Why are the formulas for the sphere so weird? (major upgrade of Archimedes' greatest discoveries)
In today’s video we’ll make a little bit of mathematical history. I'll tell you about a major upgrade of one of Archimedes' greatest discoveries about the good old sphere that so far only a handful of mathematicians know about.
00:00 Intro to the baggage carousel
01:04 Archimedes baggage carousel
04:26 Inside-out animations
04:59 Inside-out discussion
10:38 Inside-out paraboloid
12:43 Ratio 3:2
13:28 Volume to area
18:40 Archimedes' claw
20:55 Unfolding the Earth
29:43 Lotus animation
30:38 Thanks!
Those fancy conveyor belts are called a crescent pallet conveyors, and sometimes "sushi conveyors" because they were originally designed for carrying sushi plates. en.wikipedia.org/wiki/Conveyor_belt_sushi Andrew also dug up an American patent dating back to 1925 patents.google.com/patent/US1757652A/en
Great wiki page on Archimedes
en.wikipedia.org/wiki/Archimedes
In "On the sphere and the cylinder"
en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
Archimedes derives the volume and area formulas for the sphere. The proofs used in this work are quite complicated and conform to what was acceptable according to Greek mathematics at the time. His original original ingenious proof most likely involved calculus type arguments. Marty and I wrote about this here www.qedcat.com/archive_cleaned/99.html and here www.qedcat.com/archive_cleaned/100.html Also check out this page en.wikipedia.org/wiki/The_Method_of_Mechanical_Theorems
Why is the formula for the surface area the derivative of the volume formula? Easy:
V'(r) = dV/dr = A(r) dr / dr = A(r).
A nice discussion of the onion proof on this page I'd say check out the discussion of the onion proof on this page en.wikipedia.org/wiki/Area_of_a_circle
B.t.w. this works in all dimensions the derivative of the nD volume formula is the nD "area" formula. en.wikipedia.org/wiki/Volume_of_an_n-ball
Wiki page on Cavalieri's principle
en.wikipedia.org/wiki/Cavalieri%27s_principle
Includes both hemisphere = cylinder - cone and paraboloid = cylinder - paraboloid
Video on the volume of the paraboloid using Cavalieri by Mathemaniac
kzhead.info/sun/iJaYntqejnt3jH0/bejne.html
Henry Segerman: en.wikipedia.org/wiki/Henry_Segerman
Henry's video about his 3d printed Archimedes claw: kzhead.info/sun/fqiYmLamlmtthYE/bejne.html
Henry's 3d printing files: www.printables.com/model/651714
Andrew Kepert: www.newcastle.edu.au/profile/andrew-kepert
Andrew's playlist of spectacular video clips complementing this Mathologer video:
kzhead.info/channel/PL9JP5WCX_XJY9GmMO-kotRR5bRYOPOtn9.html
All of Andrew's animations featured in this video plus a few more (actual footage of a fancy baggage carousel in action, alternative proof that we are really dealing with a cylinder minus a cone, paraboloid inside-out action, inside-out circle to prove the relationship between the area and circumference of the circle, etc.)
There is one thing (among quite a few) that I decided to gloss over at the end of the video but which is worth noting here. At the end it’s not straight Cavalieri. Before you apply Cavalieri, you also need to put some extra thought into figuring out why the flat moon that runs along the semicircular meridian can be straightened out into something that has the same area (straighten meridian spine with interval fishbones at right angles). Here I was tempted to include a challenge for people to figure out why the red and blue surfaces in the attached screenshot have the same area: www.qedcat.com/ring.jpg
Funniest comment: Historians attempting to reconstruct the Claw of Archimedes have long debated how the weapon actually worked. The sources seem to have trouble describing exactly what it did, and now we know why. Turns out it was a giant disc that slid beneath the waters of a Roman ship, then raised countless eldritch crescents which inexplicably twisted into a sphere, entrapping the vessel before dragging it under the waves, all while NEVER LEAVING ANY GAPS in the entire process. No escape, no survivors, fucking terrifying. No wonder that Roman soldier killed Archimedes in the end, against the Consul's orders. Gods know what other WMDs this man would unleash on the battlefield if he were allowed to draw even one more circle in the sand. The Roman marines probably had enough PTSD from circles.
T-shirt: One of my own ones from a couple of years ago.
Music: Taiyo (Sun) by Ian Post
Enjoy!
Burkard
P.S.: Thanks you Sharyn, Cam, Tilly, and Tom for your last minute field-testing.
