But why is a sphere's surface area four times its shadow?

[@albericseydoux]

Can we just take a moment to admire the quality of all animations in these videos ? It's just insane

[@wilfroberts637]

I'm pretty sure all the animations are written in their own coding language as a part of latex if I remember correctly

[@ammyvl1]

@@wilfroberts637 not quite - he uses latex for notation, but the animations are made with his own library called manim (which I assume comes from math anim).

[@pranavmishra9366]

Hum bhi bnane baithe the manim se gand fat gya mera

[@wesleymays1931]

well he did make a Python library to animate these videos for him

[@omkar9285]

IIT

[@noorulmubeens2306]

7:53 "sometimes easy to forget why we're doing this".. its amazing how you pay attention to where one might loose track, after having mastered the topic.

[@52m2]

and your face makes me pay attention to your saying

[@AdamsAppleseed]

Yeah I noticed how well placed that was!

[@windywednesday4166]

A well planned video.

[@aarontongon3248]

This is why this is one of my favorite channels.

[@AlexandrBorschchev]

I also do that when i use tutor my friends. i think in their own perspective and think of problems they might not be getting and i tutor them base on self-explanatory conditions and not those complicated concepts that no one can even logically answer like "how did the long division method got formulated", it should be self-explanatory.

[@lakeemon]

The production quality of these videos never ceases to amaze me: The fluiditiy of the animations seamlessly demonstrating the ideas as they are being narrated. The impeccable pacing in the script that dives into the real unexpectedness and wonder of math. The passion and care that are woven through it all. You really are today's Feynman to me! Thank you!

[@captain_clark868]

well said.

[@Lcxruns]

also he makes it understandable to people watching even if theyre like in middle sch who didnt learn advanced maths or something

[@THEELEMENTKING]

The craziest thing about the animation is that it’s all computer generated from code

[@SuperFerz]

While there are tools to help visualise geometry, the amount of work to produce such material is still so enormous!

[@900bpm]

@@THEELEMENTKING fr, i find that more impressive than doing it by hand. he managed to create a script to make fluid and clean animations for him

[@shravs3776]

This guy knows how to explain every detail and knows exactly what questions will be asked and immediately answers them. This guy is truly amazing!

[@clearthinking5441]

An absolutely brilliant video! I don't understand how the visualisations get better and better through each video, simply superb. I particularly enjoyed the alternating flashing technique to emphasise complicated parts of the video. You are the Shakespeare of Maths please never stop I am sure you will inspire some great minds in the future.

[@rickliles2460]

I'm already inspired - kind of addicted to math now The quaternions and projection series explaind the connection between s -doamain and z domain in DSP with out mentioning either. this guy is a genius at teaching. Maybe the next Plato or something like that

[@matthewburson2908]

github.com/3b1b/manim

[@garryiglesias4074]

@Matthew Burson - Thank you. I didn't know he had a github repo. This guy is very nice and smart. Indeed, may he inspires a new generation which will have better tools, better understanding, and solid foundation to continue the "science work".

[@danielsouza2129]

Dude, Y the name 3blue1brown?

[@flimflam6652]

@@danielsouza2129 His eye color

[@Greywander87]

It helps a lot if you remember that Oscar Had A Hold On Arthur. Answers below the fold: Q1: Let's call the radius of the ring d. We have a right triangle with an angle of theta and a hypotenuse of R. In this case, d is opposite from theta. Using the above mnemonic, we can remember that O/H = sin(theta), ergo d = R sin(theta). The circumference of a circle is 2 pi R, so the inner circumference of the ring is *2 pi R sin(theta).* Thus the ring's area is approximately *2 pi R^2 sin(theta) d(theta).* Q2: The good news is that our inner radius d is the same as it was for the ring on the sphere, ergo the inner circumference will also be the same: 2 pi R sin(theta). What we need to figure out is the thickness of the ring's shadow. By drawing another right triangle where the hypotenuse is the thickness of the ring, R d(theta), we can see that the thickness of the shadow is adjacent to theta in our new triangle. Using the mnemonic above, we can see that A/H = cos(theta), ergo the thickness of the shadow = R cos(theta) d(theta). To finish off, we multiply these two to get an area of *2 pi R^2 sin(theta) cos(theta) d(theta).* Q3: Using the identity that 2 sin(theta) cos(theta) = sin(2*theta) reveals that we can rewrite the area of the shadow as *pi R^2 sin(2*theta) d(theta).* This is the same as the area of the ring except that we've dropped the 2 from in front, signifying that we've cut the value in half, but we've also doubled the value of theta. This means that the shadow at a given value of theta has half the area of the ring at double that theta value. For example, the shadow at theta = 30 degrees has half the area of the ring at 60 degrees. Thus, as we go to the next shadow, we skip past one of the rings and jump two rings ahead instead of one. Q4: Partially answered above, but as we compare each shadow to a ring on the sphere, we have to skip every other ring, jumping two rings ahead for each shadow. The other half of this puzzle is to remember that we only generated shadows from one hemisphere rather than the entire sphere. Since we skip one ring for each shadow, that means we need to use all of the rings from the entire sphere (except for the ones we jump over), instead of just using the rings from one hemisphere. An easy way to see this is to think about the last shadow, at theta = 90 degrees, which corresponds to half the area of the ring at 180 degrees, which is the last ring on the sphere. Q5: The area of the shadows sums up to the area of a circle of radius R. However, each shadow is only _half_ the area of one of the rings, and only half of the rings have been accounted for. A half of a half is one quarter. Ergo, a circle of radius R only has half the area of one hemisphere of the sphere, which in turn only has half the area of the whole sphere, and thus the area of the circle is one quarter that of the entire sphere.

[@mantisshrimp9637]

Dis is the best

[@AnonymousAnonymous-ht4cm]

Edit: Other comments explained my mistake. Thanks! That agrees with what I got, except that my area formulas has R instead of R^2. I've seen a few other commenters using R^2, so it is probably correct. I got the area as circumference*width where width is dø for the ring and cos(ø)dø for the shadow. To me this also makes sense that I multiply two distances to get area, rather than 3 (R, R, dø). How did you get the other R?

[@deeptochatterjee532]

@@AnonymousAnonymous-ht4cm the width is linearly proportional to R. For example, at a fixed central angle, the arc length of an arc will increase by a factor of R as you increase the radius. Same reason why circumference is 2πR and not just 2π

[@albericponcedeleon2696]

Bravo sir. I'm actually glad that my intuition was leading me along a similar path. If half of the rings from one hemisphere gets us a circle of equal radius to the sphere, then the other half of the rings gets us two circles, and if we do the same on the other hemisphere we end up with four circles in total.

