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.
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.
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 😊
@@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.
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!
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!!
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...
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!
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.
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 RUclips videos of all time now for explaining simply such a seemingly complicated topic.
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
Congrats on 100 videos, mate. You really made it as a maths educator and content creator on RUclips, 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.
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.
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.
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.
@@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...?
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.
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. 😊
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
@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.
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!
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
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.
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.
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?
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
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 :)
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 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
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
24:00 Well; a radius (R), of that target disk, is exactly as long, as the diameter of the sphere (and its shadow disk); so, twice the radius (r) of 1 of those; then, that 2:1 -ratio also applies to the diameter and the circumference; so, all the 1D-quantities. To get the ratio of the areas (2D-quantities), we need to scale by a factor of 2, in 2 dimensions; for a total ratio, of 2²:1² = 4:1. Thus, the target disk has 4 times the area of the shadow disk. Since the unfolding of the sphere into the target disk is area-preserving, the area of the sphere must be exactly 4 times the area of its shadow disk.
Another master class video from mathologer❤. Sir, recently I came across magic squares. They were fascinating. I watched several videos on RUclips 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.
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.
@@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. 😃
@@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
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.
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 ;).
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.
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.
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 :)
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.
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.
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.
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.
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.
Un libro grandioso de Arquímedes es "El metodo", es increíble realmente, con unos Axiomas que el crea encuentra nuevas formas de hallar el volumen, fue el primero, y único en hallar el Area de un segmento de parabola en. 1400 años con ese método. Otro libro que me encanta es "Sobre lineas espirales", donde describe la espiral de Arquimedes, hallando propiedades diferenciales de ella sin cálculo. Un genio tranquilamente
Love it! I think what is a fantastic migraine occurs when looking at the transformation from the area of the Earth to a 2d projection. The lines that sketch the Earth, when traced, travel from Antarctica to the outskirts of infinity. But never touch back to Antarctica. Appears like a great analogy to the idea of infinity by reminding us that the limit of a Cauchy Sequence, as beautiful and simple as the equivalence appears, will only ever be a limit. Ever closer, but nothing more than approximate. As though the Transcendentalism of a solution, like knowledge of Pi, becomes clearest to clever when seen to the change of 3d, through time (watching it morph), into a 2d object. Reminds myself that no matter the dimension we change to acquire equivalences (be it time or other dimensions), does not remove the Transcendental number. So then it must be True that the Infinity we chase to acquire a limit, is then equivalent to the infinity of perspectives possible to shape one proof of a solution into another?
Worth also considering the work by Stephen Hawking on the Big Bang when he compared the question "What was happening before the Big Bang?" to the question "What is there on Earth when you go further south than the South Pole?" (paraphrasing - i.e. there was no time before the Big Bang). What would the projection of the Universe look like if you transformed it in the same way that you did with the globe? I think there might be some food for thought there... (By the way, I used to get the similar migraine auras without the pain, but now I wear photosensitive glasses I have stopped getting them.) @@Mathologer
8:42, i was going to comment this: we did learn this at school in Italy, where the construction is known as the 'scodella di Galileo'( Galileo's bowl). I am not sure where Galileo comes into this, but i thought of adding this for the maths history afficionados. The proof we learnt is that one that uses Cavalieri's principle . :)
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.
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.
@@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.
I was so excited when I first noticed that the area and volume of a ball in any dimension are related via derivative...and then disappointed this doesn't work for a square unless you carefully pick how you measure the length.
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.
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.
RUclips 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😂
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 :)
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.
Very true 😁.
Congrats on your 100th video! Very special number, since it's the fourth octadecagonal number, amongst other things 😆
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
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!
Congratulations on 100 videos! Your channel is awesome ❤
Thank you so 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 :)
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...
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!
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.
My favourite Mathologer video thus far. Props to Archimedes et al.
That's great :)
As always sir, I appreciate the free Educational Videos.
You are keeping the love for Numbers alive.
Gratulation zum 100. Video!
Alles ist wirklich ein Hochgenuss, einfach perfekt, vielen Dank!
👍
Vielen Dank!
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 RUclips videos of all time now for explaining simply such a seemingly complicated topic.
That's great :)
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
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
Congrats on 100 videos, mate. You really made it as a maths educator and content creator on RUclips, 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.
Congratulations Mr. Polster (and Marty) for your 100th video! The more I watch, the more I ❤ it.
That's great :)
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 :)
One of my favorite channels on RUclips. Congratulations to Burkard and his team. Ausgezeichnete Arbeit! I always look forward to your new videos.
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 :)
Congratulations on 100 videos. Your videos are impressive, to say the least.
Thank you very much :)
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. :)
Congratulations on hitting 100 videos! That’s quite a milestone! Love your content, Mathologer!
Thank you very much :)
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.
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. 😊
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.
*Wow, just WOW!* Especially the final animation, The Lotus!!
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 :)
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
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!
🎖🎖🎖🎖🎖🎖🎖 incredible again!!! Those transformations and metamorphoseses... always open a totally new view on a 'known' issue.
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 :)
It's always lovely to watch people actually excited about what they're talking about, instead of just smiling for the camera.
Your enthusiasm for maths is really infectious 😊
Glad you think so!
Happy 100th. Again I love you videos and always watch them Sunday afternoon. Relaxing.
Glad you like them!
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 :)
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?
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 videos! Love your work ❤
Thanks you very much :)
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 :)
Congrats on 100 videos! Here's to many more!🎉
Well, as long as enough people keep watching I'll keep making these videos :)
Happy Anniversary Mathologer! Thanks for you've done to educate us.