Archimides was doing calculus without algebra. No wonder he's your favourite. That's pretty much the spirit of this channel if you think about it.
Exactly :)
Don’t…..blow my mind like that
There was no calculus ... it's just a manifest coincidence that the algebra makes it look so!!!!
But... if Pyramids were built by aliens I am sure either he was one of the aliens or had access to a super computer :D
Calculus is a happy accident we find along the way.
1:36 "circle gets turned inside out" **"smiles and frowns" flashbacks intensify**
That video lives rent-free in my brain.
Different turning inside out ;)
Is this it? Is this the hemisphere turned inside out? That wasn't easy to follow, was it?
@@stapler942 Ah, yes I remember that :)
HAHAHA
Thanks Burkard for putting together such a nice presentation and filling in so many connections. It has been a load of fun working with you and Henry on this.
KZhead tells me that you've already been subscribed for 5 years. This project was definitely a lot of fun and probably my favourite this year :)
Hmm. Now I’ll have to figure out another bit of fanciful geometry….
@@andrewkepert923 I can think of a couple of other projects if you are keen :)
@@Mathologer uh-oh. reverse nerd-snipe.
@@andrewkepert923😂
I remember when I first noticed the derivative relationship between volume, surface area, and circumference of the sphere. It’s been over a decade since my first calculus course but it’s still so satisfying. When you consider that the derivative is a rate of change, the relationship begins to make perfect sense. I think what really made it click was when we started doing integrals. If you integrate the circumference from r=0 to r=R, you wind up with a sweep of concentric rings that cover the whole area, meaning that the rate of change of the area is given as the circumference as r is varied. Similarly, as you integrate the surface shells across the range of r- values, you wind up with the whole sphere! It’s brilliant once it all clicks and you can see those animations in the mind 😊
And that derivative relation works in every number of dimensions; so yes, it's no coincidence. Fred
This is discussed at length in the Tau Manifesto, as part of their argument for why A=pi*r^2 is less "natural" than A=(tau/2)*r^2.
@@TheShadowOfMars I love that. At first the 1/2 looks ugly, but when interpreted as the byproduct of the power rule it looks natural. Reminds me of kinetic energy E = (1/2)mv^2 which is the integral of momentum p=mv
@@emuccino the kinetic energy form meaning (AB^2) /2 is ubiquitous in both maths and physics and is therefore a powerful argument for Tau instead of Pi to be the base circle constant.
Congrats on your 100th video! Very special number, since it's the fourth octadecagonal number, amongst other things 😆
If you make a PDF I will buy it, and probably I will not be the only one. This really associates the formal math and the intuition in a striking way. Well designed, brilliant accomplishment.
Well, definitely check out Andrew's extra material linked in from the description of this video :)
Awesome video (im a 33m into the past time traveler)
To the person above, alright.
@@DontReadMyProfilePicture.273I don't see your profile picture. I think it's for the same reason I don't see ads
Congratulations on 100 videos! Your channel is awesome ❤
Thank you so much!!
Hey, I won gold because of this channel (long story lol), and have a suggestion for a small addition to the part two. The correspondence between the map of Lambert and Kepert is done by taking circle segments between two points, and varying the "angle" of the circle segment. 0° is a line segment, and 90° is a halfcircle, like used in the video. One of my own proofs, of the cylic quadrilateral angle theorem (that has undoubtedly been found by someone else as well) is that given a quadrilateral, you can look at the line segments like they are circle segments. The "circle segment quadrilaterals" have invariant α-β+γ-δ. Since you can merge two pairs of circle segments, you basically directly get the theorem. This even works for hyperbolic geometry! It is a nice proof, using unorthodox techniques, so it probably has to show up on this channel eventually, although it likely won't fit. It also has a dual theorem, where the opposing sides of a quadrilateral sum to the same length if and only if the quadrilateral has an inscribed circle.
It's actually really enlightening to see the reason why exactly Lambert's and Andrew's maps do the same in terms of latitudes (knowing that one is area-preserving then implies that the other is too). Also, really nice proof for the cyclic quadrilateral theorem :) If you've written this up, would you mind sharing this with me to be included into my to-do folder :) burkard.polster@monash.edu
This is exactly the idea behind circle inversion, you may be interested in that Pole and polar reciprocity is very very very cool tlo
The real key here is that we now have a transformation in which all the CURVED longitudes in 3-space map to a single flat surface with the same length - no distortion. The curved longitude really is a flat 2-d curve if you just look at it from the right perspective but now ALL of them map to the same plane! Then the area stuff follows by INTEGRATING over closer and closer longitudes. For me, this elusive mapping is a real game changer. I mean it really is. Curved arcs on the sphere are now lying on the same flat plane with no distortion from which it follows (by integration) that curved area of sphere = flat area of circle (like adding up the arc lengths of an infinite number of undistorted semi circular longitudes to get the area). With radius of 2R to boot! No stretching or squeezing. Well done and thank you ALL so, so much! This has resolved a long time major conflict in my own mind trying to understand what curved area really means and how we can transform a curved area into an equal flat area with no distortion - which what area really is defined as all along - flat.