[@StankyPickle1]

@@AnonymousAnonymous-ht4cm You are close. The widths are actually R*dø and R*cos(ø)*dø for the ring and shadow, respectively. That is where the extra R arises that you are missing. This should make intuitive sense because a sphere with a larger R will have wider strips for a given dø. Similarly, smaller spheres will have thinner strips for a given dø. You can also roughly think about this in terms of units. dø and cos(ø)*dø are essentially unitless. One needs to multiply these quantities by R to get a unit of length.

[@uhSighLimb]

See most people find maths stressful and anxiety inducing... but somehow, this man has made it relaxing and beautiful.

[@jakub.kubicek]

Compulsory schooling is to blame

[@JohnSmith-kj2od]

I don't think it's the concepts people find stressful, if 3b1b started keeping quizzes and grades for his viewers, a lot would run away

[@fannyalbi9040]

still stressful to me. i can’t accept cylinder equal to sphere. i can accept it is close but not equal.

[@prumchhangsreng979]

@@jakub.kubicek stop blaming school. Even after watching this video, there are still people who doesnt even understand what is going on. People hate school becuz require real serious test. The test result determ if u are qualify for next grade or not up until the final grade which then u take college entrance exam. School need to prepare u for college entrance exam, it is more intense which is why it is more stressful. But on a youtube video, u are free to not understand anything u just watch but at the same time , u arent gonna pass any entrance exam anyime soon. These kind of video are good for inspiring new ppl to love math or for math entusiastic to learn more about math that they alr learn at school.

[@AL_saoud_the_century_of_satan]

@@fannyalbi9040 the concept is simple, compare an arc to its projection

[@_Fox]

Brown: *explains* Blue: But why? Brown: *angry noises*

[@dayzimlich]

you just helped me make the connection. The name of the channel is 3blue1brown, and in the animated classroom there are 3 blue pi student creatures and 1 brown pi teacher creature

[@_Fox]

@@dayzimlich Yeah you are welcome

[@KenWasHere]

Lol

[@stev009]

as the 464th like that's hilarious

[@haipingcao2212_.]

​@@dayzimlichcorrect.

[@isaacbastos2762]

Congrats on such an amazing video, omg it's 3AM but here's how I did the exercise proof: 1. The circumference length of each ring is 2*pi*R*sin(theta), since the distance from the ring to the axis is R* sin(theta) (trigonometry). Hence, the area of a ring is 2*pi*R*sin(theta)*R*d_theta = 2*pi*R^2*sin(theta)*d_theta (1) 2. To calculate the area of a ring's shadow, I used some trig relations as well. In this case, the thickness of the ring shadow is R*d_theta*cos(theta). Therefore, the area of the ring shadow is 2*pi*R^2*sin(theta)*cos(theta)*d_theta (2) 3. Multiplying by both sides expression (2), we get 2*pi*R^2*2*sin(theta)*cos(theta)*d_theta I've put the number 2 right next to sin(theta)*cos(theta) to explicit the trigonometrical relation: 2*sin(theta)*cos(theta) = sin(2*theta) That specific angle is 2*theta. 4. So, this means that the area of the shadow of some ring with a corresponding angle of theta is equal to the area of a ring which has double of this angle. 5. Notice that, when theta ranges from zero to pi/2, we get to form all of the rings related to their shadows in the superior hemisphere. The total area of the shadow is the sum of all the thin ring shadows. Thus, that is equal to half the area of the hemisphere. That is, the area of the shadow (pi*R^2) is 1/4 of the surface area of the sphere. A(sphere) = 4*pi*R^2 Q.E.D.

[@Maki-ng4jk]

Why didn't you use the appropriate symbols (i.e. π )? >.>

[@joshuaronisjr]

I'm confused... When we sum up the areas of the each of the shadows of the rings, for all the rings the top hemisphere, we get the shadow of the sphere itself. That's equal to (pi*r^2). I get this step The shadow of a ring at an angle of theta has the same area as a ring at an angle of 2theta itself. Therefore, summing over all the shadows in the top hemisphere is the same thing as summing over the area of every other ring in the top hemisphere...this step I don't get at all. Why is this? Why does the fact that for any ring A, there's a ring B at double the angle of A that has the same area as the ring A's shadow, imply that when we sum over all the shadows if the rings in the top hemisphere, we get half the area of the top hemisphere...? Assuming that we do, we need to multiply this area by 4 to get the surface area of the sphere...so this step I get again. But I'm lost in the in-between step. Thanks!

[@EvilAxelord19]

Thank you for this, I was able to get most of the math right, but mistaking the cosine for a sine messed everything up, and figuring out that the result of sin(2theta)=sin(theta) was talking about two different angles was super helpful

[@Pranavtubecom]

Great explanation for the process, it is hard to explain that good.

[@yashagnihotri6901]

Well thats a mere calculus proof... No offense and of course i respect the fact that you tried it out all by yourself , but yeah its too easy.

[@alfianfahmi5430]

Normal people : Why? 3Blue1Brown : But why?

[@giantrunt]

3blue1One?

[@alfianfahmi5430]

@@giantrunt Fixed

[@leif1075]

Whonactually understood any of this video..its convoluted and impossible tonunderstand and turns me off from bring curious to understand things because itntakes so damn long and is so confusing tonunderstand one simple thing..sheesh..

[@aviralsood8141]

@@leif1075 The point is to gain a general understanding of the principles and techniques at work. You may know the formula from another method of derivation, but Grant's videos emphasize the methods as well as the result they talk about. It helps to cultivate better mathematical thinking.

[@jitendrapandey1085]

@@aviralsood8141 hello aviral Aryan here

[@ExaltedDuck]

A couple years ago, I was trying to estimate the amount of paint needed for an airliner. I based it on some simplifying assumption with regard to shapes and was surprised to find an equivalence between a hemisphere capped cylinder and open ended cylinder of the same overall length. Made me double check my math for the generalized case and indeed, (2 pi r)(l + 2r) describes both cases. So for my estimate paint usage, I just used the open ended cylinder with the length and radius for the fuselage and engines, plus a rough combination of triangles for the tail. (Wings, too, but their numbers stay separate since they use different products to prevent adhesion of ice)

[@matthewrigby6089]

Sooo, how close were you to getting it right with these approximations?

[@puppergump4117]

@@matthewrigby6089 His paint usage was much lower than expected and he now has a striped truck

[@egg1645]

the guy from the textbook problems is real 0.0

[@-_deploy_-]

Bro gets his name called out on the math text book and it isn't just coincidence

[@haipingcao2212_.]

1+¢=€¢-1

[@tripathi26]

I remember trying to prove this in highschool. It seemed impossible given the knowledge of a child but I wonder if I had a teacher like you that time, it would have been an enlightened day of my lifetime. Thanks for these elegant proofs ❤️

[@WritingMyOwnElegy]

him: unwraps a circle into a triangle me: you CaN dO tHaT?

