@@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
***** 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 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.
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.
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.
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!
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!
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
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
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 UA-cam Video. Thank you so much!
@@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?
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!
@@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
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!
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.
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 ......
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)
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 :)
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
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.
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.
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.
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
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.
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
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...
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.
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.
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.
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.
Hey, I love your videos! They make everything so much clearer about math! I actually do not quite get proofs for the law of cosines, so I was hoping you could do a video on it. Thanks!
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.
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?
@@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.
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?
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?
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
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.
An explanation that is easy for students to grasp is the physical size relationship of the inscribed circle to the square or cube. So this is for students looking to work out how to calculate the volume or area of a sphere. A ratio of a square to inscribed circle is approx. ¾ or 0.75 or π/no edges = 4 … So, if your circle is 2cm diameter, then the square is 2cm wide x 2cm high and its perimeter is 2cm * 4 edges (8cm). The corresponding circle that inscribes the box is therefore about 6cm or (8cm * (π / 4)). The area of the box is 4 cm2 so the area of the circle is about 3 cm2 or (4 cm2 * (π / 4)). ratio = π / 4 ratio * perimeter(square) = circumference(circle) cm ratio * area(square) = area(circle) cm2 Similarly constructed ratio of a cube to inscribed sphere is approx. ½ or 0.5 or π /no faces = 6 … So, if your sphere is 2cm diameter, then the box is 2cm wide x 2cm high x 2cm deep and its surface area is 4 cm2 * 6 faces (24 cm2). The area of the corresponding sphere that inscribes the box is therefore about 12 cm2 or (24 cm2 * (π / 6)). The volume of the box is 16 cm3 so the volume of the sphere is about 8 cm3 or (16 cm3 * (π / 6)). ratio = π / 6 ratio * area(cube) = area(sphere) cm2 ratio * volume(cube) = volume(sphere) cm3 I realise that this may be obvious to everyone here, but the reason I mention this is just that ratios seem to be a much simpler way for students to grasp the concept that an inscribed circle has a linear relationship to the area and perimeter of the square that bounds it … as does a inscribed spheres area and volume to the bounding box. Students easily grasp the volume of a box, by counting blocks, and knowing the relationship of a corresponding sphere is a fixed ratio, allows them to explore how that ratio was derived.
To determine any ratio from a regular polygon for a circle inscribed within the following formula can be applied to both perimiter and area: n = number of sides eqn = pi/n.tan(pi/n) tri = pi/3.tan(pi/3) sq = pi/4.tan(pi/4) hx = pi/6.tan(pi/6) oc = pi/8.tan(pi/8)
BEAUTIFUL presentation! Clear. concise, organized, with good graphics and pacing.
Thumbs up and subbed!
how am i supposed to understand this stuff?
@@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
@@sam-ui5lc ok sam thnks a lot
@@sam-ui5lc and limits too
An elegant method to derive the formula for the area of the surface of a sphere without using calculus.
***** 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.
+Dan Kelly he used formulas for a cone and a frustum. How are you supposed to do it without calculus?
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.
+Dan Kelly i agree on what he did.. i can understand this... but if he used calculus i coud not
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.
One word: "Perfect!"
This presentation couldn't have been done better.
Thnks
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.
Check out the 3b1b episode too
I relate 100% to your comment!
The better way to spend your time is to learn calculus.
@@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
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!
I can't believe this channel is not that popular omg it is precisely amazing
Thnks
Thnks
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!
I totally agree.
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
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
If you were expecting a simple answer...you were wrong.
That's why people are watching this video: the formula is so simple.
Hermes Mercury simple mind?
Simpler than integral calculus.
Well it wasn't THAT hard to understand :P
Hermes Mercury If the diameter is the same volume of the circumference, then it'd have a ratio of 4, am I wrong?
your channel is amazing, thats true mathematics! thank you! the world really needs your!
Fantastic - beautifully clear explanation.
