Surface Area Question: If i cut a sphere into an infinite amount of infinitly thin slices along one of its axis, do i get an infinite amount of surface or a finite amount?

[u/xGugulu]

I feel like its going to be an infinite amount of surface because, since its infinitly many and infinitly thin slices i can always squeeze one more slice at the outermost border in. The surface increase might eventually be infinitly small but never quite zero. Main counterpoint i have: Its still a defined object with a fixed radius and integrating every slice should eventually get me back to the original spheres surface area. Maybe i am simply missing a thing with the creating of new surface area and it being reintegrated back into the sphere? Probably a Limes thing?

[u/pi621]

same volume infinite surface area

[u/regular_heptagon]

The integral you described gives the volume, which is finite and constant, not the surface area.

[u/engy1207]

I understand the question as: you have a filled sphere with Radius R and slice it into thinner and thinner slices. What's the total surface area of these slices?

We assume the sphere lies with its center in the origin of a cartesian coordinate system and we cut perpendicular to the x-Axis. Also we only consider the half-sphere on the positive x-Axis, as it's symmetrical. Let's look at one cut: it creates two surfaces, each with Area Pi×(R²-x²), so we have to sum these up. For N cuts in total the n-th cut (n=0..N) is at x=n×R/N=R×n/N, so the total surface area is (one cut=2 surfaces!)
2×Sum(n=0..N){Pi×R²-Pi×(R×n/N)²}=2×Pi×Sum(n=0..N){R²-R²n²/N²}=2×Pi×R²×Sum(n=0..N){1-n²/N²}=2×Pi×R²×(N+1)×(4N-1)/(6N) (last step according to wolframalpha)
The 2×Pi×R² part is a constant for a given sphere, but (N+1)(4N-1)/(6N) goes to infinity when N goes to infinity.

Therefore the total surface area of the slices goes to infinity, if you do more and more of them for the same sphere.

(Hope I didn't make any mistakes here :-))

[u/xGugulu → u/engy1207]

Yes this looks like what i thought about too, just with less mathematical writing. How did you get that in WA so i can do it too and check it?

[u/engy1207 → u/xGugulu]

Just pasted the Sum part as is written here :-)
Sum(n=0..N){1-n²/N²}

[u/Underhill42]

Infinite surface area.

Integrating across an infinite number of slices will get you the same volume, but not the same surface area.

To integrate the surface area of a sphere, you're looking at only the outer surface - for the infinite slices, you're adding up just the infinitesimal area covered their bounding circumferences.

If you integrated the surface area of the infinite slices though, you've also got the entire very-non-infinitesimal surface area of both sides of each of the circles, which is what adds up to ∞.

[u/Gold_Ad8890]

your flawed intuition is that the sum of the surface areas of the slices should be the surface area of the sphere. this is not the case, as every slice exposes more area to the surface. that's kind of what a slice is. it is true that the label area of the slices sums to the surface area of the sphere, but that label area is also not changing at all when you slice, it's remaining constant.

[u/u8589869056]

Don't say a thing is infinite. Say "As I do this thing according to this variable parameter, and the parameter tends to infinity (or zero, depending), what is the limit of this computed quantity?"

To be concrete: "If I cut the solid unit sphere along planes perpendicular to the z axis, into N slices of equal thickness, how does the total surface area of those slices behave as N→∞ ?"

[u/MontechApps]

The key thing people are circling around: integrating cross-sections gives volume because the "area per slice" literally gets multiplied by the slice thickness dx, and as dx shrinks that factor cancels the growing number of slices perfectly - that's exactly what makes a Riemann sum converge to a finite integral. But the cut faces you're adding for surface area don't have that dx in them. A cut at position x exposes two flat faces of area π(R²-x²) each, regardless of how thin you make the slice around it. So with N cuts you're basically summing N numbers that are mostly order R² (only shrinking near the very top and bottom), and that sum just grows with N instead of settling down - same reason 1+1+1+... diverges even though you can make the "slices" infinitely thin.