[@Deafeatist]

Kayle Needler you can unwrap a circle into a triangle by cutting a perfect spiral from the edge to the exact centre

[@MohaMMaDiN55]

Actually you can’t do that unless I think if the circle is made up of an elastic material like rubber.

[@Deafeatist]

Theoretically

[@RieMUisthegoaT]

@@MohaMMaDiN55 this is mathematics not physicsm, but your logic calculus is invalid because in calculus you have to have continuous matter and there's NO continuous matter in real life. talking about you know the the normal derivative and such things

[@MohaMMaDiN55]

abdullah almasri This actually has nothing to do with what I said. I wasn’t talking about either calculus nor continuous matter.

[@imyasharya]

I'm watching this video after a year and it makes much more sense to me than ever. I remember when I first learnt about the surface area of sphere, it itched me and I searched KZbin for that but it didn't made that much sense to me but now, I'm satisfied. Thanks 3Blue1Brown!

[@maxthexpfarmer3957]

The most amazing thing is that Archimedes found the surface area of a sphere over a thousand years before calculus was even invented. While he didn't prove it to modern rigor, he can hardly be blamed for that.

[@peteiiseasy3217]

Yup

[@wordup897]

How did he work it out?

[@yatesfletcher1424]

Actually, his proof was bullet proof even by modern standards. He proved one miraculous lemma which underlay not only his proof that the area of a sphere is 2/3 the area of its circumscribed cylinder (lateral area + top & bottom) but also that it's volume is 2/3 the volume of this same cylinder! Legend has it that the sphere and its circumscribing cylinder were inscribed on his tombstone.

[@tarnopol]

Absolutely!

[@unknownkingdom]

He didnt

[@RedLuigiE]

I love this channel. It answers the WHY, the question no one ever seems to answer. All my math teachers have just re-explained the formulas, without ever saying how it actually works.

[@krishnachattopadhyay3251]

Make a video on how 3 cones make a cylinder

[@xicad1533]

Fill 3 cones with water, pour them all in a cylinder of the same height, thats how my geometry teacher taught it

[@realeyes8199]

@@xicad1533 and same radius.

[@chrisding1976]

Mr. Virtual 𝕟𝕠 𝕠𝕟𝕖 𝕤𝕒𝕚𝕕 𝕒𝕟𝕪𝕥𝕙𝕚𝕟𝕘 𝕒𝕓𝕠𝕦𝕥 𝕡𝕣𝕠𝕠𝕗𝕤

[@twicepilled]

@@chrisding1976 *_how do you make that font teach me the ways master_*

[@onetwo7191]

Tʜᴇ ғᴏɴᴛ ᴀᴘᴘ

[@abdellahdany1689]

But why on earth maths isn't taught like this?

[@Thepiecat]

Because good teachers are extremely rare, even more so in maths. And the technology to do this is finally becoming nice enough to do this kind of thing. I hope we see a big push in the next 5-10 years with this kind of teaching.

[@SkateGeneva]

After watching this you may have the general idea of how the maths behind this work, but have you tried doing it yourself ? You probably won't be able to do it. That's why mathematics aren't taught like this. You need to do it yourself, sit in front of a blank piece of paper and try the demonstration on your own.

[@Absilicon]

Because you'd have to wait for somebody to animate everything 😓

[@zeqizhang5860]

you dont have 1000 mintues every class...

[@dsdsspp7130]

Cause there are hundreds of formulas and teacher cant spend 20 minutes for each one.

[@romawang9212]

I love 0:40 when he tried to cover the surface of the sphere with circles

[@Sciencedoneright]

And how he gets frustrated lol

[@elixir662]

Yes! It somewhat scratches an itch.

[@fnersch3367]

I grew up in the "primitive era", learning math was murder. We've come a long way in the last 60+ years.

[@Perririri]

OK, Boomer!

[@puppergump4117]

@@cdfactory I really don't wanna hear their hour long story about how they got to school

[@bringonthevelocirapture]

You should hear old scientists complain. Basically, everything that took 5-10 years and/or tons of money can be done in a week to a year for a fraction of the relative cost.

[@GodzillaGoesGaga]

@@bringonthevelocirapture Except when they can't. Try designing an aircraft. Still damn hard.

[@Pao234_]

@@Perririri trianon

[@MatematicaRio]

Your channel is just ABSOLUTELY AMAZING! I love your videos! Thank you for producing such nice Math content for the world. 👏🏻👏🏻👏🏻

[@MGRambo-mi4hg]

Procopio por aqui hein???? Tudo pela matemática!

[@NidaraSnil]

Concordo, esse canal é um dos melhores que já vi.

[@ffc2632]

3Blue1Brown dinamita na matemática

[@claudioestrada2774]

Realmente es un excelente video, la claridad de la explicación, la nitidez y belleza de las imágenes, la sencillez del lenguaje, este video deberîa ser usado por muchos profesores en sus clases..

[@luccaassis2148]

E aí Procopio!! Concordo, muito bom mesmo.

[@EidosGaming]

I'm so smart I did the exercice all by thought without a piece of paper And I also got it all wrong

[@ViguLiviu]

It's not that hard to not use a piece of paper to do this if you are careful.

[@whynotanyting]

The risk I took was calculated, but man am I bad at math

[@Danilego]

I took the time to actually get the paper and pencil and do it But still got it wrong

[@EidosGaming]

@@Danilego I got this as the area of the shadow 2π*R^2*deltaø*cos(ø)*sin(ø). It's a complete mess, i'm not even sûre about the cos thing. I tried replacing the whole 2sin*cos thing by sin(2ø) but I didn't found any way out

[@pursuitsoflife.6119]

@@EidosGaming same dude I got stuck right there

[@evaristegalois6282]

3Blue1Brown: *Explaining mathematical concepts better than school ever could*

[@MrMctastics]

Evariste Galois not everyone can be like him. Also, sadly, there is a ton of stuff they are required to cover so that they have barely anytime to do stuff like this

[@MrMctastics]

I'm majoring in math, so I might be a math teacher some day. I'm planning on also completing a computer science degree, so hopefully I won't be.

[@MrMctastics]

I value proofs a lot more than my peers

[@zXToThaFaceXz]

My cal 4 professor talked about this when we did double integrals to find surface area, so it's not entirely fair to say none of them do this

[@dannygjk]

Good luck getting a high school diploma merely by watching KZbin videos for a few hours.

[@Suburp212]

I am not interested in math. And yet, your videos are amazing, one of the few ones on KZbin I watch fully. They are clear, calming, Super simple, a great way to spend time, always make sense and make simple things that looked like magic and wizardry back in school look like Kindergarten math. Wow. Thanks.

[@gopikrishnamaji536]

*ADMIT* *IT* The *_Beauty_* *_of_* *_mathematics_* is the most satisfying thing ever..

