4πr^2 video explanation ua-cam.com/video/6EzQEdBX_30/v-deo.html At first the sphere isn't rounded because it is made of a few pyramids. The trick is to increase the number of pyramids while maintaining the same radius therefore the bases of the pyramids become smaller. By increasing the number of pyramids and the bases are getting smaller, so the collective shape of the pyramids gets rounder and it resembles more the shape of a sphere. If we have an infinite amount of pyramids then the volume of the pyramids eventually becomes equal to the volume of a perfect sphere.
You seriously in a nutshell said: "a quick explanation why the surface area of a sphere is 4pir^2 is because of the amazing fact that surface area of a sphere is actually 4pir^2"
If we have an infinite amount of pyramids then the volume of the pyramids eventually becomes equal to the volume of a perfect sphere. And I do have a video response explaining the surface area of a sphere, thanks for watching.
Anime Movies and series This is explaining what the integration is doing. most people have no idea what integration is, and even if you do, just showing an integral isn't helpful or intuitive
Great video. But even if we have a lot of "pyramids", they are never real pyramids because there is always a very small round part at the base. Where is the sum of these tiny round parts in the formula ?
@@emmanuelkuzniak4534 I think that's why calculus eventually gets involved, to add up all those little pieces (pretty much integration, but not precisely)
Volume and Surface of sphere with Integrals: Koords of a sphere: x=R*sin(a)*cos(b) a=[0,2*Pi] b=[-Pi/2,Pi/2] y=R*cos(a)*cos(b) z=R*sin(b) derivatives of a: x=R*cos(a)*cos(b) y=-R*sin(a)*cos(b) z=0 derivatives of b: x=-R*sin(a)*sin(b) y=-R*cos(a)*sin(b) z=R*cos(b) crossprodukt from derivatives of a and derivatives of b: x=-R²*sin(a)*cos²(b) y=-R²*cos(a)*cos²(b) z=-R²*cos²(a)*cos(b)*sin(b)-R²*sin²(a)*cos(b)*sin(b) =-R²*cos(b)*sin(b)*(sin²(a)+cos²(a)) =-R²*cos(b)*sin(b) norm n of this vector: n²=x²+y²+z² =R^4*sin²(a)*cos4(b)+R^4*cos²(a)*cos4(b)+R^4*cos²(b)*sin²(b) =R^4*cos²(b)*(sin²(a)*cos²(b)+cos²(a)*cos²(b)+sin²(b)) =R^4*cos²(b)*((sin²(a)+cos²(a))*cos²(b)+sin²(b)) =R^4*cos²(b)*(cos²(b)+sin²(b)) =R^4*cos²(b) n=R²*cos(b) calculation of surface area A and volume V of the sphere with integral: A=int(int(n,a=0..2*Pi),b=-Pi/2..Pi/2) =int(R²*cos(b)*2*Pi,b=-Pi/2..Pi/2) =R²*sin(Pi/2)*2*Pi-R²*sin(-Pi/2)*2*Pi =2*Pi*R²*(sin(Pi/2)-sin(-Pi/2)) =2*Pi*R²*(1-(-1)) A=4*Pi*R² V=int(A,R=0..r) =4/3*Pi*r^3-4/3*Pi*0^3 V=4/3*Pi*r^3 Yea, i'am bored and have no gf
Philipp Petsch Have u copied or written it yourself??? OMG !!! I was just shocked. I think you could be the boss of mathematics hmmm... 👍👍👍👍😊😊😊😊😊😊 But I haven't understood even a bit(Lol)😕😕😕😕😕😕😕😕
Christina DC thank you for the compliment. :) Didn't copy it, but when you know how integrals work, then it's not that hard. Had to learn this staff last year and so i know it. But please don't be sad, at first, for me it was hard to unterstand too. It takes some time and explanation. I only gave formula here. I can try my best to explain some more of what i did here if you want, but I can't guarantee that it helps, because for me it's not easy to explain something in english. It's not my main language and google translate sucks sometimes. :D
Philipp Petsch did you seriously just calculated how to calculate the area of a sphere with trigonometry in a UA-cam comment section? Man you must be really bored
This is what real mathematics is. We were so unliky that our teachers never demonstrated these concepts while were were learning at college. I love maths from my childhood, but our teachers were unable to show us the beauty of mathematics, and I lost the chance to build a career based on engineering subjects, and ended up becoming management graduate. However, there is a saying that you are never too late to learn something...so now I am learning these amazing techniques and teaching students these beautiful techniques of mathematics...
I was really ashamed inside that I didn't know why the formula of sphere's volume is that. I thought they taught but I forgot how they derived it but it turns out, they did not. They just made us memories the formula. This is so dope thank you.
I don't know a lot of english myself and i understood your lesson 1000000000000x times better than any other video in my language, 1000^999999 times thanks!
