/sci/ - Science & Math

Um, not enough info..?

68 cubic metres

>>2034524

i can't be arsed to calculate the general formula for this

>>2034524
whats the radius or volume of the whole and volume of the initial sphere?

Obviously it would be the weight, minus the amount lost due to the drilling. This could be represented by Y=X-Y; where Y= the final weight and X equals the original weight and Y equals the amount lost due to the drilling... Problem Aristotle?

Put the sphere in a graduated cylinder half full of water, measure change in water level.
Why is this hard?

How many centimeters in six inches ?

>>2034599

you have no liquids at your disposal, that is why.

>>2034605

15.4 centimeters aprox

It would remain the same

>>2034587

If the hole was drilled all the way through the sphere then the sphere has a diameter of six inches and a radius of 3 inches. That means the volume of the sphere is 63.61. I would assume that I would just have to calculate the volume of the cylinder and subtract it from the volume of the sphere but I don't know the radius of the cylinder. I also suck at math.

>>2034524

1,92 liters, approximately.

>>2034524

It's easy :

V(remaining in the sphere) = V(whole sphere) - V(cylinder you drilled)


Duh.

is the hole 6 inches i ø too?

r = 3 inches durr durr
you need to integrate sphere using cross sections too for cross sections greater than the radius of the hole. Fucking be a hoss and only integrate over the radius of the sphere, then double that fucking result

>>2034631

1.853 Liters to be more precise.

It depends.

>>2034612
Melt the material removed during drilling and measure the volume!

Also sage for homework!

>>2034657

actually, it doesn't.

>>2034648

nope, 6 inches is the depth of the hole, not the diameter of the sphere.

>>2034524
37.7 inches cubed

I was really hoping this was a lead in to a joke about sticking a penis shaped cylinder into your original image.

"a cylindrical hole 6 inches long"

inb4 people substract the volume of a cilinder to the volume of a sphere because that is wrong

>>2034675

Ok then, what's the volume of our sphere of undisclosed size with a hole of an unknown radius that goes 6 inches deep?

>>2034696

The wording of the question suggests that the hole started one side of the sphere, passed through the center, and ended the other side.

>>2034524

the thing is that there are infinite holes and spheres that match those requirements.

>>2034713

Indeed, but we're measuring the hole, not the portion of the sphere that was removed.

>>2034715
yes, but do they all have the same volume?

>>2034711

113.0973 inches cubed.

I like this problem! I will solve it shortly.

36pi

>>2034744
Cubic inches, obviously.

>>2034735

No, both the cylinder and sphere's volume change depending on the cylinder radius.

>>2034747
then what are those two guys who posted 36pi on about?

>>2034753
They all have the same volume, it doesn't matter how big the original sphere is.

>>2034754
Given that it's at least 6 inches in diameter since otherwise the problem would be impossible.

>>2034754
You're saying if I bored a hole through the center of the earth, removing nearly all of the earth, and all that was left was a giant ring, the cross section of which looks like a D, where the straight edge is six inches long, then the volume of the remaining portion of the earth would be 36pi cubic inches?

>>2034696
>straight through center
>through center
>through
but actually it does say the hole goes through
>>2034713
>agreed

>>2034648
This still stands by the way, just scratch doubling the result part, integrate from 0-6inchs, and then add to 6-Dinches
but really this is a troll thread
because its poorly worded, indeterminate, and has no general or discreet answer. op is laughing

>>2034760
If the earth was a perfect sphere, yes.

>>2034753

the volume of the sphere and cylinder change, but the difference between them remains the same.

It's a bit of a mind-twister.

This problem reminds me of Mathcounts.

This is just a control thread to see how much faster /sci/ gets the answer than /a/. I wouldn't reply, but イか娘 compels me.

>>2034787
Ah damn. I got the answer in about 10 seconds, but didn't see the thread until it already had 70 replies.

If the hole is 6 inches deep, then the diameter of the sphere is 6+X, with X being the height of the curved bit removed by the drilling but not part of the hole itself.

With this in mind, everything else can be solved.

>>2034787

actually, this came from /v/ right now.

>>2034798

I can't believe so many people missed this bit.

>>2034810
No one missed it. You're just thinking about it in an overly complicated way.

>>2034814

No, neglecting this property of drilling a sphere is imprecise.
That shit doesn't fly here.