[@ssuriset]

Well.. it's Mathematics and Physics

[@gopikrishnamaji536]

@@ssuriset yeah

[@dumb_tan2539]

Ya

[@black_jack_meghav]

@@ssuriset well, it's science in general

[@ESPkenner48]

You havent tried extasy

[@bliz85]

I have an intuitive geometric interpretation for the second part. I'm not sure if it qualifies as a proof or not: Using x, y, z axes in the right, up, and inside directions respectively. Consider a top hemisphere on the xz-plane, and its y projection to a circle (call it PC). Now using the rectangles (call it OR) from the first part, we ignore the width of the OR and just consider its height, H, defined as the tangent of a circle along the xz-plane. Let the length of the projected rectangle (call it PR) be P. It follows that P = 0 at the equator, and P = H at the pole. Now consider a length on OR called E (for extra), such that E + P = H. It follows that E = H at the equator, and E = 0 at the pole. So we have 0

[@dayzimlich]

Whoa mind blown. Your explanation helped me understand. Thanks!

[@kevinnguyen4055]

Yes this exactly what I did, but I used cylindrical slices and then separated them into rectangles to justify my calculations. Shadow to surface ratio is height (h) / radius (r). The central ring area is x*2pi*r, where x is the height increment. All other areas are this area multiplied by the ratio. So ring area = hx *2pi. Factor out the x and multiply by r/x getting a factor of r. And summing all areas from 0 to r you get pi*r^2 . Obviously rectangles and angles make a quick and simple solution that can more easily be justified, but I’m happy with what I can do with my mental math. Edit: errors in explanation!

[@ahmadness6995]

Please write a book on the beauty of mathematics!!!

[@ASIGOO]

He's recommended a book before called Measurement by Paul Lockhart. It's a really good book and I would also recommend it. That said, this guy should definitely write a book of his own

[@muhammadputera6593]

Lots of good books on math already. 3B1B's strength is making videos on the beauty of math, so I just hope he keeps doing this!

[@AuxiliaryPanther]

A flipbook

[@sharinabhamid9955]

Yes.i would buy it

[@einSteppenwolf]

A movie. Even better: a 3D movie.

[@ksbalaji1287]

Grant, your math, narration and animation come together seamlessly like a Swiss watch. These videos are the very definition of 'Excellence'.

[@michaelneugebauer7063]

This is the best mathematics channel on KZbin. There is literally no competition. I want these videos played at my funeral.

[@klausbrinck2137]

hopefully not very impatient...!

[@gaurangagarwal3243]

'At my funeral' #Atruemathlover

[@pachov]

lol

[@kernoelnerd]

Wouldn't an endless loop of one of his animations on a screen on your tombstone be yet cooler ;-)?

[@SeiberGraff]

The math class is extended because the 3rd pi is still asking “why”. Whole class hates that pi.

[@SeiberGraff]

@Sanchi Sharma but WHY

[@raghvendrasingh4421]

I am one of the pi material

[@VictorRodriguez-zp2do]

But that pi will be the one which will end up learning the most.

[@aadithyahrudhay2269]

I AM THAT PI!!

[@chrisjfox8715]

Sanchi Sharma that pi believes 9/11 never happened

[@kacee3472]

These videos are so beautiful, I've only just started pre-calculus in high school this year but I love learning more in my free time and it's so cool to see this video touching on some ideas I actually know some stuff about- and also it's presented in such an elegant, intuitive way that's so much easier to understand than anything my complete mess of a pre-calculus teacher could come up with.

[@ingwermoschus5630]

Consider yourself lucky discovering 3b1b before leaving school :D would have helped me a great deal .. but instead it awoke a new interest in math for me :) and it still might be useful

[@Inspirator_AG112]

*@[**06:24**]:* This is due to the omni-directional symmetry of a circle, as well as another calculus fact: • For any point on a curve residing on that curve's line of symmetry, if said curve is C¹ continuous (meaning continuous first derivative) at that point, the tangent is perpendicular to that line of symmetry.

[@SavageEnder]

I really love how he explains WHY certain concepts work the way they do, instead of what they do at school, shoving them in our face with the reasoning ‘just because.’

[@Spractral]

This is exactly what I needed. Im going back over precalculus now that Ive finished trudging my way through calculus one (barely got a b+; havent been to school in 15 years and never really had a proper education ). I was going over the proof for the area of a circle and of cylinders spheres etc by extension and it finally clicked yesterday afternoon. This video has helped parse out all the details and cement it in. Perfect job.

[@williamcaton3409]

I have a different take that I think is relatively simple. After answering the questions 1, 2 and 3: Question 1: C=2πRsin(θ) C×Rdθ=2πR²sin(θ)dθ Question 2: R(e):exterior radius R(i):interior radius cos(θ)=(R(e)-R(i))/Rdθ =>R(e)-R(i)=Rcos(θ)dθ C×(R(e)-R(i))=2πRsin(θ)×Rcos(θ)dθ =2πR²sin(θ)cos(θ)dθ Question 3: 2πR²sin(θ)cos(θ)dθ=πR²sin(2θ)dθ since sin(2θ)=2sin(θ)cos(θ) Therefore, the area of the shadow of a ring is half the thickness of the ring at twice the angle of the ring that casts the shadow. Questions 4 and 5: I decided to combine these 2. First of all, question 4 is partly answered by the conclusion of question 3. We see a shadow and when comparing it to a ring in terms of area we must skip a ring and look at the 2nd one as our answer. However, regarding the initial goal of proving the the area of a sphere is 4πR², we can take those rings and put them up against each other. Now, the shadow that's being cast will be a circle and as we proved earlier the area of a shadow is half the area of the ring at twice the angle of the ring that casts the shadow. However, since now the rings, since we added them all together, have become a hemisphere we can make the conclusion that the circular shadow that is cast by the collection of rings, now the hemisphere, is half the area of the hemisphere formed by the collection of rings. In addition, we know that 2 hemispheres make up a sphere, therefore the area of a circle, since the hemisphere casts a shadow of a circle, is a quarter, 1/4, of the area of the whole sphere: Sphere: S Area of sphere: AS Hemisphere: HS Area of hemisphere: AHS Shadow: Sh Area of shadow: ASh ASh=1/2×AHS 2×HS=S=>HS=1/2×S =>AHS=1/2×AS Therefore, ASh=1/2×AHS=1/4×AS And since a shadow is 2-dimensional and it's a circle since it was cast by a hemisphere, THE AREA OF A CIRCLE IS A QUARTER OF THE AREA OF A SPHERE.

[@andrewgjennings]

Your videos are stunning in their simplicity, a perfect blend of math, computer, programming, and speech. Well done!

[@matematikaadit]

This youtube comment section is gold. Thanks 3b1b for making such a wonderful community.

[@15october91]

17 minutes of heaven.