I'm amazed that the paraboloid & the parabola were even known and studied that long ago. How did they define it without coordinate geometry?
A parabola is one of the conical sections. People have been obsessing about these curves for a long, long time :) en.wikipedia.org/wiki/Conic_section
I'm not a math historian but I believe that it is because the Greeks studied sections of the cone. If you cut the cone parallel to the base, you get a circle. If you cut the cone oblique to the base, you get an ellipse as long as the cut is not parallel with the sides. If you cut the cone parallel to the side, you get a parabola. If you cut it more oblique than the sides, you get a hyperbola. en.wikipedia.org/wiki/Apollonius_of_Perga
...and, if you cut it straight through the vertex perpendicular to the base, you get an isosceles triangle. 😉
@praveenb9048 You have it backwards. Even today the definition of the parabola is the conic section you get when you cut the cone parallel to a side. A defining property of the parabola is that all lines parallel to the axis of symmetry of the parabola cross at a certain point when they're reflected on the parabola. Ancient Greeks knew this too (as it's a purely geometric property). The fact that second degree polynomials' graph is a parabola is not a definition. You have to prove it using the defining property above.
Take a tank of water and spin it as a whole; the surface becomes a paraboloid.
hey Congratulations on your 100th video!~ given your number of subscribers, it just reflects on the quality per video. thank you ❤
As always sir, I appreciate the free Educational Videos. You are keeping the love for Numbers alive.
It makes sense that the area is the derivative of the volume, if you think that the volume is created by adding all the surfaces of the spheres with smaller radiuses. Basically like blowing up a perfectly spherical balloon. That trick should work for any shape, not only spheres. For example a cube defined by the 8 points (+/-r,+/-r,+/-r) has a volume of 8r^3. If you drive that by r, you get 24r^2 and that is exactly the surface of such a cube with side length 2r.
Yes - I like to think of it as adding a new layer of paint, and adding up (integrating) all the layers. For some curves/surfaces the tricky part is making sure the new “layer of paint” is the same thickness at all points. Measured normal to the surface, that is. It can be a challenge for some curves (surfaces) where the offset curve (…) is a different type of curve to the original, such as for a parabola.
It works *if* the surface "moves outward at the same speed" everywhere when the parameter is increased. So the perpendicular thickness of the shell (the dV) is equal *everywhere* (the dr). For instance, the area of an ellipsoid is NOT the derivative of its volume (for common parametrizations). Even simpler, it doesn't work for non-regular polyhedra. Like, the volume of a pyramid with square base with side x as well as height x equals ⅓x³, but its area is (1+√5)x²...
One of my favorite channels on KZhead. Congratulations to Burkard and his team. Ausgezeichnete Arbeit! I always look forward to your new videos.
Congratulations Mr. Polster (and Marty) for your 100th video! The more I watch, the more I ❤ it.
That's great :)
I now understand why the edges of 3d shapes of constant width look just like the animation at 27 minutes! It's a sphere being mapped between the vertices, so elated! Thank you, Mathologer, for another wonder full lesson!
8:00 Just filling in the details of this proof because i havent seen any other commenter do so yet: (unless i missed them, idk) Set R = outside radius of sphere= outside radius of cylinder. h = height of our cut plane r = radius of the circle x-section And a = inside radius of ring. Area of the circle cross section: A1 = pi*r^2 = pi(R^2 - h^2) via the Pythagoran theorem. Area of the ring: A2 = pi (R^2 - a^2) A1 will equal A2 if we set a = h, so then the subtracted cone will have straight sides with rise equal to run, and therefore the cone is actually a cone.
Thank you this helped :)
My favourite Mathologer video thus far. Props to Archimedes et al.
That's great :)
Congratulations on hitting 100 videos! That’s quite a milestone! Love your content, Mathologer!
Thank you very much :)
I've been following your channel for years and have always loved the way you've explained math. I get excited every time I see a new video from you. Congrats on the 100th video milestone!
Thank you very much :)
Happy Anniversary Mathologer! Thanks for you've done to educate us.