Thnkx
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 UA-cam Video.
Thank you so much!
Glad to hear you enjoyed it!
@@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?
@@leif1075 It is an ancient proof by Archimedes, it gives us insight to early mathematics.
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!
What do you use to edit the video? The animations are so clear and helpful.
Superb proof!
luca nina Archimedes proof.... 200 years before JC!
Thanks
@@stephyelle1 Amazing to think how Mathematicians used to derive this stuff back then when Maths wasn't this advanced
@@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
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!
This is the definition of a perfect video
Great. I have never seen a clear explanation like this!
It's crystal clear. I can use this way to enhance students understanding. The way I like that
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.
dekippiesip As UNBELIEVABLE as it looks, if U use derivatives, 4*Pi*(r^3)/3 turns into 4*Pi*(r^2)!
dekippiesip its*
dekippiesip approximation*
dekippiesip gets*
I think volume is derived using SA itself! By integrating SA for all r from 0 to R. So u can't use that.
complex concept, but brought forward in a simple and understandable manner. thanks a bunch man
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.
The best explanation over youtube. Thank you very much.
Thanks for this - I expected a very complicated explanation,but actually it all made sense. Great video.
To be honest with yall, this guy's explanation is excellent fr
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 ......
Beautiful!
Simply...
Beautiful!
Thanks a lot for this simple explanation to the otherwise seemingly complicated problem.
Thank you!!!
Thank you. I was always wondering but never got such an explanation.
Very good comprehensive video. I always tend to take these formulas for granted.
Mathematics basics are explained very clearly . Great work nicely done. Thank you
wow. my jaw is on the floor. I loved how it all simplified so nicely in the end. Great video, btw!
Animations and explanations are best... thanks for making this types of videos.
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)
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 :)
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
Very nice bro
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.
@@joshuaronisjr True, but the distance is constant, and it's equal to pi * r , as you move it "Half the sphere".
@@gligoradrian784 What distance is constant?
@@joshuaronisjr I mean, the 180* around which you rotate the circle, and also pi.
Math really does builds upon itself
interesting how the area of a circle is pi*r^2 but the (surface) area of a sphere is pi*d^2
That's an even simpler mnemonic tool.
thanks Ryan.
Ryan Bell thnx
The surface area of a sphere is 4*pi*r^2......
please explain someone
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.
U just made ur life harder bro good job
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.
Very intuitive. Maybe a comment that this derivation also applies to polygons with more than 8 sides, would be perfect.
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.
Thanks. Excellent, logical and easy to follow.
The best explanation I ever seen thanks buddy I'll be your subscriber forever
Appreciate it!
Wow. This is like, proofs to the max. I've never seen such a complicated proof about spheres; great job!
Hi like im dad
Thank you for your help.. 😊. Was looking forward for such theory and I guess I got what I wanted to see!
mind blown ! i've never thought of this before it's a master piece
That was just... BEAUTIFULLY done! Thank you!
Beautiful explanation!!
One of the most helpful answers
Well done. Classic proof with great explanation and illustration.
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.
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
That's just "reducing" a simpler problem to a harder problem.
Nyx Avatar what is calculas
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.
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
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...
this was a very elegant and simple way to solve it, thank you!
You have made me do my homework.
Thank you very very very ....much.
We can prove it by integrals too. And I think it's better! But your proof is pretty good too!
Best explanation i ever seen on youtube.
Amazing! I had no idea it was this complex!
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.
Perfect ❤👏
Greetings to you from Egypt !!
Awesome video. Animations were clear and helpful and the proof was simple and beautiful. Liked and Subbed!
Your videos are awesome and very informative and are on a different level from most explanations, Thank You.
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.
AMAZING. MAN OF THE PEOPLE RIGHT HERE.
Conceptual answer. Good explanation
Simply beautiful, great video!
Very elegant solution. Thanks for posting!
I hadn't thought it's so complex
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.