>>2034822
That shit gets drilled out though. It's not part of the remaining sphere.

>>2034524

It's a constant, but I'm not sure of the value.

>>2034524
actually, its just the volume of a sphere with a 3 inch radius.

In which case, the cylindrical hole has an infinitesimally small radius.

>>2034845

not necessarily.

What about a sphere with an 60 inch radius?

>>2034810
The problem is still undefined. The problem, as it is stated, could refer to any one of these three situations. In all cases, a cylindrical hole of side-length a (in this case a=6 inches) through the center of a solid sphere. Clearly, they will have separate volumes.

The "right" answer is for a certain special case, with assumptions made about the positioning of a cylinder and what it means to "drill through."

The mathematical problem, though, is interesting, though not surprising. Given the special conditions, we can see that as the scale of the ball grows, the "drilled" section grows as r^3, exactly the same as the volume of the sphere. Hence a constant result is somewhat expected.

>>2034845

see

>>2034798
>>2034810

IKA MUSUME STOP TEASING THE MATH NERDS AND GET BACK TO SUCKING MY DICK

>>2034869
The problem should have been phrased better, but I think it's fairly clear that the hole was drilled from the surface of the sphere and that all volume that was drilled is eliminated.

>>2034867
Well, my answer was based on the assumption that there was a solution to the problem, in which case it doesn't matter the size of the sphere, and the simplest form of the problem is a sphere with an infintessimilly small radius to the hole drilled through it.

>>2034884
Tell me this, then. If one of the cups remains, barely hanging on, while the other one is "drilled out," how does the answer change? Explaining without using calculus or an exact result.

>>2034524
it isnt a sphere any more now is it?

lol i don't fucking know you didn't give a radius of either the sphere or the cylinder, its whatever the fuck i want it to be with the information given.

>>2034869
Straight through however shows that it goes from one side to the other through the exact center of the sphere.

>>2034825

That's not the problem. The problem is that the sphere's diameter is actually wider than the length of the bore, which throws any calculations off if you presume they're the same.

>>2034869

Actually, It would be like this.

>>2034906

actually, it's not.

>>2034922
>The problem is that the sphere's diameter is actually wider than the length of the bore, which throws any calculations off if you presume they're the same.
I'm not presuming they're the same though. The fact that they're not the same is the reason the solution is always 36pi for any sphere of radius greater than 3.

>>2034985
>no information about the cylinder radius

mmkay

>>2034524
What is the threading on the drill? Any answer I give assuming a solid cylinder must be wrong since a drilling motion has spiral threads which would cut into the sphere's insides.

>>2034993
The cylinder radius is determined by the size of the sphere. For a given sphere there is only one possible cylinder radius that makes the cylinder 6 inches long.

>>2034906
The radius of the sphere is obviously 3 inches. Since the cylinder "drilled" through is 6 inches (the diameter of the sphere). HERPADERP

>>2035008
Only if you assume that the drilling is done entirely along a diameter of the ball AND that the maximum possible length is also the one drilled. Too many assumptions.

>>2035046
this is a beautiful logic puzzle because everything but the essential has been eliminated. Most problems can only be solved once you assume there is an answer. That's the beauty.

>>2035046
The problem states that the hole is drilled through the center. The problem also states that the length of the cylinder is six inches. Christ you're retarded.

>>2035057
You can drill through the center without drilling all the way through. I'm not retarded, I just think outside the box.

cant you just solve it with r = radius of sphere and r2 = radius of cylinder

triple integral and all that jazz

>>2035069
"straight through the center"
Pretty sure that means all the way through.

>>2035069
I'd say that's fine, so long as you see what the problem is pointing at. But you're trying to hard to find fault with the question.

>>2035082
The problem is that the question was written with a specific answer in mind. Real problems and questions are open-ended unless sufficiently defined. Some are actually OVER-defined and therefore have NO answers, as well. We are taught to abstract real-world problems into text-book like ones we can solve. However, anything new, be it knowledge or technology or the like, only comes about by dropping an assumption or two and then finding where you end up. I assure you, the person who first solved this problem didn't look for some constant answer, but stopped assuming that all you could do with a ball and cylinder was intersect them. He (or she) thought about a dynamic path, and suddenly a novel result was born...