Happy 100th. Again I love you videos and always watch them Sunday afternoon. Relaxing.
Glad you like them!
Gratulation zum 100. Video! Alles ist wirklich ein Hochgenuss, einfach perfekt, vielen Dank! 👍
Vielen Dank!
Congratulations on 100 videos. Your videos are impressive, to say the least.
Thank you very much :)
I remember learning that the volume of a sphere equals the volume of the respective cylinder minus that of the cone back in school, but it completely blew my mind to learn the equivalent is also true for a parabola. Also, it's amazing how the derivative of the volume of a shpere is its surface area. Made me realize how I've been taking things for granted withhout actually analyzing them. Thank you for providing me with all this insight! Keep making great videos like this!!
I'd expect a LOT of people watching this video to feel the same :)
Congrats on 100 videos! Here's to many more!🎉
Well, as long as enough people keep watching I'll keep making these videos :)
yes, very special! I wondered about the Chinese (or Japanese, as it turns out) flavoured music, but I guess it was suggested by the sinuous line of the baggage carousel. I was really skeptical about the preservation of the area through the meridional lay-down, but the travelling circle argument convinced me. So much to think about in this video!
Great video. I'm new to this channel I gotta say I am blown away how you're able to talk about the intricacies of mathematics for a half hour and at the end I was disappointed that it was over.
Welcome aboard!
Gratz on 100 videos 🎉 Thank you ❤
Thanks for being a subscriber for seven years :)
So many thanks to you mathologer for your tireless work. we all salute you.
Thanks you :)
Archimedes' claw, seriously? What a shame not to name it Archimedes' pumpkin 🥲 Or Archimedes' Kabocha to be even more accurate.
In Andrew’s playlist (link in description) he actually shows a physical model in orange, which looks like a pumpkin.
Well, unless it's a possessed pumpkin it doesn't do much clawing :)
It is a riff on "The Claw of Archimedes", a super weapon created by Archimedes, also called the "Ship Shaker". Look it up.
Archimedes' bunch of bananas 😀
Archimedes' SUN LAZER
Brought my jacket and a tie for the grand premiere :) Thank You for the great format of Your videos. They help me stay sharp long time after university studies.
That's great :)
At 27:20 I was expecting another _"No, we're not quite at a proof yet,"_ and was very surprised that we didn't get it. It's not enough to note that the areas of the red and blue skeleton of the moons tend to each other as the number of slices go to infinity. They also have to tend to each other *fast enough*, because as you add more slices you also add up more differences. This is not a trivial thing to check, in fact, checking this rigorously is pretty much the impetuous for defining calculus formally!
You are absolutely right there. In fact, originally I had a couple more "not so fast"s at the end of the video but then ended up cutting a lot of it out :)
FWIW I think it’s nice that we can get so close to a proof without calculus. The original motivation wasn’t to prove but to visualise - Grant Sanderson put out a challenge and I had a go at it. Any complete proof along these lines requires a lot of baggage* such as properties of cyclides or inversion in the sphere. By the time you have all of that as prerequisites you may as well start using coordinates, trig and either some calculus or pre-calculus ideas such as small angle approximations. Then with this toolkit there are much better proofs that skip the cyclide construction. Anyway, see my supplementary playlist if you care for some more background. * and a baggage carousel to carry it
I love your content, hope you have a nice day!
I'm mathematically challenged, but this was one of the most entertaining videos I've seen in a while! Job well done! 👍
Glad you enjoyed it!
Congratulations on 100 episodes! May you make many more ...
That's the plan!
What’s amazing is how this represents the three aether modalities. Dielectricity, magnetism and electricity.
100th video, first collab, and an obscure topic?! We're eating good today! Great video, man.
Extra, extra special :)
If only we’d reached out to Vi Hart we could have timed it to match her recent croissant video. Too much cyclide-shaped food is never enough.