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.
Nicely explained.
Wonderful explanation
wow what an awesome explanation 😊
From a calculus standpoint, the surface area is the derivative of the volume, 4/3pi(r^3)
Beautiful explanation.
The guy who cane up with this clearly had a love for geometry
this is the best proof i've seen.
Thank you so much for this wonderful presentation.....
Just excellent
Hey, I love your videos! They make everything so much clearer about math! I actually do not quite get proofs for the law of cosines, so I was hoping you could do a video on it. Thanks!
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.
Indeed. The formulas are probably true when there are more than 2 frustrums, but that needs to be demonstrated, which it wasn't.
Super explanation
Thank you
Not only with octogon
We can also try it with hexagon
Great explanation sir 😊😊
A double integral in polar coordinates works best!
What about surface Integrals?
+VivzStudioSs "Double integral"
VivzStudioSs Surface integral turns out to be a double integral
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?
@@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.
this derivation was shocking fr me
well done 👍
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?
BRO YOU ARE HEAVENLY!
BEAUTIFUL JUST BEAUTIFUL
Thnx a lot!!!
It really helped me out in my seminar.
U da BEST!!!!
Awesome thought process! Someone was much smart.
Holy hell, this was masterpiece.
Thnks
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?
Thank u so so so so much sir......😁😁😄...
Keeep solving problems of us
Thanks so much for your sharing. It’s crystal clear.
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
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.
An explanation that is easy for students to grasp is the physical size relationship of the inscribed circle to the square or cube. So this is for students looking to work out how to calculate the volume or area of a sphere.
A ratio of a square to inscribed circle is approx. ¾ or 0.75 or π/no edges = 4 …
So, if your circle is 2cm diameter, then the square is 2cm wide x 2cm high and its perimeter is 2cm * 4 edges (8cm).
The corresponding circle that inscribes the box is therefore about 6cm or (8cm * (π / 4)).
The area of the box is 4 cm2 so the area of the circle is about 3 cm2 or (4 cm2 * (π / 4)).
ratio = π / 4
ratio * perimeter(square) = circumference(circle) cm
ratio * area(square) = area(circle) cm2
Similarly constructed ratio of a cube to inscribed sphere is approx. ½ or 0.5 or π /no faces = 6 …
So, if your sphere is 2cm diameter, then the box is 2cm wide x 2cm high x 2cm deep and its surface area is 4 cm2 * 6 faces (24 cm2).
The area of the corresponding sphere that inscribes the box is therefore about 12 cm2 or (24 cm2 * (π / 6)).
The volume of the box is 16 cm3 so the volume of the sphere is about 8 cm3 or (16 cm3 * (π / 6)).
ratio = π / 6
ratio * area(cube) = area(sphere) cm2
ratio * volume(cube) = volume(sphere) cm3
I realise that this may be obvious to everyone here, but the reason I mention this is just that ratios seem to be a much simpler way for students to grasp the concept that an inscribed circle has a linear relationship to the area and perimeter of the square that bounds it … as does a inscribed spheres area and volume to the bounding box. Students easily grasp the volume of a box, by counting blocks, and knowing the relationship of a corresponding sphere is a fixed ratio, allows them to explore how that ratio was derived.
To determine any ratio from a regular polygon for a circle inscribed within the following formula can be applied to both perimiter and area:
n = number of sides
eqn = pi/n.tan(pi/n)
tri = pi/3.tan(pi/3)
sq = pi/4.tan(pi/4)
hx = pi/6.tan(pi/6)
oc = pi/8.tan(pi/8)
that is a beautiful derivation...
Please also make a video on formula of (A3-B3)=
A great and clear explanation. Thank you.
I like your explanations
that's what we call a good and perfect explanation!!!!!!< ;
Thank you. It's a wonderful and a lot benefiting video. Please, can you also make videos to explain surface areas of cube and cuboid?
Beautiful explanation,thank you very much!