[@josketobben]

12 minutes of heaven, then he gives homework! YOUR NOT MY REAL DA.. uhm.. TEACHER

[@15october91]

Joske Tobben lol

[@shuvamchatterjee8611]

15october91 17 min 1 sec

[@tomwilson8176]

Ended No Nut November on a good foot.

[@classifiesconfidential4330]

I like how your explanation is mainly geared towards students in “pre”- calculus.

[@rajdeepnaha1242]

Yes calculus is op

[@CocoTheTortie]

Man, I was just on a binge of shorts and this really feels like a great exit point for me into a longer format video ❤

[@ruchirrawat8804]

Geography teacher : doing 4 questions in 4 minutes is the same as doing 1 question in one minute Calculus : *am i a joke to you?*

[@h-Films]

yes explain

[@viriyo4958]

​@@h-Films Let: 1. number of questions done = f(t); unit = no of questions 2. time = T; unit = minutes The function f(t) does this: 1. when t = 0, x = 0 2. when t = 4, x = 4 The history teacher concludes that the gradient, f'(t) or df(t)/dt, is a constant value, which is 1 question per minute, obtained by this formula: f'(t) = [f(4)-f(0)]/(4-0) While it is valid if f(t) is a straight line, it is usually inaccurate otherwise since f'(t) often changes based on t (e.g. quadratic f(t)). This method is in a way similar to trapezium and newton-raphson approximations. Or, he's just saying that difficulty variations of calculus Q is very high, so you can't just say that all questions take the same time (1 minute) to complete.

[@chuanyinglin3926]

Chopin?

[@Un-homme-avec-raison]

Chopin I love u

[@kronigen]

@@alwaysseverus741 my homie's tryna flex out here man

[@sangjunechoi4369]

In general, it is 4^k/(nCk). It seems that the integer multiple case appears only for k=1. Assume there exist an integer p such that 4^k/(nCk)=p. Since 4^k/(nCk) is monotonically increasing as k increases, we consider p>4 cases. Then, k*Log(4/p) = Log(nCk). In addition, LHS is negative for p>4 while RHS is positive, for positive integers k. Hence, only for p=4, LHS=RHS=0. This interesting case seems like following the law of small numbers.

[@sangjunechoi4369]

I am considering only the odd dimensional cases. But it is also suprising that the transcendental number Pi does not appear for odd dimensional cases, while it is not for even dimensional cases.

[@NTdredd]

Actually, there is another way to find surface area of circle, I actually noticed it when I was in my high school, if you differentiate the volume of sphere w.r.t. radius, then you get total surface area of sphere... The same case applies for Circumference of circle and area of circle, the circumference of circle is derivative of area of circle w.r.t. radius. I don't know if this is just a coincidence or there is actually some relation. You can also apply this rule to total surface area of cube and volume of cube etc.

[@vihanduudawatta322]

@@NTdredd woah that's actually cool, never knew this

[@LakshitSinghBisht]

​@@NTdredd Actually, this is not a co incidence. Derivative is good way to find it. Maybe you can find more on quora or just google it.

[@cracknblast8247]

@@NTdredd Yeah, that follows from the definition of the derivative. When you know why things work the way they do, that is precisely when math starts getting interesting.

[@Asdayasman]

Good lord how do you do these animations? You're incredible.

[@onyxiris]

I love 3b1b so much. 18 months out from the last time I sat in a maths classroom, I happened to see a picture of a problem on a whiteboard in the background of a photo shoot that was solving for the area of a sphere. I got curious and decided to look up the maths (because I thought it was wrong, yikes), and came upon this video. Whereas another video easily could have delved straight into calculus that I definitely no longer remember, this video ended up not only answering my question in detail, but left me saying, out loud, "How cool is that?". You are such a fantastic communicator and orator, and I'm so excited to see where you go as a new subscriber.

[@orange422]

I'm torn. Is the math more beautiful or is it the animations?

[@kalebbruwer]

The animations describe the math and are created with math, so I'd argue that they are the same thing.

[@cleitonoliveira932]

You need much more time to imagine all that without 3b1b. Years. Imagine a student learning the first time.

[@Lorden1516]

The math _is_ the animations.

[@andymcl92]

Maybe the brain behind the animations

[@yuvrajgarg3921]

It's all Mathematics

[@beactivebehappy9894]

This has been literally the exact question I had in my brain since I learned this topic, 7 years now, and finally YT algorithm has found my inner self!!

[@garydunken7934]

I'm now in my 40s and I find math more interesting than how I felt during my high school learning.

[@maheshm5463]

G Dunken I am 55 and feel the same

[@burnerjack01]

@@maheshm5463 I'm older still. I find calculus astounding. In school I found math nearly useless and when asked, no one could tell me why I should lean it except "you'll need it if you go to college." If anyone had just told me "basic math and algebra are the language of straight lines and planes (not always true, I know) but calculus is the language of curves and therefore, its the mathematical language of the Universe." It would have changed my perspective and therefore, my life. Tragic, really. Just 1 hour of the concept of integral calculus would have sparked my interest and I would have gone from "Learning because I have to" to "Learning because a WANT to".

[@EidosGaming]

Welcome to maths. Unlike what it may have seem to be, it's a wonderful universe !

[@dennisbauer3315]

Lol, read my comment above. I am 70.

[@marcussparticus8380]

I'm 60 and. I wish I had a maths teacher like this in school, instead we had a loud mouthed aggressive bully that terrified everyone in class, and no one was good at maths in my class, I wonder why ? He died of a Hart attack the year after I left school aged 40 and thought it's a pity he didn't die the year before I started. Now I'm disabled and at home most of the time and due to KZbin I've found a new love for maths, I would like to start all over again from scratch as a beginner. If any one out there knows of some good videos on KZbin with the same inspiring content as this please post a link. Thanks in advance 😀

[@david203]

This is just what I wanted to know over 60 years ago, thanks!

[@vegetasama493]

U can easily integrate it to get the answer

[@david203]

@@vegetasama493 Please go ahead and do so. State carefully exactly what you are integrating, in what direction, and what the limits are.

[@sreedar13]

@@vegetasama493 but integration only gives the answer (4.Pi.R.R). We all know that but why is it true is what we are trying to understand

[@Sanjay-ub7eq]

we did a simple experiment long ago. So take a Sphere and cover it completely by thread. Then make as many circle as you can with that thread with the same radius. And you'll get exactly 4 circles by that thread if done precisely.

[@david203]

@@Sanjay-ub7eq That is an excellent experiment. How did you cover the sphere perfectly with thread, without twisting it and thereby changing its length? Figuring out how to do that must have been a real challenge.

[@jezer8325]

Could someone check if what I have is right? So the area of the rings for any angle θ is 2π(r^2)sinθdθ And the area of their shadows is 2π(r^2)sinθcosθdθ = (1/2)π(r^2)sin2θd2θ So for the shadow of a ring at an angle θ, there is a corresponding ring at angle 2θ having a quarter of the area. θ should range from 0 to 90, and not all the way to 180, because otherwise the shadow will be cast twice. Also the new sphere made from the corresponding rings has courser iterations. Although, now I think that (1/2)π(r^2)sin2θd2θ could have been expressed as π(r^2)sin2θdθ with alternate rings which I guess is what the video was going for.