Congrats on 100 videos, mate. You really made it as a maths educator and content creator on KZhead, and I'm looking forward to seeing you do even better in the future. Hope you blow our minds again with each video you make That said... CHALLENGES! 11:18 I have no army of middle school minions but I am still ready to attack Same reasoning as before. This time we start with the second paraboloid, the one carved from the cylinder. Makes the maths a little easier. The paraboloid has radius R and height H. Cutting it at height h will leave a ring with outer radius R and inner radius r. The paraboloid is modeled after a parabola y=ax^2, and so we should have H = aR^2 and h = ar^2. So it's possible to solve for r and get r = Rsqrt(h/H). The ring thus has area pi * R^2 - pi * R^2 * h/H, or piR^2(1-h/H). The first paraboloid should also have that area. Thus its radius should be Rsqrt(1-h/H). Now the inverted paraboloid can be modeled by another quadratic, but the important takeaways are H = bR^2 and H - h = br^2. Solving for r this time gives Rsqrt(1-h/H), exactly the same as what was predicted by the circles area argument. Or you can use integrals. Whatever floats your boat 17:42 The layers of the onion look almost like surface areas stuck together. That can be written as: V(R) = the integral from 0 to R of SA(r)dr By FTC1, we can also write this as V'(R) = SA(R) And so the derivative of the volume is the surface area. Even in 420 dimensions. 18:04 The base of the cylinder has area piR^2. The height is 2R, and the circumference is 2piR. In total, the surface area of the cylinder is 6piR. With the surface area of the sphere being 4piR, the ratio of the surface areas of the two shapes really is 3:2.
Very good :) Full marks.
You know what would i love? A series of videos where you take the greatest mathematicians in the history and show us some of their contributions in a brief but in that marhologer level of explainability. I love to hear about the mathematical advances that the great minds performed and How these impacted maths and the humanity in general.
Congrats on the centenary! Cant wait for the next 100.
15:15 Hahaha...My kind of joke! Congrats on the 100th video...and thanks for the amazing content.
Thanks :)
Christmas has come early. What a mathematical gem!
This is the ultimate introduction to Maps Projections.
@@DontReadMyProfilePicture.273 What a convoluted way to say "ignore me". 😀
Already seen Henry Segerman's video and left a comment there. :) Fantastic to see this explained in great detail and crossing my fingers for that part two. I really like the map projection. Instead of Antarctica, you can take your home location as the centre of the map and your homeland will then be the least distorted country on the map.
But have you seen Vi Hart’s video on edible cyclides? (This one *wasn’t* a collab - just a coincidence.)
@@andrewkepert923 I have seen the latest Crescent Rolls one a couple of days ago, yes. :) Not sure if that's the edible cyclides one you mean though...?
@@willemvandebeek yes - the banana / croissant shapes are formally known as “horn cyclides”
@@andrewkepert923 kk, Vi Hart also made other edible math videos with cyclide shaped food, like string beans, hence my confusion. :)
Fascinating. Congratulations ❤
Glad you enjoyed this one :)
Just yesterday I was memorizing the formulas for the volumes of 3D objects for a quiz on centroids for my Statics subject and I thought about how fascinating it was that the volume of a hemisphere, paraboloid, and cone of similar dimensions turn out to be 2/3, 1/2, and 1/3, respectively, of the volume of the cylinder that you can exactly fit all of them in, though I never really understood why until now, just a day after the start of my query. Thanks so much! This has got to be one of my most favorite KZhead videos of all time now for explaining simply such a seemingly complicated topic.
That's great :)
Congratulations on 100 videos! Love your work ❤
Thanks you very much :)
What an amazing video. Also sending a salute to Andrew for the great ideas ;)
Thank you :)
salute gratefully acknowledged.
It's always lovely to watch people actually excited about what they're talking about, instead of just smiling for the camera.
Congratulations on 100 videos!
Thanks :)
🎖🎖🎖🎖🎖🎖🎖 incredible again!!! Those transformations and metamorphoseses... always open a totally new view on a 'known' issue.
You know I've always wanted to see a cone turned insideout ever since I was a kid. Impressive as always! I thought when I was a kid that the animation shown would have made some kind of other cone, but I guess its a sphere
Yes, quite surprising and beautiful, isn't it?
This is a bit bizarre. I was just reading about this in your QED book, a few days ago.😮 Congratulations on 100 videos!
People are still reading QED, good to know :) Also, good to know that people still read :) :)
Congratulations for the 100 video. 100 more will come. Thanks for all your insights.
Thank you! "100 more will come." Fingers crossed. There is certainly no shortage of great topics :)
*Wow, just WOW!* Especially the final animation, The Lotus!!
Your enthusiasm for maths is really infectious 😊
Glad you think so!
Congrats on 100!