A simpler method is as follows... Draw your sphere, centre (0,0). Allow sphere radius to be r. Select a value of x to the right of (0,0). Erect a perpendicular (perp) of height y. Rotate that perp about the x axis to form a disc. Allow that disc to have width dx. The incremental volume of that disc is its area A = pi*y^2 multiplied by its width dx.... dV = pi*y^2*dx The perp height y is related to x by the classical equation of a circle... y^2 + x^2 = r^2 make y the subject... y^2 = r^2 - x^2 It will follow that... dV = pi*(r^2 - x^2).dx To determine the full volume of the sphere, integrate that last equation -r to +r... V = integral of pi*(r^2 - x^2).dx between -r and +r V = pi*( r^2*x - x^3/3 ) Insert the limits.... -r and +r V = pi*( r^3 - r^3/3 - (-r^3 + r^3/3) ) = pi*( 2*r^3 -(2/3)*r^3 ) V = pi*r^3*(2 - 2/3) = pi*r^3*(6/3 - 2/3) = (4/3)*pi*r^3 V = (4/3)*pi*r^3
TroyaE117 You are right but your method relies on differential calculus. The method and reasoning shown in this video is that which was used by the ancient greeks long before Newton and Liebnitz.
+Edward Suleski Wikipedia - history of calculus credits Eudoxus who predates Archimedes by 100 years. Its called the method of exhaustion. Everyone has heard of Archimedes and Pythagoras so they get all the credit while poor old Eudoxus is flagged as a spelling mistake. Not fair eh?
+TroyaE117 AS jack hodges pointed out, your solution requires calculus. To most people, calculus is *not* simpler. Therefore, your solution, though correct is not simpler than the one presented in this video. No offense intended. Just pointing out what I would have thought to have already been obvious.
+Chris Sekely Chris, If you think the method in this video is simpler, I suggest you brush up on your calculus.You will find it very interesting, transparent, and very much faster.Best...Troya
This is why I failed geometry in high school. I watched the video twice and it still makes no sense to me. I never had a problem with algebra, but geometry just makes my mind shut down.
Elementary Foundation Stone Of Differential Geometry Given a 120 centimeter Diameter multiplied by three, the length of the circle will be 360 centimeters. Given 360 degrees to the circles length, each degree will be one centimeter in length. And not one or any number of mathematicians, geometer's or geniuses can alter or change the fact that; A circles length is exactly three times that of its diameter length. Area of a 120 centimeter diameter Circle 3 X r2 (Three times the radius squared) 120 cm x 120 cm; Square = 14. 400 sq cm 1 right angle = 120 cm x 120 cm ÷ 2 = 60 cm radius 60 cm x 60 cm = 3,600 sq cm 3,600 sq cm x 3 = 10, 800 sq cm to the area of the Circle Which is three quarters of the area to the 14, 400 sq cm square. In sum A Circles Area is always three quarters of the area gained by squaring its diameter length. A circles length is always three quarters of the perimeter length gained by squaring its diameter length. Twelve Steps From The Cube, To The Sphere (Volume & Surface Area) Calculating the surface area and volume of a 6 centimeter diameter sphere, obtained from a 6 centimeter cube. 1. Measure the (a) cubes height to obtain its Diameter Line, which in this case is 6 centimeter’. 2. Multiply 6 cm x 6 cm to obtain the square area of one face of the cube; and also add them together to obtain the length of perimeter to the square face = Length 24 cm, Square area 36 sq cm. 3. Multiply the square area, by the length of diameter line to obtain the cubic capacity = 216 cubic cm. 4. Divide the cubic capacity by 4, to obtain one quarter of the cubic capacity of the cube = 54 cubic cm. 5. Multiply the one quarter cubic capacity by 3. to obtain the cubic capacity of the Cylinder = 162 cubic cm. 6. Multiply the area of one face of the cube by 6, to obtain the cubes surface area = 216 square cm. 7. Divide the cubes surface area by 4, to obtain one quarter of the cubes surface area = 54 square cm. 8. Multiply the one quarter surface area of the cube by 3, to obtain the three quarter surface area of the Cylinder = 162 square cm. CYLINDER TO SPHERE 9. Divide the Cylinders cubic capacity by 4, to obtain one quarter of the cubic capacity of the Cylinder = 40 & a half cubic cm. 10. Multiply the one quarter cubic capacity by 3, to obtain the three quarter cubic capacity of the Sphere = 121 & a half cubic cm, to the volume of the Sphere. 11. Divide the Cylinders surface are by 4, to obtain one quarter of the surface area of the Cylinder = 40 & a half square cm. 12. Multiply the one quarter surface area by 3 to obtain the three quarter surface area of the Sphere = 121 & a half square cm, to the surface area of the Sphere Confirmation by Weight Given that the 6 Centimeter Diameter Line Sphere was obtained from a Wooden Cube weighing 160 grams, prior to it being turned on a wood lathe into the shape of a sphere The Cylinder of the Cube would weigh 120 grams The waste wood shavings would weigh 40 grams Given that the Cylinder weighed 120 grams The waste wood shavings would weigh 30 grams. Note: And ironically you can also obtain this same result by volume, using Archimedes Principle. www.fromthecircletothesphere.net
Very nicely presented! Visuals are quite awesome, illustrative, and creative. Spheres are quite fascinating, even eerie, which is why they are sometimes called SphEeries. Similarly, circles are often called Quirkles, because they are quirky! Of course, Calculus provides more rigor in solving this vital volume problem, but the visuals here have plenty of helpful VIGOR to help people get to more RIGOR! Thanks for all your dedication to edication (oops, education)!