[@3blue1brown]

Nice!

[@jezer8325]

Thanks! I love all your videos btw!

[@ViguLiviu]

Math on your head is not as simple as on paper, but basically that's almost what i got thanks. Although that trick of chaining dθ into 1/2*d2θ i didn't thought about.

[@ViguLiviu]

Although to be honest in my mind it first appeared for cosθ, sin(90-θ).

[@deeptochatterjee532]

@@3blue1brown God damn I was so close. I just didn't get how 2θ related to every other ring

[@rewrose2838]

If only the pen paper drawings were just as easy to play around with~

[@madhoshyagnik3679]

I think that particularly was the biggest bottleneck for me back then!

[@annoyyourmouse2451]

10pm: 1 last video and I will go to sleep! 3am:

[@joet840]

I'm falling asleep from boredom just watching it.

[@pulakdas8998]

Wow

[@johnjonjhonjonathanjohnson3559]

6am:

[@ariat3381]

@@joet840 thats probably because you dont get it

[@cowboybob7093]

Opposite affect on me, but I came back and glad I did.

[@eekee6034]

I've always thought geometry is the best way to introduce many mathematical concepts. And why haven't I watched 3blue1brown before? This is very much my kind of explanation. However, as tired as I am, I might have to skip getting the paper out. I'll just have to watch this one again some time. :) Edit: Optical illusion at 12:08 -- when separated, the rings appear to shrink latteraly.

[@DrDineshKumar]

great explanation with graphics! the best use of technology in teaching.

[@marcelb4889]

true

[@sarthakgirdhar2833]

I am from India. The IIT JEE is considered the toughest exam here, and probably in the world for 17 year old students. And I don't think that out of thousands of students who crack it with amazing grades, actually know anything with this precision.

[@muskansiddikee2171]

You Said the correct thing Man math and physics is not a headache it's amazing if thought in the right way

[@sarthakgirdhar2833]

​@@youtubeshorts2911 The standard method Grant used was new to me, but to be honest I solved this problem using the method he has devised on his own, way back in class XI. Do not underestimate anyone. And Grant is not a Ph.D. He is a graduate from Stanford University. Ph.D doesn't make you knowledgable, hunger and patience do. Peace.

[@youtubeshorts2911]

@@sarthakgirdhar2833 Graduate ka matlab google kar le bhai. All the best with your GRE preparation.

[@critisizerr245]

@@youtubeshorts2911 shivansh joshi , i agree with u what u think bout sarthak bro I am in 10 preparing for jee While I was reading about traingle I got the angle bisector theorem that it divides the opposite side by the same ratio of the two other sides First i proved it pratically by with the help of goemetry(meansurment ) and the i tried to prove it theoretically by using properties Although i wasn't able to prove it theoretically but then i saw it and i found it, i was very happy that i proved it pratically. Just telling because sarthak bro thinks that students dont understand concepts in depth. Bless me for jee 🙏 I want to secure AIR

[@critisizerr245]

And sarthak is it not precise

[@homeschoolofficial8709]

This is an amazing illustration. Great job. I m so impressed by your clear explanation of such a difficult topic, that I have subscribed to your channel right away.

[@shelbythecyborg4399]

Thank you dearly! I'm not a student, I accidentally got into engineering. I just needed to replace my speakers power supply/cord, which a new cord would be expensive in relevance to the cost of the speaker. I decided to look up a little bit about electrical, math, and some science to repair it. 6 months later and some help from amazing sources like you I was able to work with an Arduino UNO! Math is a rabbit hole and so is all the subjects its beautifuly displayed in. It helps us to understand this abstract world. I see that school and I were not very capable. In fact I did not understand any of it. Failed alot but I have my GED. There is a lot of ways of learning and I found that I do not "space out". Im a visual person and that works for me when i let myself apply that type of learning and problem solving style. I find engineering is pretty visual and I don't think I'll ever be able to stop asking questions upon many subjects. These videos are very informational and insightful... especially into various ways of looking at something. Thank you! -cyborg with brown eyes 👀

[@spaceguy20_12]

I noticed that the 4 circles in the thumbnail is 3 blue ones, and 1 brown one, placed in spots to kinda resemble 3blue1brown

[@shaikhmohdhamza3126]

Thanks for this video,i remember when i was in 10th class i ask my teacher about surface area and volume of sphere , he said no need to know that just learned the formula , so thanks for this .And one more thing that can make a video of volume of a sphere

[@itolukibami725]

shaikh mohd Hamza that's bad your teacher was only teaching you for marks not for true education

[@RobertsMrtn]

​@@itolukibami725 He probably did not know the derivation. This video is superb!

[@TheBasikShow]

The channel Think Twice has the best explanation for the volume of a sphere that I’ve ever seen. It uses Cavalieri’s principle (which it explains), and the face that the area of a pyramid with height h and base area A has volume hA/3 (which it does not explain, but is clear with some elementary calculus, and has some cute visual proofs). Go check it out!

[@antonofka9018]

1. Circumference: 2pi*sin(th)*R; Area of the ring: 2pi*sin(th)*R^2*dth 2. Area of the shadow: 2pi*sin(th)*cos(th)*R^2*dth = pi*sin(2th)*R^2*dth 3. For them to differ by a factor of 1/2, sin(th) must be equal to sin(2a) (where a is the other angle). So th=2*a. 4. Mapping area of each ring on the top of the sphere on to the shadow (halving the angle for each) we get a circle of shadows, whose radius is R/2, and whose area is 1/4*pi*R^2. After doing the same for the bottom portion, the total area is 1/2*pi*R^2. It is exactly 2 times less the area of the rings, so the area of half a sphere is pi*R^2 and so the area of the whole sphere is 2*pi*R^2, which means there's a hole in my argument but the general idea is correct I guess. * Correction * Actually, when I said that the radius of the mapping is equal to half the radius of the sphere I was wrong as it must be equal to √2/2 since the angle is 45° and cos(45°)=√2/2. And so the area of the portion of the shadow that we get after the mapping is equal to π*(√2/2*R)^2 = π/2*R^2. This way we get the right answer if we proceed with my steps sketched above, in the main part of the comment.

[@TheFlue2000]

Can you explain how you got to 2?

[@nischaldhungana7014]

@@TheFlue2000 the circumference 2pi*sin(th)*R is the same and the projection of thickness, 2pi*R*dth, is 2pi*Rcos(th)*dth. multiply them together for the result.