Thank you very much :)
Congrats & thanks for 100 videos! 100 videos with not a 'bad apple' among them! Say, the "baggage carousel" transformation of the hemisphere to cylinder-minus-cone, reminded me of a method I saw in a book about 60 years ago, for finding the volume of the sphere (well, technically, the 3-ball), which I believe was due to Cavalieri. At 2:36 - 2:49 & beyond, in your "hemisphere = cylinder - cone" graphic, Cavalieri, using his eponymous theorem, had doubled both figures by reflection in a plane containing both bases, resulting in a complete sphere and a cylinder with twice the height of yours, minus a two-branch cone with its vertex at the center of the (now taller) cylinder. Next he passed a plane parallel to the base of the cylinder, from top to bottom of both solids, cutting the sphere in a circular disk, and the (cylinder - cone) figure in a circular annulus, with constant outer circle and continuously varying inner circle. It is then a simple matter to verify that the disk and annulus are always equal in area, and he concludes (by his theorem) that the volumes of the solids are equal. Resulting in the now-familiar formula for the volume of a sphere. Having now gone further into your video, I see you bring Cavalieri into this at 8:24. Bravo! Fred
I'd say keep watching :)
@@Mathologer I always do, because you never get boring.
26:48 - Cavalieri works for parallel slices, but those slices you depicted aren't parallel, even in the limit. I think there's a small gap in the proof here.
Well, spotted. This is one of the things (among quite a few) that I decided to gloss over at the end of the video but which is worth noting here. At the end it’s not straight Cavalieri. Before you apply Cavalieri, you also need to put some extra thought into figuring out why the flat moon that runs along the semicircular meridian can be straightened out into something that has the same area (straighten meridian spine with interval fishbones at right angles). Here I was tempted to include a challenge for people to figure out why the red and blue surfaces in the attached screenshot have the same area: www.qedcat.com/ring.jpg
Very kind of you to call it a proof. 😆. There are ways to make it more rigorous, the simplest involving calculus. But that’s not the intention here. It was to get some understanding / feel / intuition of why the lune on the sphere (between meridians) matches the lune on the plane. When folded down, it’s in the wrong direction, which is disappointing. What was needed was something that linked the two areas, and the “channel surface” property of cyclides does this. I say a bit more on this in my playlist.
@@Mathologer Oh! I know a general principle (also ancient) that can prove that challenge! en.wikipedia.org/wiki/Pappus%27s_centroid_theorem
@@teolopezpuccio739how did i not know that theorem, wow!
@@theo7371 If I'm understandung you correctly this isnt quite right, its because integrating along a curved line is the same as integrating along a straight line, but only as long as you extend equal directions inside and outside the curve. For example, the volume of a torus is the same of the volume of the cylinder you get by "unwrapping" it, the inside of the torus is squished but the outside is stretched and the two effects cancel out exactly.
This is a really great presentation, but at every point I can think of an architectural example from history built and still existing today which demonstrates the points of the presentation. For example think of Bernini’s Baldacchino at St. Peter’s. The twisted ‘rope’ columns have rational and calculable areas and volumes. This was built a few years before the mathematics presented by Cavalieri. Imagine what the ‘3D printer’ used to build it was like, - 400 years ago. 😊
Very nice! It seems to me that, by putting these things together, Mathologer has just added one more proof of Pythagoras' theorem to the long list: Because can use Pythagoras' theorem to prove that Cavalieri's areas cut through the sphere and the cylinder-minus-cone are the same, you should be able to use the conveyer belt alternative prove with the same result to backwards proof Pythagoras' theorem.
Good idea, that should work :)
I really enjoyed that. As a follow-up, could I suggest that you become the Maptologer for one video only to reveal the Math behind various map projections and the various possible compromises when moving from a sphere to a subset of the plane?
Sort of on my to-do list (as many, many other topic :)
Great work 👍
1:22 I never realized that's why the belts had that shape. Never saw one up close. this projection of the sphere is so much better than the one at the start of the lotus animation.
Really love this video, and Archimedes is my man as well!!
As I said, link can be found in the description of this video :)
You, Professor, are a brilliant orator. Thank you for sharing your gift of training the mind to better think.
Amazing "upgrade" of the classic relationship. I also liked the OpenAI logo that references this cone/sphere/cylinder relationship and includes the following shape-to-letter mappings: A -> cone G -> sphere I -> cylinder This sequence references AGI, an acronym for artificial general intelligence.