(4/3) * pi * 3in^3 - pi * r^2 * 6in = V

36pi - 6pir^3 = V

that's as far as I got.

I don't think you can answer it because we don't know the radius of the hole... unless I missed something

Not enough information. Need the radius of the cylindrical hole.

V of a Sphere formula is (4(pi)r^2)/3 where r is radius iirc. The question implies that the radius of this sphere is 3in. 12(pi) is our volume of the sphere. Now we need to subtract the volume of the hole. Which ends up being (pi)hr^2 where r is radius of the base and h is height. Of course, this still won't be the right answer, you need to calculate exactly how much volume the cylinder and sphere would share, and make sure not to subtract the volume that is just the cylinder from the sphere's volume.

>>2035142
Shit. It's r^3 isn't it? Not r^2. Volume of solid sphere should be 36pi in that case.

And incase I didn't explain well enough about that last bit, just realize that the "volume" of the hole will have a 3D base, and not a 2D circular base. If you try to base it off of the 2D base, you'll be subtracting too much volume from the 36pi.

>>2034535
Most intelligent post in this thread, sadly enough.

No radius or diameter given for the "cylindrical hole", you cannot calculate its volume to subtract from the sphere

Also, you cannot consider it a traditional cylinder because a cylinder is flat on both ends. The segment cut from the sphere will be rounded and only have a length of 6 inches directly through the middle. Thus, supposing they did give you the radius of the "cylinder" you would have to calculate its volume and then subtract the material missing from it (i.e. from the rounded ends) before applying that volume as missing from the sphere.

The fact that people are considering inscribed cylinders just shows me that /sci/ is full of armchair scientist retards... go back to studying for that derivative test in calc 1

>>2035178

see

>>2034810 and >>2034798

>>2035126
You're trying too hard. The problem is sufficiently clear. "Drilled through the center of a solid sphere." The word "through" means that the drill comes out the other side. The word "into" would have been used if this wasn't important.

>>2035178

If you excavate a mountain and create a crater where it was, would you consider the space previously occupied by the mountain as part of the hole, or just what is below level?

>>2035178
>armchair scientist retards
Yup. Lrn2math. The answer is 36pi.

dun dun dun....

c>a

x = (c+acosv)cosu
y=(c+acosv)sinu
z=asinv

>>2035232

whereas

since it is undefine it believe it's a ring torus.

>>2035226
The final answer is not 36pi.

36pi only represents the volume of the solid sphere.

Expertly shopped this.

what is the radius of the cylindrical hole..?

>>2035185
>see >>2034810 and >>2034798

My favorite part is the linking of two fallacious posts =]

Firstly, the diameter of the sphere is already given, 6 inches... otherwise the "cylindrical hole" would not be a hole, at the longest parts of the "cylinder" it is 6 inches long, there is your diameter of the sphere.

You just linked two inscribing examples which, despite this problem itself being full of holes, just don't work without first figuring out dimensions of the "cylindrical" object including the rounded ends, which they are preemptively assuming as determined.

Step up your game, douche.

now I just need to figure out Cartesian coordinates really quickly, it's my first time give me a couple minutes.

HEY

DUMBASSES

YOU NEED TO KNOW THE RADIUS OF THE CYLINDER, WHICH YOU DON'T

/THREAD

>>2034524

>ITT: Half of /sci/ is dumbfounded

why it doesn't surprise me?

>>2035240
This is all you really need to understand. A little bit of calculus will tell you what the area of the blue sections is. Don't forget you're applying this to a 3D object, and not just something in 2D like that picture. Calculating just the visual blue area won't be enough.

>>2035248

see

>>2034928

Those are cylindrical holes of 6 inches of depth on differently sized spheres.

Your move.

>>2035240
We know the cylinder is 6" long, that's measured at the edge.

>>2035252
do you?

>>2035252
You don't need to know the radius of the cylinder. Dumbfucks can't into math.

>>2035265
That is true, my mistake. Too lazy to update image now though.

>>2035296
And that makes the answer 36pi.

>>2035260
You just used the same argument again, any inscribed cylinder is a wrong way to go about trying to solve this problem, which cannot be solved anyways. You need to consider not the ends as a cap, but rather an inverted area of the total cylinder.

I attached an image so that you can visualize this, rather simple since it is just alternating the inscribe into a reversed partial inscribe

>>2035308
Samefag here, the image is not to say that the blue section is irrelevant to the total problem, but rather the problem of finding the volume of the curved areas.