What’s amazing is how this represents the three aether modalities. Dielectricity, magnetism and electricity.
Christmas has come early. What a mathematical gem!
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.
So many thanks to you mathologer for your tireless work. we all salute you.
Thanks you :)
Congratulations on 100 episodes! May you make many more ...
That's the plan!
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
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.
You, Professor, are a brilliant orator. Thank you for sharing your gift of training the mind to better think.
Congrats on the centenary! Cant wait for the next 100.
24:00 Well; a radius (R), of that target disk, is exactly as long, as the diameter of the sphere (and its shadow disk); so, twice the radius (r) of 1 of those; then, that 2:1 -ratio also applies to the diameter and the circumference; so, all the 1D-quantities. To get the ratio of the areas (2D-quantities), we need to scale by a factor of 2, in 2 dimensions; for a total ratio, of 2²:1² = 4:1. Thus, the target disk has 4 times the area of the shadow disk. Since the unfolding of the sphere into the target disk is area-preserving, the area of the sphere must be exactly 4 times the area of its shadow disk.
hey Congratulations on your 100th video!~ given your number of subscribers, it just reflects on the quality per video. thank you ❤
Numero uno della divulgazione matematica mondiale ❤ congratulation
Glad you that you think so :)
Another master class video from mathologer❤.
Sir, recently I came across magic squares. They were fascinating. I watched several videos on RUclips 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? ruclips.net/video/FANbncTMCGc/видео.html
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.
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.
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
What an amazing video. Also sending a salute to Andrew for the great ideas ;)
Thank you :)
salute gratefully acknowledged.
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
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 :)
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 :)
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 :)
You're simply AWESOME
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 :)
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.
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 :)
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 :)
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
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.
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 :) :)
Cool way to do it. It is actually similar to using the integral calculus concept of a solid of revolution - revolving a semicircle about its diameter.
An even deeper dive on this would be great
15:15 Hahaha...My kind of joke!
Congrats on the 100th video...and thanks for the amazing content.
Thanks :)
I love your content, hope you have a nice day!
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.
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.
Congratulations on 100 videos!
Thanks :)
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.
Un libro grandioso de Arquímedes es "El metodo", es increíble realmente, con unos Axiomas que el crea encuentra nuevas formas de hallar el volumen, fue el primero, y único en hallar el Area de un segmento de parabola en. 1400 años con ese método. Otro libro que me encanta es "Sobre lineas espirales", donde describe la espiral de Arquimedes, hallando propiedades diferenciales de ella sin cálculo. Un genio tranquilamente
Gratz on 100 videos 🎉
Thank you ❤
Thanks for being a subscriber for seven years :)
This is the ultimate introduction to Maps Projections.
@DontReadMyProfilePicture.273 What a convoluted way to say "ignore me". 😀
Love it! I think what is a fantastic migraine occurs when looking at the transformation from the area of the Earth to a 2d projection. The lines that sketch the Earth, when traced, travel from Antarctica to the outskirts of infinity. But never touch back to Antarctica. Appears like a great analogy to the idea of infinity by reminding us that the limit of a Cauchy Sequence, as beautiful and simple as the equivalence appears, will only ever be a limit. Ever closer, but nothing more than approximate. As though the Transcendentalism of a solution, like knowledge of Pi, becomes clearest to clever when seen to the change of 3d, through time (watching it morph), into a 2d object. Reminds myself that no matter the dimension we change to acquire equivalences (be it time or other dimensions), does not remove the Transcendental number. So then it must be True that the Infinity we chase to acquire a limit, is then equivalent to the infinity of perspectives possible to shape one proof of a solution into another?
I get migraine auras without the pain. What I see there is even more fantastic than this transformation :)
Worth also considering the work by Stephen Hawking on the Big Bang when he compared the question "What was happening before the Big Bang?" to the question "What is there on Earth when you go further south than the South Pole?" (paraphrasing - i.e. there was no time before the Big Bang). What would the projection of the Universe look like if you transformed it in the same way that you did with the globe? I think there might be some food for thought there...
(By the way, I used to get the similar migraine auras without the pain, but now I wear photosensitive glasses I have stopped getting them.)
@@Mathologer
8:42, i was going to comment this: we did learn this at school in Italy, where the construction is known as the 'scodella di Galileo'( Galileo's bowl). I am not sure where Galileo comes into this, but i thought of adding this for the maths history afficionados. The proof we learnt is that one that uses Cavalieri's principle . :)
Galileo and Cavalieri definitely knew each other very well but it beats me how they came up with this name.
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 :)
Burkard's little giggles are do endearing.
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 :)
Thanks so much once again! Let me get my popcorn.🍿
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
@@teo_lphow 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.
Wow what great animations. Chapeau🎩🤠
3:2 fantastic ! And what a beautifull pumpkin shape .
I was so excited when I first noticed that the area and volume of a ball in any dimension are related via derivative...and then disappointed this doesn't work for a square unless you carefully pick how you measure the length.
For a guy who is unable to straighten his elbows, Mathologer has really contributed a ton of good content to youTube.
Congratulations on #100!
Thank you :)
Congratulatios, 100 !!! 👏
Thanks !
Only a mathologer video to cheer me up ^^
That's a very satisfying final animation
Andrew really knows what he is doing :)
Here in the Kingdom of Mathologica we put lunes on our lunes. 😎