Understanding the Surface Area of a Sphere Formula

[@learnerlearns]

BEAUTIFUL presentation! Clear. concise, organized, with good graphics and pacing. Thumbs up and subbed!

[@flower_girl4983]

how am i supposed to understand this stuff?

[@sam-ui5lc]

@@flower_girl4983 learn the basics first for example the properties of triangle and other basic shapes, then go for average (this video) and finally the difficult ones. That's how you can master the art of learning mathematics

[@mikasaackerman3946]

@@sam-ui5lc ok sam thnks a lot

[@JustAgreenBoy6969]

@@sam-ui5lc and limits too

[@dannygjk]

An elegant method to derive the formula for the area of the surface of a sphere without using calculus.

[@dannygjk]

***** Just intuitively, humans have been doing that in math long before calculus. One thing that is cool is that mathematicians were close to inventing calculus back in the ancient historical times. I forgot the details but I think I read it in a Lancelot Hogben book.

[@Kemerover]

+Dan Kelly he used formulas for a cone and a frustum. How are you supposed to do it without calculus?

[@dannygjk]

Kemerover He didn't use integrals did he? I only watched the surface area of a sphere part. Oh when you said cone I started to think about volumes sorry. Anyway not all formulas require calculus to derive them. Some formulas were derived about 1500 years before calculus. Some of the concepts that are used in calculus were developed about 2,000 years ago.

[@ivanereiz1533]

+Dan Kelly i agree on what he did.. i can understand this... but if he used calculus i coud not

[@dannygjk]

Ivan Ereiz Yes he just used one of the concepts used in calculus but he didn't use the 'language' of calculus so this is a nice piece of work that everyone who has high school algebra can understand.

[@Westkane11]

One word: "Perfect!" This presentation couldn't have been done better.

[@mikasaackerman3946]

Thnks

[@hermesmercury]

If you were expecting a simple answer...you were wrong.

[@levi4328]

That's why people are watching this video: the formula is so simple.

[@stephyelle1]

Hermes Mercury simple mind?

[@maacpiash]

Simpler than integral calculus.

[@TheZMDX]

Well it wasn't THAT hard to understand :P

[@landlord112]

Hermes Mercury If the diameter is the same volume of the circumference, then it'd have a ratio of 4, am I wrong?

[@kagaminefever]

Thank you very much! This video is very helpful for students like me who have yet to learn calculus, but still want to understand what they're doing. I can usually come up with my own "proofs" for most formulas, but when it came to spheres I was completely lost. Now it makes sense to take a polygon with infinite sides, just as you do with the circular(ish?) part of cone. Thanks! Really hits home the beauty and creativity of math, especially for a subject that most people assume is dry with no room for creativity.

[@mujtabaalam5907]

Check out the 3b1b episode too

[@IsaacAsimov1992]

I relate 100% to your comment!

[@zachansen8293]

The better way to spend your time is to learn calculus.

[@kagaminefever]

@@zachansen8293 yeah that would have been really easy during o levels man thanks bro should have just done that on top of studying only 2 months

[@davidsica8996]

Beautiful! Raw simplicity & beauty of mathematics presented with clear & concise explanation and graphics. It doesn't get much better than this! Thank you, thank you, thank you!

[@anthonygopeesingh7645]

Damn who came up with this -_- i understand but i would never think of something like this. Imagine you were in a time where all you had was a sphere in your hand and someone was able to think of this AMAZING

[@kazukawasaki97]

manipulating pre existing resources and bending it to fit to a new problem was my introduction to calculus,its not easy,but does not take a genius to do this, fascinating regardless

[@shampadutta7322]

By far, the most elegant and unique derivation of the formula, without calculus, which makes it understandable to a larger number of students. A mathematical elegance presented in clear and concise graphics and a truly immaculate approach. It can't get any better. Thank you, on behalf of all the students who are not yet introduced to calculus! Beautiful. Subbed instantly!

[@IsaacAsimov1992]

I totally agree.