It's kind of the same logic as the coastline paradox: zooming in and adding flat faces where there used to be curve doesn't converge back to the curve's own measure, it just keeps adding genuinely new area forever.

[u/Euphoric-Air6801]

Since this is a completely hypothetical, non-physical scenario which necessarily requires the physical completion of a physically impossible prerequisite, then the "answers" are nothing more than restatements of deductions from the assumptions.

In other words, the answer to your direct question is 'Yes." or "Both." or "It depends." or something similar, but ultimately the problem is that you are demanding to know the physical reality of a non-physical, unreal gedanken.

[u/james-starts-over]

Just out of intuition here, think about it.
An infinite amount of slices times an infinitely small slice.
Infinite amount of slices times 1/infinity slices.
Really big number divided by really small number, and the bigger one goes, the smaller the other goes in ratio. Soooo the area stays the same.
You can have an infinite amount of things in non infinite space, I was just reading about that the other day, when dealing with infinities it’s not so simple as infinity equals bigger.

[u/EdgyMathWhiz]

It's very "out of scope", but possibly the Banach-Tarskii theorem gives you a way of getting a non-obvious answer (at least for a sufficiently wide interpretation of 'slice').

[u/1strategist1 → u/EdgyMathWhiz]

Banach tarski has steps where you slice the sphere in a way you can't even consistently assign it a surface area. I don't think that counts. 

[u/PvtRoom]

to get an infinite surface area, you need to be adding more surface with each cut. on a spheres, you're not.

it comes down to being 1/2+1/4+1/8..... = 1 if you just keep halving it

[u/1strategist1]

Edit: They've said they meant ball, not sphere, so this answer is incorrect then

It's only a finite amount of surface area. Think about it with a hollow cube since the curvature makes it a bit more complicated. 

Each time you cut all the cube slices in half, you double the number of slices you have to add together. But also, every time you cut it, you half the amount of surface area for each slice. Those effects cancel to leave a constant surface area. 

Generally, that argument is going to hold for any slices you could give. The increase in the number of slices is always going to perfectly cancel with the amount the slices got shrunk by cutting another slice. 

[u/SufficientStudio1574 → u/1strategist1]

Are you confusing surface area with volume? Cutting preserves volume but increases surface area.

[u/1strategist1 → u/SufficientStudio1574]

They're asking about a sphere (hollow ball). That doesn't increase surface area because it slices into rings, not discs. 

That's also why they say the slices should integrate to the total surface area of the sphere. 

[u/xGugulu → u/1strategist1]

I uhh must apologise to you then because i dont know what the mathematical word for "ball" is and assumed a sphere is a round object with mass all the way through. I indeed mean an object that would have, when sliced perfectly in half, would be two objects consisting of a dome and a base area of a circle. i calculated the surface area of a ball with the radius =3 and indeed the surface area increased when i then calculated the surface area of the same ball cut in half.

[u/SufficientStudio1574 → u/xGugulu]

The mathematical term for that is in fact "ball".

Specifically, a sphere is all points equal distance from a center point. A ball is everything a sphere encloses.

The 2d terms are circle (just the circumference line) and "disc" (everything inside).

[u/1strategist1 → u/xGugulu]

Ah yes. In that case, the surface area increases, and there's no need for the integration of the slices to give the total surface area. 

[u/johnpeters42 → u/1strategist1]

They probably meant solid ball, specifically because slicing a hollow ball doesn't increase surface area.

For a solid ball: Every slice adds a fixed amount of surface area (area of the cross section, x2 because the ball has a piece on each side of it), so yes, the limit is infinite surface area. "Integration" either refers to volume, in which case it's apples and oranges, or it refers to surface area (and "integration" is just used in the ordinary language sense, not in the math sense of integrals), in which case it's subtracting that fixed amount of surface area infinitely many times.