no hole drilled in it.
Therefore, the radius is 3" and the area is 4*pi*3*3 = 36pi sq. inches. The volume is 4/3*pi*3*3*3 = 113 cubic inches.
Mods - pls put this in a spoiler. Thanks.
Last edited by: unnamed administrator on Dec 3, 2017
Casino Enemy No.1
December 3rd, 2017 at 4:59:15 AM
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I didn't ask the question very well. I edited the OP to say I'm looking for the radius of the original sphere that maximizes the area left of the sphere left over after the drilling. Something the shape of a napkin ring.
I agree that with a 3" radius the hole would be infinitely thin, leaving a sphere of (4/3)*pi*3^3 in volume. However, can you prove that is the largest volume left over?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 3rd, 2017 at 6:25:51 AM
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I'll try. The drill hole becomes a circular cylinder passing through the centre of the sphere. Therefore, the largest volume left over occurs when the cylinders height equals the diameter of the sphere and therefore the remaining volume outside the circular cylinder equals the volume of the sphere. At this point the circular cylinder does not exist.
Volume of circular cylinder = (pi*h*h*h)/6
Volume of sphere = (4/3)*pi*3*3*3
Therefore: (pi*h*h*h)/6 = (4/3)*pi*3*3*3
h = 6 inches.
Therefore, the largest volume left over occurs when the sphere has a diameter of 6 inches.
Volume of circular cylinder = (pi*h*h*h)/6
Volume of sphere = (4/3)*pi*3*3*3
Therefore: (pi*h*h*h)/6 = (4/3)*pi*3*3*3
h = 6 inches.
Therefore, the largest volume left over occurs when the sphere has a diameter of 6 inches.
Casino Enemy No.1
December 3rd, 2017 at 7:25:22 AM
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Quote: WizardI didn't ask the question very well. I edited the OP to say I'm looking for the radius of the original sphere that maximizes the area left of the sphere left over after the drilling. Something the shape of a napkin ring.
Wiz,
Are you trying to maximize the surface area or the volume of the resulting "napkin ring"?
Answer for Area:
After the cylindrical shaft of height h=6" and unknown radius r is drilled through the sphere of radius R, the surface area A of the "napkin ring" is the original sphere's surface area plus the interior surface area corresponding to the circumferential area of the shaft minus the external surface area of the two "caps" removed in the drilling process, so (let p = pi)
A = 4pR² + 2prh - 2p[(R-h)² + r²] = 2p[R² + 2Rh - r² + rh - h²]
The "cap" area formula can be found at, for example, https://math.stackexchange.com/questions/431627/archimedes-derivation-of-the-spherical-cap-area-formula
Now the variables R, r, and h/2 form a right triangle with R as the hypotenuse, so
R² = r² + (h/2)², so r = (R² - ¼h²)^(½)
Plugging this into the A equation and simplifying gives
A = 2p[2Rh - ¾h² + h(R² - ¼h²)^(½)]
with h and p constant.
Thus, in the limit as R goes to infinity, A goes to infinity.
A = 4pR² + 2prh - 2p[(R-h)² + r²] = 2p[R² + 2Rh - r² + rh - h²]
The "cap" area formula can be found at, for example, https://math.stackexchange.com/questions/431627/archimedes-derivation-of-the-spherical-cap-area-formula
Now the variables R, r, and h/2 form a right triangle with R as the hypotenuse, so
R² = r² + (h/2)², so r = (R² - ¼h²)^(½)
Plugging this into the A equation and simplifying gives
A = 2p[2Rh - ¾h² + h(R² - ¼h²)^(½)]
with h and p constant.
Thus, in the limit as R goes to infinity, A goes to infinity.
Dog Hand
December 3rd, 2017 at 8:31:27 AM
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Quote: DogHandWiz,
Are you trying to maximize the surface area or the volume of the resulting "napkin ring"?
Sorry, I'm looking to maximize the volume.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 3rd, 2017 at 8:36:59 AM
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Quote: davethebuilderTherefore, the largest volume left over occurs when the cylinders height equals the diameter of the sphere and therefore the remaining volume outside the circular cylinder equals the volume of the sphere.
You haven't convinced me this is a true statement. Why is it true?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 3rd, 2017 at 11:45:35 AM
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The volume is 36 PI, and this is regardless of the radius of the sphere
Let r be the radius of the sphere.
Since the cylinder has length 6, it extends 3 in opposite directions from the center of the sphere.
The radius of the base of the cylinder is sqrt(r2 - 32), and the area of a cross-section of the cylinder = PI (r2 - 9).
Each cross-section of the sphere parallel to the base of the cylinder is a 2-dimensional torus (I think that's what it is called in 2 dimensions) with outer circle radius sqrt(r2 - x2) and inner circle radius sqrt(r2 - 9), which has area PI (r2 - x2) - PI (r2 - 9) = PI (9 - x2).
The volume = 2 * the integral over x = 0 to 3 (the length of the cylinder in each direction) of (x2 - 9) dx, which equals 2 * PI (27 - 27/3) = 36 PI.
December 3rd, 2017 at 12:19:39 PM
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Quote: ThatDonGuy
The volume is 36 PI, and this is regardless of the radius of the sphere
Let r be the radius of the sphere.
Since the cylinder has length 6, it extends 3 in opposite directions from the center of the sphere.
The radius of the base of the cylinder is sqrt(r2 - 32), and the area of a cross-section of the cylinder = PI (r2 - 9).
Each cross-section of the sphere parallel to the base of the cylinder is a 2-dimensional torus (I think that's what it is called in 2 dimensions) with outer circle radius sqrt(r2 - x2) and inner circle radius sqrt(r2 - 9), which has area PI (r2 - x2) - PI (r2 - 9) = PI (9 - x2).
The volume = 2 * the integral over x = 0 to 3 (the length of the cylinder in each direction) of (x2 - 9) dx, which equals 2 * PI (27 - 27/3) = 36 PI.
I agree. Please add one to the count of beers I owe you. I think we would need a long weekend in Munich to pay off my debt.
I apologize for the bad wording of the problem.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 3rd, 2017 at 12:23:14 PM
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Just buy him a keg now.
DUHHIIIIIIIII HEARD THAT!
December 3rd, 2017 at 12:23:43 PM
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Quote: WizardI agree. Please add one to the count of beers I owe you. I think we would need a long weekend in Munich to pay off my debt.
I'm penciled in for Vegas the four days immediately after Father's Day - I'm not much of a beer drinker, but I might just take you up on one of those.
December 3rd, 2017 at 1:54:27 PM
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Quote: ThatDonGuyI'm penciled in for Vegas the four days immediately after Father's Day - I'm not much of a beer drinker, but I might just take you up on one of those.
I will clear my debt one way or the other.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