[@math2693]

I can't believe this channel is not that popular omg it is precisely amazing

[@mikasaackerman3946]

Thnks

[@mikasaackerman3946]

Thnks

[@danmarino900]

interesting how the area of a circle is pi*r^2 but the (surface) area of a sphere is pi*d^2

[@commentercommenting328]

That's an even simpler mnemonic tool.

[@acpf4f]

thanks Ryan.

[@purushottammotwani3082]

Ryan Bell thnx

[@Bluedragon2513]

The surface area of a sphere is 4*pi*r^2...... please explain someone

[@BouncingHope]

A= 4pi*r^2.... r= d/2....... so 4*pi*(d/2)^2 => 4*pi*(d^2/ 4).......4's cancel and all you have left is pi*d^2. Hope this helps.

[@lucanina8221]

What do you use to edit the video? The animations are so clear and helpful. Superb proof!

[@stephyelle1]

luca nina Archimedes proof.... 200 years before JC!

[@mikasaackerman3946]

Thanks

[@mustafahassan3584]

@@stephyelle1 Amazing to think how Mathematicians used to derive this stuff back then when Maths wasn't this advanced

[@mujtabaalam5907]

@@stephyelle1 his "proof" was more of an experiment test by comparing the volume of a cylinder with the volume of a sphere plus a bicone. There's a numberphile video about this

[@sushantnair2584]

I loved the video! It made the concept clear. If I watch this video 2 times to understand the problem, then, without this video, I would have understood the concept only after 200 times of reading the textbook!

[@markhatton6449]

Fantastic - beautifully clear explanation.

[@mikasaackerman3946]

Thnkx

[@dekippiesip]

Another elegant method is using the volume of the sphere to deduce it's surface area. The volume is 4 pi/3 r^3, curiously the derivative is 4 pi r² or the surface area. This is no coincidence. Take the function V(r) = 4 pi/ 3 r^3 and take the derivative. That is, (V(r+h)-V(r))/h as h goes to 0. Geometrically this represents the difference in volume between a sphere and a slightly bigger sphere. Then divide that by the difference in the radius, intuitively it's clear that you get better and better aproximiations of the surface if that difference get's smaller, so the derivative must be the exact surface area and there you have it. Very intuitive.

[@tonybidwell2582]

dekippiesip As UNBELIEVABLE as it looks, if U use derivatives, 4*Pi*(r^3)/3 turns into 4*Pi*(r^2)!

[@JorgetePanete]

dekippiesip its*

[@JorgetePanete]

dekippiesip approximation*

[@JorgetePanete]

dekippiesip gets*

[@kartikraj1779]

I think volume is derived using SA itself! By integrating SA for all r from 0 to R. So u can't use that.

[@curs3d975]

Beautiful! Just a random question emerged in my mind when I was solving physics problems turn out to has one of the most fascinating explanation about math that I've ever watch. The video is clean and smooth, didn't expect this quality from a 2013 KZbin Video. Thank you so much!

[@mathematicsonline]

Glad to hear you enjoyed it!

[@leif1075]

@@mathematicsonline Thanks for sharing but I just don't see how or why anyone would cone up with this at all? Especially since it's so convoluted and unintuitive. My idea for a proof is 4 pi r squared is 4 times the area of a circle so you can think of a sphere as having four "faces" like a box has four faces. So you can thinkof a sphere as made up of four 2d circles projected into 3d space and hence the area is 4 times the area of a circle. This seems to me like a valid alternative proof?

[@mathematicsonline]

@@leif1075 It is an ancient proof by Archimedes, it gives us insight to early mathematics.

[@rarebucko]

Theres a much simpler proof: To form a sphere, you must rotate a circle around its diameter. And, if you look, you can see that the surface area of the sphere is equal to the circumference of the shadow times the distance it was rotated. So we plug in: “SA=2πr*o” in which o is the distance the circle was rotated around. Now, if we look AGAIN, we can see that the distance it was rotated around was actually equal to the diameter. So next we plug in: SA= “2πr*2r”. Simplifying, we get “SA=4πr^2

[@Singh-be2qn]

Very nice bro

[@joshuaronisjr]