Still none of this really matters due to lack of information originally stated

>>2035308

Protip: Circumscribe it.

a=0

R,r,c = 3

V = 2 pi^2 a^2 c

V = 0

so when you drill a hole through a sphere it loses all it's volume and surface area? fuck math is cool

Mr. not grad student

where "r" is the radius of the sphere in inches, and "x" is the radius of the cylinder in inches

volume = ((4/3)(pi)(r^3))-((pi)(x^2)(6))

>>2035338
fuck i think I fucked up hold on

Yellow is the six inch long cylinder drilled through the center of the sphere. Blue is all eliminated in drilling through the sphere. Red is the remaining volume of the sphere, with volume 36pi cubic inches.

>>2035378

/thread.

>>2035378
Why is blue eliminated? This is not just a mere circle, you know. How can you drill through and yet still retain the caps of the tunnel? Are you drilling from the inside?

>>2035408
You're drilling from the top down, through the blue, through the yellow, through the blue, and out the other side.

>>2035408

fuck I had it right the first time and I second guess myself, very stupid of me.

assuming r=0, since the diameter of the whole was never given.

a = c
R = 3
r = 0

2 pi^2 * 3^2 * 3
2 pi^2 * 3^3
2 pi^2 *27
2 (3.14)^2 * 27
2 (9.8596) * 27
19.7192 * 27

V = 532.4022 in^3

V=V(sphere)-V(cylindre)
V=(3/4)π*r_sphere^3-r_cylindre*π^2*(6*0.0245)
in m^2

>>2035426
shit I'm solving this formula

V = 2 pi^2 a^2 c

>>2035427

you forgot to subtract the curved bits above the cylinders.

>>2035416
In your diagram, I noticed that the 6 inch measurement is attached to the diameter of the cylinder. Shouldn't it be the height? If this is not what you intended, then I find your diagram to be misleading. And I still don't see why blue would be eliminated. Are you actually cutting off a large portion of the upper and lower hemispheres?

>>2035445
The 6 inches IS the height. The cylinder has a much greater radius than height.

>>2035445

6 inches is not the width of the cylinder, is the height.

>>2035451
>>2035455
Ah, I see now. Basically, the remaining volume would have a D-shaped cross section.

iit: calculus nerds troll people that don't know calculus.

make more paint peons.

>>2035470

here you go, better visualized.

>>2035481

he whole is to big, it's undefined so it equals 0.

>>2035583
Not sure if troll...

>>2035583

not troll but I'm probably not using the correct terms.

This is my first time seeing doing a problem like this. At first I did the whole cinlder thing but since the diamter of the hole is not defined we can't find the answer. no if we assume that the hole does have a diameter but it is so small we can record it.

So I could be incredibly wrong but then there is this website: http://mathworld.wolfram.com/Torus.html

which describes what a sphere with a hole in it looks like.


Here: http://mathworld.wolfram.com/Torus.html

>>2035740
It's not a torus.

>>2035740
fuck, please explain then?

>>2035762
It's a sphere with a cylindrical hole drilled straight through the middle, so it's going to have "sharp" edges on the circular ring of the hole, as well as being overall shaped like a sphere with the top and bottom cut off. I'd throw something together in Blender but I'm on my laptop, maybe someone else can.

>mfw 120 posts and 10 image replies omitted. Click Reply to view.

>>2035779

Jayzus, it's been solved here, just look.

If there is a single solution, then it must be the case that there is a solution as the radius of the cylinder approaches 0. When it is 0, the radius of the sphere is 3.
volume <span class="math">= \frac{4}{3}\pi r^3[/spoiler]
<span class="math">= \frac{4}{3}\cdot 27\pi[/spoiler]
<span class="math">= 4 \cdot 9\pi[/spoiler]
<span class="math">= 36\pi[/spoiler]

>>2035772
I think one of us has a fundamental miss understanding of how the diameter of the hole effects the shape of a torus.

>>2035858
The shape created in the problem isn't a torus. Tori are irrelevant to the question.

itt: trolls trying to convince people the remaining black area in these two circles is the same

Hint: Find the cross-sectional area at a given height.

>>2035952
Doing it wrong. Length of the cylindrical hole in both those cases is less than 6.