[@LCJRdeSouza]

That's because you weren’t counting only on only the top ring, but every even (or odd) ring in the whole sphere. So in the end, the area of the odd rings happen to be 2 times that of the shadow (a circle), so getting both the sum will be (2+2) times the area of the shadow, hence Asphere=4*pi*R^2

[@HebaruSan]

@@TheFlue2000 For #2, you can subtract the areas of the circles around the inner and outer edges of the shadow rings, one of which has radius R sin θ, and the other has radius R sin θ + R cos θ dθ. (That + might be a - depending on which exact triangle you use, but it works out the same.) Remember that area of a circle is πR², and (dθ)²=0.

[@antonofka9018]

@@bjornfidder at that point I'm counting the area of only half of the rings. And the total area at the end must turn out to be 4*pi*R^2.

[@TimpBizkit]

Assume rings of width r*dx for small angle dx. Area of ring on surface is 2π*r*r*sin(x)*dx (horizontal distance to center is opposite the angle in a right triangle with R as hypotenuse) Area of shadow is 2π*r*r*sin(x)*cos(x)*dx (projected ring thickness is adjacent the angle in a right triangle with dx as hypotenuse) Using trig identity sin(2x) = 2*sin(x)*cos(x) Area of shadow = π*r*r*sin(2x)*dx So the area of the shadow is half that of the ring with twice the angle from the top. (E.g the shadow of the ring 10 deg from the top is half the area of the ring 20 deg from the top.) The area of every other ring from 0-180 deg (half the sphere area) corresponds to twice the area of the shadow of all rings from 0-90 deg. So the sphere area is 4 x the area of the shadow.

[@johnkinx4387]

The animations are absolutely brilliant.

[@bobbyt9431]

"I mean viscerally feeling a connection between this surface area, and these four circles." And that's why this channel is so popular. If math was taught this way going all the way back to introduction of the base 10 number system, we'd probably be colonizing Mars already.

[@alasdairwinter8723]

I find all these videos pretty interesting but I kinda disagree with them being useful learning tools. The best way to learn maths is to embrace things becoming abstract and stop trying to relate everything to what you can visualise. Maybe I'm biased by being a physicist (try doing quantum mechanics in a way that's relatable rather than abstract!) but I think making people rely on intuitive ways to picture problems becomes very limiting as eventually you'll get stuck when you reach a problem that can't be simplified to something intuitive

[@alexkfridges]

@@alasdairwinter8723 You are right. Everyone online wants the "fun" side of academic fields without the associated work involved to become independently good at it. That's fine, it's awesome entertainment, but is should not be confused as "the proper way to learn ". The truth is, you have to grind the technical/abstract/tedious/difficult aspects of these things if you want to reach a high level. It can't all just be fun animations that make it cool and intuitive.

[@user-ht6ql1rn3w]

You cannot learn all mathematics by visualising, only a tiny fraction of t

[@prdoyle]

Personally, I'm not a fan of the exercise format. I don't watch these because I'm enrolled in a math course and this is how I learn the topic. I watch them in my spare time because I enjoy seeing beautiful proofs presented well.

[@AirF0rce11]

Yeah, me too. I was genuinely interested in the answer to the question posed in the title, but I still do not have the answer. I don't have a piece of paper handy, and I honestly don't want to do it. As much as I don't like to, I had to dislike the video.

[@rishimehta4486]

Mathematics is not a spectator sport. You have to do it to learn it.

[@afewofmyfavoritethings]

I didn't mind the exercises themselves... but I do think this format only really works if he then works through the questions he presented (PLEASE?). This allows for both the individuals willing and interested in working the exercises themselves, those that get stuck trying to do so, and those that have no desire whatsoever to do so but still find value in the videos. :)

[@cygnontango4010]

If education is required to improve, then I will vote for this channel. The animation is superbly great which properly matches the movement of the eye, a great way to learn even with beginners and non-mathematicians. Moreover, to create videos like this, it takes a trench-level of understanding of the topic. What this channel is teaching probably isn't being taught in some schools and universities. It dives into the most fundamental concepts/roots and answers the derivation of formulas we learned in schools. You cant call math a beautiful subject instantly, but in this way, you can see that it is indeed extremely beautiful and interesting. Kudos to this channel and I am thankful that I am born in this era of technology.

[@renagonpoi5747]

10:09 YES this... all of this... I broke my head trying to wrap it around the idea of surface and volume integrals... But due to sheer coincidence I happened to think the other way around and everything just fell into place. I'm proud to say that I now have a solid baseline knowledge of calculus thanks to that, even though it is 7 years after my university.

[@SJ-dc3ei]

I've never seen such beautiful animations in my life!

[@-sui-]

The animations on this one were hypnotic.

[@duif4b]

水 -sui- Yes, smooth and like clockwork, and the equally smooth piano music supports that further. There is more, search YT for satisfying animation and similar videos....

[@raghavkumar00]

My first video watched of your channel. . Liked. Subscribed. Awesome pedaogy. Very intuitive Animation. Thanks a million from India.

[@jf8442]

I always used to love maths in school. And now that I saw mathematics in such beauty as you present it, I really start to miss it

[@MateusKaterp]

these animations are astonishing. congrats.

[@nirbhaykumarchaubey8777]

02:20 The transition from Circles to the Triangle was OP! 🤟😻

[@taboosaboo]

You Are an extrodinary animator! Lessons from You being highly valuable. Learning math and geometry from You is a pleasure. Animation im sure would be too.

[@HackedPC]

Isn't math and geometry the same thing?

[@DanBurgaud]

I've seen this video many times. Yet each time, its as exhilarating as the last one.

[@sudhindrakopalle7071]

The animation is spectacular and the explanation is so well done!

[@harshulkhanna]

Yayy i figured out the proof (challenge mode) ^_^ and what a great video Grant. Display of mathematical elegance. Love your work

[@khikhi6704]

This video explanation more easy to understand than my lesson on my main school, teacher on my main school just talk but never proof it !!!!, thanks for the creator for making this video.

[@tommyinckemann2273]

Me at 3am: I don't need sleep, I need answers

[@KedViper]

9:43 Looks exactly like how video game objects have gained more and more polygons over the years. In the same way as they are there to aproximate the surface area of the sphere and become more and more accurate the more polygons you have, polygons in video games have always been an aproximation of real life objects or rather of 3D models made with relatively unlimited detail.

[@ganondorfchampin]

The secret to good graphics isn't increasing polygon counts, it's using better interpolation techniques that generate to smoother curves. Just compare a sphere-approximation colored with flat shading to one with Phong shading.