Hey...I'm losing you. "To form a sphere, you must rotate a circle around its diameter." Okay, that makes sense. "The surface area of the sphere is equal to the circumference of the shadow times the distance it was rotated." Again, that makes sense - and the circumference of the shadow would be equal to the circumference of the circle. "So we plug in: “SA=2πr*o” in which o is the distance the circle was rotated around." Got you. "Now, if we look AGAIN, we can see that the distance it was rotated around was actually equal to the diameter." Wait a sec...why is the distance it was rotated around equal to the diameter? If I have a circle, and I rotate it by 180 degrees with a diameter of that circle as its axis, and let the points on the edge of the circle trace out a surface, points on the part furthest will have moved pi r, and points closer will have moved less...how did you get that the distance rotated around was equal to the diameter? Different points on different parts of the circumference of the circle rotate by different amounts.

[@gligoradrian784]

@@joshuaronisjr True, but the distance is constant, and it's equal to pi * r , as you move it "Half the sphere".

[@joshuaronisjr]

@@gligoradrian784 What distance is constant?

[@gligoradrian784]

@@joshuaronisjr I mean, the 180* around which you rotate the circle, and also pi.

[@sylvainpilon9496]

wow! With demonstrations like this, the schools would keep attention of students instead of them losing interest because they don't understand where the formulas come from ! Bravo!

[@gavincraddock5772]

Thanks for this - I expected a very complicated explanation,but actually it all made sense. Great video.

[@jackmack1061]

Nearly lost me for a moment but I'm very glad I stuck with it. November 2021. You just wouldn't believe what's been going on.

[@HollywoodF1]

The area formulae for the surface area of a cone and a frusturm are presented as though they are trivially obvious. If that's the case, then so is surface area of a sphere. But given that we are attempting to derive the latter, we should derive the former.

[@ShreyShidhaye]

This is the definition of a perfect video

[@Junnnn___]

To be honest with yall, this guy's explanation is excellent fr

[@사니-f6f]

It's crystal clear. I can use this way to enhance students understanding. The way I like that

[@VReinthal95]

Maybe it would be a good thing to tell, that the small formula r_1 + r_2 + r_3 = AE * AD / 2s works in a similar way for all polygons you choose. Otherwise the resulting area of the polygons might change while increasing the number of vertices.

[@davidbrisbane7206]

Indeed. The formulas are probably true when there are more than 2 frustrums, but that needs to be demonstrated, which it wasn't.

[@lyrimetacurl0]

That was amazing but I would have thought there would be a simpler explanation? Like using a hemisphere:- Surface area of a circular strip = pi * (r1+r2) * l As it goes to infinitesimal, r1 + r2 become the same, so 2r So 2 pi r * integral of all the ls would give the hemisphere. All the l's are straight lines along the radius, added up for the hemisphere gives r So 2 pi r^2 The multiply by 2 for the sphere: 4 pi r^2 Or is this insufficient proof?

[@tylerpruitt9572]

if you wanna look at it using a different calculus approach then it's the derivative of the volume which makes sense if you think about how the surface area is pretty much the rate of change of the volume

[@swn32]

That's just "reducing" a simpler problem to a harder problem.

[@QwertQwert-qo3le]

Nyx Avatar what is calculas

[@mr.moodle8836]

You're right, however, as someone who's curious but not up to calculus yet, I really appreciated this proof. It was simple and only required a decent understanding of geometry and manipulating equations, making it more accessible to a far wider audience.

[@tiscojack]

But how do you derive the volume? Btw what you stated isn't always true, for example in a cube the rate of change of the volume is only half of the surface area, cause increasing the side only affects one direction, which would be analougous to the derivative of the sphere volume with respect to d

[@connoribbotson1337]

I always get slightly confused when I think of it this stuff using derivatives. Like if you differentiate a circles area (pi r^2) then you get 2 Pi r - the circumference. Differentiate that and u get 2 Pi, the amount of radians in a circle. But what happens when you differentiate that? What’s that? And when you differentiate a spheres volume, you get the surface area, differentiate that and u get 8 Pi r - the circumference of a sphere??? It just leaves to many loose ends...