I think the transformation of the hollow space within the claw's sphere-like configuration would be interesting to see in animation too. Just off the top of my head, I believe it should correspond to the divets between the lunes in the disc-like configuration, but I've no solid proof for that right now
Actually, I also have a version of Henry's claw which only features half of the moons. Really nice to see how exactly both the inside and outside of the claw transform :)
7:20 Of course; assuming that the missing circle of negative space has equal area to the circular cross-section of the hemisphere (c); in order for the ring and the small circle (c) to have the same area, the large circle (C), that is, the ring plus the circle missing from it, must have exactly twice the area of the small circle (c). This amounts to the radius (R) of the large circle (C) being exactly √2 times the radius (r) of the small circle: R = √2r -> A(C) = (π(√2r)² / πr²) * A(c) = (√2)² * A(c) = 2A(c). Then, removing A(c) from A(C) leaves: A(C) - A(c) = 2A(c) - A(c) = (2-1)A(c) = A(c); and the ring and the (cross-section) circle have the exact same area.
Congratulatios, 100 !!! 👏
Thanks !
Thank you for your videos sir, very relaxing, interesting and informative, this channel is one of the reason i'm a math major
That's great :)
Another master class video from mathologer❤. Sir, recently I came across magic squares. They were fascinating. I watched several videos on KZhead on them. All of them were just about tricks to make magic squares. There was no explanation behind them about how and why these tricks work. I also checked the internet but without much success. I request you to make a detailed video on magic squares and math behind them.
Have you seen this Mathologer video on magic squares yet? kzhead.info/sun/eaWHktKbi395eps/bejne.html
It makes total sense that the derivative of the area of a circle with respect to radius is the circumference of a circle, using "onion reasoning". If you took a circle of radius r and increased its radius infinitesimally, you would basically be adding a "line" of area to that circle, which would have a length of the circumference of that circle. Do this many times and you will find that the area of a circle must be equal to the antiderivative of the circumference of a circle.
Yep, that's pretty much it :)
Grats on 100 ^_^
Finally enough I tutor second semester calculus and kinda intuited something similar to this about two months ago when I helped someone do an integral similar to the cone cylinder one.
Great video!! Would love to see a video on fractal dimensions (i.e. Hausdorff and Packing dimension) with some mathologer animations. The tubes arguments I see in papers are a bit dull on the page at times.
On my to-do list :)
Logically, the volume of a cone must be the volume of a cylinder of the same size minus a hemisphere. Very inspiring video!
Yes! Well deduced. If you look in my playlist, down the bottom there is exactly this animation. Challenge for everyone else is to find other shapes where the lune mapping turns it into something recognisable.
Did you also know that you can unfold a sphere's surface area into the area under a sine wave with important values that are directly related to the important values of a sphere? The amplitude should equal the circumference of the great circle of the sphere, and the period should be the diametre.
this was one of the things I had in mind before I found the cyclide construction. There is a traditional map projection called "interrupted sinusoidal" that decomposes the sphere's surface into "wedges" that are then flattened out into lens shapes. These are usually displayed side-by-side, although if you put them radially you get the "flower" configuration that features at the end of this youtube video. Anyway, the "sinusoidal" part of this comes from the same place as your observation.
The moon argument is one of the most beautiful proofs I've ever seen
Excellent! I love it.
Glad you like it!
16:41 That’s a great way to, in a simple way, illustrate Green’s theorem in 1,2 and 3 D. 😊
You should really elaborate on this comment a bit :)
@@Mathologer You could also see this way of looking at Green’s as an illustration of the fundamental theorem of calculus: the value of the integral over an “interval/area/volume” is equal to the value of the derivative on the boundary points/line/surface. Makes it easier to remember stuff in multivariable/vector calculus. 😊 Now, I’m no longer an instructor at any university, but the few students I tried this on sitting in the student coffee shop ten years ago seemed to like the analogy. 😃
@@MathologerOh, sorry, my original comment made no sense without the time stamp. I thought I put one in before. Fixing that right now. 👍🏻
@@martinnyberg71 Your coffee-shop discussion is a manifestation of a deeper truth: the Stokes-Cartin theorem. This is a theorem that generalises the fundamental theorem of Calculus, Green's theorem, Stokes theorem, Gauss's theorem and a lot more in one simple equation. Sometimes mathematics gives us results of indescribable beauty. en.wikipedia.org/wiki/Generalized_Stokes_theorem
Congratulations on #100!
Thank you :)
The ‘inverted’ sphere is the void space of a cylinder of heigh r, where the increasing radii of the sphere from pole to equator are the cross sections of the cone/void cut from the cylinder. It’s like an anti-integral. Remove from a solid cylinder whose radius is r, the shape of changing radii (0 to r) integrated across the height equal to the radius. I could go one further and fully derive the equations of geometric cosmology (A=pi(r)2 is to e=mc2 as V=(4/3) pi(r)3 is to … 😉), but you’re on the right track, and I enjoy watching these and don’t want to ruin the surprise. If this blew your mind, just wait.