[@mihailazar2487]

well, m8 good graphics ain't all about poly count ... it's all in the DISPLACEMENTS ... basically, whenever you want your fundamental shape to be rendered, the computer will choose a light according to a scene enviornment map (basically a 360 panorama that stores reflections) and based on where the normal of the shape points, your shape will have it's base color and the lighting color blended BUT the computer has to calculate this for every pixel in the shape so what you do instead is you go ahead and store bumpyness on the texture itself so when the computer calculates the lighting, it might as well nudge the normal off to one side according to a value stored in the displacement map, a different amount for every pixel in that shape so it then looks like it's got extra depth information even though it's still flat the beauty of this is the fact that it takes up exactly as much processing power to render the shape without displacement as it does WITH displacement, which is why many 3d artists call displacement maps "Free detail"

Instead of using polygons you could use an algorithm that takes into account the distance between the camera and the object and approximate how it should be rendered. In theory it is a perfectly round shape, but on your screen it is actually rendered with more detail as needed (due to the fact closeness to an object uncovers smaller and smaller details). But it is always an approximation, and since pcs can't calculate to infinity..

[@ganondorfchampin]

Calamiro Asjcobeti That is part of how it’s done, but it still requires polygons to define what the shape actually is.

[@yashagnihotri6901]

Well he is a only teacher who doesn't gets angry on asking why , rather he ends up answering even the why's that were gonna spawn in future ! Genius ...

[@benitorossi1076]

Why 2πr^(2) is not true? (2πr/2*2πr)

[@greenfocus7552]

THANK YOU. All the animations superbly reflect the thought process required to understand

[@ashutoshbhatt566]

Very cool explanation

[@vishaljethva]

I've never seen such a powerful animation study in my life. Great job man.. keep going 💪

[@pursuitsoflife.6119]

13:27 who just started integrating 2πR^2(sinθdθ) from 0° to 180°?

[@kaiwalpanchal5728]

this would mean nothing, though maybe

[@dee8163]

That is actually what I predicted the proof would be! I haven't yet wrapped my head around the one without integration lol.

[@breadandbutter644]

This will give u the surface area of sphere. But we are here to prove that area of sphere is equal to area of rectangle ( of height equal to diameter of sphere and lenght equal to circumference of sphere) Not to prove that area of sphere equals to 4πr^2. But i also started doing the same thing u said...😂😂

[@JensenPlaysMC]

@@dee8163 that is why mathematics is such a powerful tool as you do not need to intuatively understand "why" for something to be proven

[@dee8163]

@@JensenPlaysMC oh god you're right. This is a scary thing to have to confront. (I'm applying to colleges for maths honours currently)

[@prototropo]

I utterly agree that visualizations of a concept, like calculating area or volume, is useful to try before learning the equations and operations of pencil-on-paper calculation.

[@livintolearn7053]

Amazing!! I've been wondering about this sort of an understanding of the formula. But I never found one... This is it! Finally!!!!! And it's beautiful!

[@FareSkwareGamesFSG]

15:15 Interestingly enough, I found this fact to hold true for shadows of 2 dimensional regular polygons onto a 1 dimensional line. If you imagine holding them perpendicular to the ground at all times, a 2d object will cast a 1d shadow. The average length of that shadow turns out to be 1/π 'th of the perimeter of the polygon. I'm not sure if this holds for all convex shapes, but it also holds for a circle, the average shadow of which is simply its diameter.

[@TheShamansQuestion]

Cool idea and sorry to be a nitpicker, but isn't the only 1-dimensional shadow for a circle just its diameter? Rotating it any other way so it's perpendicular and still projects a 1d shadow is just to spin it (there are no corners) so it just stays as diameter. Any other orientation change doesn't leave a 1d shadow.

[@zar6]

@@TheShamansQuestion Yes, the only 1-dimensional shadow for a circle is just its diameter in the exact same way that the only 2-dimensional shadow of a sphere is a circle of the same radius, due to their rotational symmetry. It's still accurate to say that the average shadow of a circle is its diameter.

[@TheShamansQuestion]

@@zar6 i think I was picking up the fact you said "the average" shadow is diameter, rather than all But I guess on a 1d line it could be two dots if we hold the circle parallel to the line (the 2d circle "intersects" the line at two points... Or its 2d shadow would)

[@nirajrajkharel5848]

The video, the explanation, the simplification and the object of the work -- all are in point. This is how we need to learn maths. Mr 3B1Br, I'm really honored to watch your videos and the way this inspires me is inexpressible. Thank you so much!

[@EngineerHunter]

The explanation is so good.

[@antoined5849]

Your animations are absolutely mind blowing ! Which software do you use ? Thank you very much for such great contains. Greetings from france

[@supreetsahu1964]

MAnim software

[@benitorossi1076]

Why 2πr^(2) is not true? (2πr/2*2πr)

[@mrnerd3143]

greetings from america

[@austinsharpe8157]

@@bamberghh1691 Thanks very much

[@LilCalebW]

Me at 12:34 I LIKE THE CYLINDER I REALLY LIKE THE CYLINDER

[@keaganwells318]

Great video as always, just finished cal 2 with an easy A thanks to some help from your vids on Taylor series!

[@williammook8041]

Easy A are the best.

[@skyli7711]

Finally!!! After 3 yrs!!!! You updated another video about the average shadow area!!!!! Thank god you still remember!!!!

[@thomasboudier1152]

3Blue1Brown: "that's negligeable" Me: *dies*

[@debonairrose]

*giggles*

[@vedant6633]

I feel you should sell more mathematical stuff on the 3b1b store, things like mechanical calculators, harmonic analyzer, some mathy visualization tools, some new version of chess u invented etc etc etc..... That represents you better than clothing with math printed on it...

[@3blue1brown]

Great point! The most honest answer here is that producing harmonic analyzers would be much harder, specifically in that getting a third party to handle logistics would be harder. I want to spend most of my time on videos, and for me, the store is a nice little way to promote expressions of a love of math while not pulling me away from the main pursuit too much. That said, I do agree with you, so will put in some more thought here... Anyone know a good harmonic analyzer supplier ;)

[@emmanuellopez3471]

@@3blue1brown MathMo deal striking!

[@markellospatras3039]

I like what you are doing here. Back when I was into maths, I wasnt as intrested in the proofs themselves as to understanding why this is true. I understand why maths and logic are suposed to focus more on proving and less on explainin the reasons behind facts, but Im glad I see more people being intrested in what I am.

[@ravirajmayapur]

However good the visualisations may be, I feel the explanations top them, because the clarity of understanding oneself, what needs to be explained, and the ability to do that is extremely rare. It comes from understanding the core principles in logic of the subject, and not just the technical aspects of the subject. This teacher has really understood this topic.

[@yashrawat9409]

13:05 At this point I was convinced enough that we might as well integrate the area of these small rings

[@DingbatToast]

"In the spirit of mathematical playfulness..." Not a phrase I will ever use! 😆

[@bryanfuentes1452]

oh my god...this was one of my questions when i was in high school taking geometry and i was asking" is it coincidence that the area of sphere is whole numbers of area of 2D sphere?"

[@vals6004]

You break my brain in the most beautiful way. Thanks