[@shynnsup8383]

Who was hoping he said R2D2?? Please tell me I wasnt the only one

[@cosmopolitan4598]

Hahahaahahaha

[@connorcriss]

C^3*PO

[@JorgetePanete]

Shynn Sup wasn't*

[@Alex_science]

Great. I have never seen a clear explanation like this!

[@portalsrule1239]

wow. my jaw is on the floor. I loved how it all simplified so nicely in the end. Great video, btw!

[@sidaliu8989]

Thank you for your patient explanation, but I think there is still one flaw in the line of reasoning: since we derived the formula of the surface area of model = pi*AD*AE only when the number of sides of the inscribed polygon was 8, how could we use it for n is greater than 8?

[@mathmaticalproblemandsolution]

brilliant explanation i think this explanation contain all procedure that we study from basic level....which is easily understandable but ....some teacher go directly to the formula and did not teach the basic concept ....i think every theorem should be taught like this way ......

[@keiichiiownsu12]

If I wanted to just, say, take a circle, measuring only its circumference, then rotate that circle an increment of dθ, then basically keep rotating that circle dθ, summing up each circle's contributory radius until I went around 2π, i.e. integral from 0 to 2π of the circumference of a circle rotating about dθ, would that give me similar results? I find calculus gives somewhat more intuitive answers sometimes

[@kshitizpoudel-jp7cc]

can't you keep a rectangle inside the circle for easy calculation of area

[@guitarttimman]

A double integral in polar coordinates works best!

[@VivzStudioSs]

What about surface Integrals?

[@AlchemistOfNirnroot]

+VivzStudioSs "Double integral"

[@kartikraj1779]

VivzStudioSs Surface integral turns out to be a double integral

[@guitarttimman]

It takes an understanding that the differential surface area of a simply bounded function in three space is linked to the sum of the magnitude of the cross product along two bounded axis. The double integral with the differentials alone is the area, and the function defines the height at each mid point. You need to understand partial derivatives to understand what I'm talking about. Deriving the surface area of a sphere is straight forward, by what if you have a simply bounded elliptical hyperboloid with specified boundary parameters?

[@singh2702]

@@guitarttimman Take two coordinates (rcosx , rsinx) , people don't realize that if sinx is increased by r times then that new point , rsinx , is increased by r in the x direction. So before integrating all points on the circumference , rsinx must be multiplied by r again giving us r^2sinx. Then it's easy to integrate to get surface area take the definite integral of 2pir^2sinx over o-pi.

[@curtisnotestine3134]

How are the surface areas of the cone and frustum derived?

[@skrd37]

The best explanation over youtube. Thank you very much.

[@shotaaizawa1888]

complex concept, but brought forward in a simple and understandable manner. thanks a bunch man

[@EDUARDO12348]

I was looking at the formula for a sphere the other day in a math book expecting myself to derive this formula in my head, clearly, my brain would have exploded if I had really tried.

[@robkb4559]

The overall approach is nice and your animation is very clear, but I don't think you formally generalised from the octagon you started with to the general 4n-gon that's required for the limit process to be valid. Or am I missing something?

[@guhaonkar]

Beautiful! Simply... Beautiful! Thanks a lot for this simple explanation to the otherwise seemingly complicated problem. Thank you!!!

[@haroonrasheed3844]

Its a beautiful proof. But I have one question. The realtionship you derived in the form of r1 + r2 + r3 is only applicable to an octagon. And so is the area formula you wrote down for the frustums and cones. So when you increase the sides of the polygon, shouldn't the area formula and the r1+r2+r3 relationship become wrong? Why is that allowed?

[@eskimo227]

can you do the surface area of the cone and frustum? I'm trying to understand this down to its beginnings

[@alxjones]

This video should be called "How to derive the surface area of a sphere (assuming you somehow know the surface area of a cone and a frustum)". If you're going to approximate the sphere with cones and frustums, why not approximate those surfaces with triangles and trapezoids? Deriving the area of those objects is actually pretty easy, so you only need to derive those simple polygonal areas and you can derive this fact. This is more useful as a derivation than assuming knowlodge of the surface area for some uncommon solids.

[@muratbilir4240]

Can you please explain why the triangle ADE is similar to other triangles.