Burkard's little giggles are do endearing.
I would love to see a Mathologer video that detailed both of Archimedes proofs of the area underneath a parabola. What he did using the Law of the Lever as a primitive form of algebra was brilliant. I also really like Galileo's intuition of how he actually used both the Law of the Lever and the Law of Buoyancy when calculating the volume of an irregularly-shaped crown during his eureka-moment when he was running down the streets naked.
I have a theory that we could make math more interesting for students if we taught it in chronological order of discovery so we could travel down the same paths as our intellectual forefathers. I would be ready to take it even farther and have them doing abacus math with Roman numerals so they could see the brilliance of Fibonacci's contribution to the world of math by bringing the Hindu-Arabic numeral system to Europe. Everyone knows his name because of the rabbit problem but his contribution of a numeric system with a built-in abacus was far more important.
I have often thought about a chronological math path as well!!! In fact, that is how I taught decimals and fractions to younger students- started with ancient man being happy with whole numbers until bartering started happening, and then needing parts of a whole. The human brain responds well to story!!!
I used to teach Archimedes proofs of the area underneath a parabola in a math course for liberal arts students. A couple more Archimedes themed topics are on my to-do list :)
There are some brilliant History of math textbooks that I've used in the past to do something like what you have in mind here. One that I like in particular is John Stillwell's book Mathematics and its history: www.amazon.com.au/Mathematics-Its-History-John-Stillwell/dp/144196052X
An even deeper dive on this would be great
Wow! 100 videos and 8 years. When I first found your channel I was a high school student learning calculus and now I’m a graduate student studying statistics and coding for public policy. Your passion for math helped me find wonder in what was being taught in class and put me on the path for success that I’m on today!
That's great :)
Numero uno della divulgazione matematica mondiale ❤ congratulation
Glad you that you think so :)
Mathologer, I have a book titled MATHEMATICAL MASTERPIECES, published by Springer Verlag, where you could find the necessary demonstration why Archimedes is Euler's dad or grand dad, where you could discover many more of his masterpieces. I want you to find out the source of these masterpieces. I think some of the Indian mathematicians or some unknown Ptolemaic mathematician may be the culprit. However, your friends Andrew and Henry made my day along with you. Thank you.
11:22 Working out the width of the cross section of such a paraboloid at a given height h yields two times sqrt(t - h) where t is the total height of the paraboloid. The corresponding cross section of the cylinder minus something has the width 2 times (sqrt(t) - x) , where x is again the radius of the inner circle we don't know yet. The respective areas are pi times sqrt(t - h)^2 and pi times sqrt(t)^2) minus pi times x^2 . Simplifying: pi times (t - h) equals pi times (t - x^2) , i. e. h = x^2. In other words, the sides of the part cut out of the cylinder are another paraboloid, and since it is described by the same parabola as the paraboloid with which we started, its volume is also the same.
Great :)
Thanks Burkhardt! Eve & I were super thrilled to see this episode. We like watching you & 3b1br. I had mentioned to her all of this, with appropriate references to mandarines, and as she doesn't know Andrew she was amazed that he would do that for her! Afterwards she was a bit surprised, I think, that mathematicians get paid for such fun! 😂 . Keep it up please!
That's great. Thank you for encouraging Andrew to get serious about making his discoveries known to the world. And, yes, we are being paid to have fun :)
So so good!
Glad you liked it :)
Nice!
This video gave me an early Christmas gift. I don't want any gift for the rest of the year.
That was pretty much my response when Andrew told me about all this :)
@@Mathologer your friend andrew is a genius
I have to watch Toroflux again ... and also think about the Archimedes wheel paradox. (Escher would have appreciated this, too.) Yes, this raises all sorts of inquiries.
As I said, more than enough material for a part 2 :)
Holy wow, that was beautiful!!
Congratulations for your 100th video. Please go through the works of Aryabhata I 5th century CE.
Sort of on my to-do list :)
@@Mathologer Very thanks to you to consider my request.
Fun :) its also fun to see the difference in relation to poisson ratio of some incompressible solid that can deform in any given direction like a fluid. Its quite intuetive, if you unfold a sphere into a circle just by bringing down all the meridians you will get a larger area, but the difference between the circles of latitude on the sphere and the circles of radius r that corresponds, gets you a number for how much you should retract the distance between each circle of some radius on the disk to be area preserving. I wonder whether the angle of the meridian turned at 90 degrees onto the disk to the radius of the disk says something important ;).
That's a very interesting thought. Will have to ponder this and the Poisson ratio in general :)