[@Elseano14]

That was cool. When you mentioned many little sides, I immediately jumped to the idea that limits were to be involved. (Technically they were, but is was phrased in a different way)

[@joy2000cyber]

Very intuitive. Maybe a comment that this derivation also applies to polygons with more than 8 sides, would be perfect.

[@Metal_Master_YT]

Well, since the formula for the surface area of the model simplified down to pi times the circumference times segment AD, that indicates that the number of sides of the polygon does not affect the accuracy of the answer, since by only filling out the circumference, (which is unchanging) and segment AD, you can calculate the surface area.

[@morgengabe1]

Any way to do this using shapes that are simpler than a sphere?

[@bobvonbuelow9983]

would have liked to see .5! on the graph and maybe points between the integers too. since 0! is 1 on the graph and sqrt(pi)/2 isn't one, what does the graph look like

[@Croftice1]

Wait a moment, why that extra step with r3 at 2:53, when it's equal to r1? We're talking about a sphere, so the left side and the right side are equally big. I like this video, but that one step is completely unnecessary. The cones on the left and right are identical, they are the same shape, the same length, even the same color (don't mind the last one), so why to label that line differently from the r1?

[@NobleTheThinkingOne678]

I had figured it out on my own but wanted confirmation that I was correct. I was. Anyways, the point of this comment is that this video was beautifully illustrated and explained. Also, that math has many avenues by which one can reach the desired answer. What I did is I drew a sphere and drew two circles in it on the x, y and z-axis. Then I drew a separate diagram of one of the circles. I know that 2(pi)r or (pi)d were my circumference. I used (pi)d. I then imagined another diameter on the z-axis coming from the first circle. I then multiply (pi)d*d. I got(pi)d^2. I then converted d^2 to r. I got 4r^2. This gave me 4(pi)r^2.

[@cba8351]

Math really does builds upon itself

[@josephprashanthbritto8349]

Mathematics basics are explained very clearly . Great work nicely done. Thank you

[@weirdshamanwizzard3156]

The guy who cane up with this clearly had a love for geometry

[@patrickfeng5066]

When you started aproxomating the figure to infinity, wouldn't all of your work with the similar triangles been rendered useless? All the work assumed that the figure was 8 sided.

[@cheongziyong8871]

They work for any polygon with an even number of sides

[@agarethy]

Super high quality and very polished! Great for people who haven't learned calculus yet! For a (much shorter!!!) Proof using trigonometry and calculus, do a youtube search for "Proof of Surface Area of a Sphere" (Not my channel, just promoting another good video :)

[@najibqunoo7232]

9:05 you mean that AE and AD are both equal to the diameter of the sphere ; so here you will have it like this SA=pi*AE*AD SA=pi*d*d SA=pi*(d^2)

[@DulksVenee]

I don't want to boast about this, but we had this assignment in the exam to prove 4(pi)r^2 to be a sphere's area, and I got the max score. Of course we were given the required formulae. I assume that's because they assumed that the genius who discovered this had his notes to help him.

[@divyanshusah8311]

Please also make a video on formula of (A3-B3)=

[@h1a8]

Good job. But someone would want to know where the lateral area of a right circular frustum comes from (which is derived from the lateral area of a right circular cone).

[@dsy9578]

The best explanation I ever seen thanks buddy I'll be your subscriber forever

[@mathematicsonline]

Appreciate it!

[@JohnDixon]

Wow. This is like, proofs to the max. I've never seen such a complicated proof about spheres; great job!

[@odysseytkl7261]

Hi like im dad

[@vekidf1]

i had one doubt that the hypotenuse of a right angled triangle can never be the same as any other side hence AE can never be equal to AD so its a bit inaccurate but it surely does make sense when u round it off

[@HecticHector]

U just made ur life harder bro good job

[@guitarttimman]

I posted some Calculus videos on my channel which is just a sample of what I know about the subject. I do an eloquent derivation using single integrals.

[@qwerty11111122]

What was the step that allowed for the approximation of the polygon's area to approach the surface area of a sphere? It went from 2-D to 3-D and I didn't see how

[@leenagupta6586]

Thank you for your help.. 😊. Was looking forward for such theory and I guess I got what I wanted to see!

[@user-rs8965grt]

Thank you. I was always wondering but never got such an explanation.

[@MegaJayPower]

Very good comprehensive video. I always tend to take these formulas for granted.

[@banajadandasena4142]

Animations and explanations are best... thanks for making this types of videos.

[@ezrapotter4631]

From a calculus standpoint, the surface area is the derivative of the volume, 4/3pi(r^3)

[@mauk2009]

Best explanation i ever seen on youtube.

[@SocratesAlexander]

how do you animate? it seems very time consuming if you do it without any automation.

[@kannusingh7003]

I like how the color of the 3d model become black at infinity which is extremely true

[@ffggddss]

IOW, the surface area of a sphere is A = 4πr² = πD² = area of a circle whose radius = the diameter of the sphere. Hmmm. That diameter happens to be a ball of 1 dimension with radius r. Rotating about one endpoint makes a circle whose area = surface area of a 3-dimensional ball. This is, rather amazingly, a pattern that holds in any number of dimensions. And knowing that the volume of an n-dimensional ball is (r/n) times its (n-1)-dimensional surface, you can derive the formulae for the volume and surface of a general n-dimensional ball, using as your "starters," the 1- and 2-D cases. Call this, the "D+2 Theorem," because it allows surface and volume of an (n+2)-dimensional ball to be found from those of an n-dimensional ball. In order to carry this out, you will need Pappus' Theorem, which says that the n-D volume or (n-1)-D surface of any n-D figure, H, formed by rotating an (n-1)-D figure, G, about an (n-2)-D hyperplane that's in the same (n-1)-D hyperplane as G, is the (n-1)-D volume or (n-2)-D surface of G, times the arc length swept out by the centroid of G as it's being revolved. (Yikes!!) Well to help clarify that, use the example above. Here, n = 1; n+2 = 3. Take the line segment (1-D ball), L, of length 2r, and rotate it about an endpoint. Its centroid is its midpoint, which sweeps out a circle of radius r, and thus, of length 2πr. Pappus' Theorem says that the area of the circular disk generated by this, is A = L·2πr = 2r·2πr = 4πr² The D+2 Theorem says that this is the surface area of a sphere (2-D surface of a 3-D ball). And the (r/n) rule says that the volume of that 3-D ball, V = (r/3)4πr² = (4/3)πr³. For n = 2, n+2 = 4, the 4-D ball, take a 2-D ball (a circular disk), and rotate it in space about a tangent line. This makes a "donut," or torus whose minor and major radii are both = r, so that its "donut hole" vanishes. The centroid of the circular disk is its center, which is at distance r from the tangent line (rotation axis). Pappus' Theorem says that the volume of that torus is V = A(disk) · 2πr = πr²·2πr = ·2π²r³ The D+2 Theorem says that this is the surface volume of a 4-D ball of radius r. And the (r/n) rule says that the volume of that 4-D ball, V = (r/4)2π²r³ = ½π²r⁴. Note carefully that while this is a handy trick, I haven't supplied a proof. That would be too involved to go into here. Fred

[@hamiltondepaula]

the best thing is when you can understand, that's proportionate by a good explanation, thank you. Muito bom, pena não haver canais assim em português.

[@TsegayMehari-y8h]

how to calculate the Spherical Dome volume, Example, the thicknes is 15 cent meter

[@chapalex1872]

We can prove it by integrals too. And I think it's better! But your proof is pretty good too!

[@monikagoyal7227]

I hadn't thought it's so complex

[@francisfellippeamorgumahin5619]

There's something that doesn't feel right --at least for me. The narrator derived the formula of SA of a Sphere by first establishing the relationships of the triangles within an octagon as shown around 6:21. However, around 8:46, they assumed a model with an infinite number of sides; hence they argued that line segment AD is essentially AE --the radius. Doesn't assuming an infinite number of sides messes with the relationships established on 6:21 (because these relationships were acquired from an octagon)? Can someone to me where in my idea go wrong? PS: Im only a high school student, go easy on me. :)

[@denzelbacarra1082]

Why is the derivative of the volume is the surface area? and the derivative of the area is the perimeter?

[@ffggddss]

Derivative with respect to the radius, that is. Because you can form a ball by piling up layers of infinitesimally thin spherical shells; or a circular disk by piling up layers of infinitesimally thin circular annuli. That is a process of integration; so to go back the other way, you just differentiate. Those processes work in any number of dimensions. If you know the n-D volume, V, of an n-D ball, you get its (n-1)-D surface, A, by differentiating wrt r. Which amounts to multiplying by (n/r), because V will always be proportional to r^n, whose derivative is n·r^(n-1) = (n/r)·r^n; so that A = dV/dr = (n/r)·V. Fred

[@darshansinghrahal2549]

You are describing by taking similar triangles in plane geometry at the same time you are finding total surface area of sphere

[@pauldifolco5736]

Awesome video. Animations were clear and helpful and the proof was simple and beautiful. Liked and Subbed!

[@WilliamMcCormickJr]

The square of the length of one side of a cube, times six gives you the surface area of a cube. A sphere that exactly fits inside that cube, having the same diameter as the length of any side of that cube. Will have an area equal to pi times, the diameter times diameter, of that sphere. The volume of a cube is length times width times height. The volume of a sphere is diameter, times diameter times diameter, times pi, divided by six. So the volume of a cube times pi divided by six will give you the volume of a sphere with a diameter equal to one side of that cube. Sincerely, William McCormick

[@geckchanhong4513]

It is a beautiful proof. I enjoy reading it. Just one comment on how the prove can be generalize to (r1 + r2 + ... rn) = AE * AD / (2s) by specifying "diagonal" lines are between two consecutive vertical lines and the triangle form are similar. I do not see how the extension is achieved when I first read the proof. May a diagram of more triangle with ... between is shown. Once again, thank for the excellent presentation. I love it! Also, one observation, the angles are the same because there subtend the arc length.

[@mathematicsonline]

Thank you for your comment!

[@joeywarren]

Well done. Classic proof with great explanation and illustration.

[@Everest314]

So if we already know how circles and rotational bodies work at the beginning, why don't we just construct the rotational body of a circle?

[@Kokurorokuko]

You counted the base area of a cone and top base of that other figure but they dont make the area of s sphere because they are inside

[@anzatzi]

awesome graphics--what program are you animating with?

[@DeathScakez]

mind blown ! i've never thought of this before it's a master piece

[@johnno4127]

This does not show the step between a 4 frustum approximation and an 'n' frustum approximation of a sphere. (Counting cones as frustums with a top radius of 0.)

[@akshitasiddhapura4626]

One of the most helpful answers

[@flower_girl4983]

i just wanted basic sphere formula and how to get an answer for it, i don't think in exam we would have any time to do any this analysis. Do i have to known all this? ( i am doing basic GCSE foundation, this looks like University stuff)

[@InDstructR]

No u don't need to know this, and when you learn calc(b4 uni) it becomes much easier. Calculus gets this done in 20 seconds.

[@ketofitforlife2917]

That was just... BEAUTIFULLY done! Thank you!

[@djokoadiredono5776]

I never expect that the formula of surface area of a sphere can be derived by a triangle .. so simple ..

[@highguardian13]

4:58 I bet r2d2 is the only reason you named the sides r and d in the first place

[@hoodiesticks]

1:03 The polygon's sides can't be an arbitrary number, it must be even. Doesn't affect the formula, but I get bugged by those mistakes.

[@DavidSwartwoodPDX]

At 2:05 he clarifies that the solution he is showing relies on an even number of sides.

[@farahjaber6024]

it's the same as perimeter of the base ( the circle ) 2πr*the height 2r = 4πr^2=πd^2

[@mewsis14]

In Calculus we DERIVED the surface area as well as the volume of a sphere. 4/3 Pi (r)3 was derived.

[@AlchemistOfNirnroot]

Archimedes didn't have integral calculus.

[@benjaminbuynay1]

Thank you. Can you put a subtitle on it?

[@SciphiMath]

What mathematical software do you use for your dynamic algebra?