I totally agree with your method, which is exactly what I was saying, the calculus (integration) method is the most complete and thorough. The limit of the length of one side of the base is 0, eliminating the curved surface effect. We agree ! :-)
Came across this video while researching for an assignment, and wow, your channel's very interesting!It seems like you explain the steps behind the formulas,and I really need that to solve math problems.
A VERY GOOD EXPLANATION FOR THE DERIVATION OF THE FORMULA OF VOLUME OF A SPHERE.IT IS THE BEST METHOD A BEGINNER CAN UNDERSTAND THE DERIVATION OF VOLUME OF A SPHERE FORMULA. EXTREMELY USEFUL FOR STUDENTS.
Circle some equations: circumference over 2 pie = radius, area over 1/2 circumference = radius, area over radius squared = pie. circumference over diameter = pie, area over 1/2 radius = circumference circumference times 1/2 radius = area, Thank you Enjoy your computation by using the formulas.
Just learned more in a few minutes than I did in all of high school math. If only teachers could have explained it like this and not to like 30 other kids at the same time
@vampiracy Just as a note: sqrt([F_φ]^2+[F_θ]^2+1)=1. This is because we're talking about a sphere of radius R. This is represented in spherical coordinates as ρ=F(φ,θ)=R. Since R is a constant, taking it's derivative with respect to any variable will yield 0, so [F_φ] and [F_θ] are both 0. When we plug this into our surface area differential, we get sqrt(0^2+0^2+1) or sqrt(1), which equals 1.
The hemisspehe volume is - circle base area times radius or (pie radius squared) X radius, The hemisphere surface area is - circle base circumference times radius or (2 pie radius) X radius, Thank you!!!
I agree with your assertion about surface area proof is 'lacking' however this is the best non calculus solution I have seen for deriving a spheres volume. "Moreover, the bases of the pyramids are not flat, but have a curved surface", That doesn't matter as the sum of all bases are substituted with 4Pir^2 which is curved, he has essentially done a limit by setting their sum equal to this. Calculating these things with calculus is commonplace even though the formula has been around much longer
Volume sphere = V = Pi * 4/3 * r^3; Surface of sphere S = Pi * dV/dr = Pi * 4 * R^2 Circumference of disc through center C = Pi * dS/dr = Pi * 2 * r Problem is that when you differentiate, the constant Pi is lost. And when you perform the operations, you have to add Pi. Notice the topology: volume, surface and circumference. Surface area of sphere-> circumference of the disc is intuitive. But volume of sphere -> surface of sphere is less intuitive, at least a priori Strikes me as a bizarre approach to the question but it just popped into my mind.
To respond to Daren Soobrayen: Good observation. This explanation might help: When I studied calculus in undergraduate school, they taught us to add together an infinite number of squares to determine the area of an irregularly shaped object. So imagine a flat surface such as a butterfly shape drawn on a piece of paper. You want to determine the area of that shape. If you draw an infinite number of squares and/or rectangles on that shape, then add together all of those areas (which are easily determined by using the formulas for a square and/or a rectangle), you can then compute the area of the butterfly shape. So looking at the sphere, imagine an infinite number of pyramids each with a base the size of a pin point; or imagine pyramids the size of a human hair. Then add all of the bases (rectangles) together, and you come up with the surface area of the sphere. Just make the pyramids' bases so small that they fill in every nook and cranny of the curved surface area of the sphere. I hope conceptualizing it like this helps out. If you make the bases (and the pyramids) small enough, they'll fill in even a curved surface.
Brilliant !.... Why haven't I got a teacher like you, when I was at school ?..... You are fantastic.... Thank you so much for that magic video... :-) From Brussels, with Love....
You could also do it by cutting the sphere into infinitely many shells (hollow spheres.) Each of which has a surface area equal to 4(pi)(radius of the shell)^2. Then the volume of any shell would be equal to the surface area of it times dr (the infinitely small width of the shell.) then the total volume of the sphere would be the summation of the volumes of these shells. This would be equal to the integral of the surface area equation with respect to r from 0 to r. This integral is equal to (4/3)(pi)r^3, the volume of a sphere. I just appreciate the fact that volume of the sphere is equal to the integral of its surface area
+wv1seahawks Infinitely thin shells have no volume. You cannot get a volume with adding 0-volume objects, no matter how many you will add up (even infinitely many). This is a common misconception that haunts Mathematics from centuries: that volumes are made of surfaces, surfaces are made of lines, and lines are made of points. This is utter bullshit, because their dimensions don't match. Points don't have length. Lines don't have breadth. Surfaces don't have volume. (Remember Euclid?) True, lines _contain_ points, but it doesn't mean that their _length_ is made of the "infinitesimal lengths" of those points, because again: *points don't have length*. True, surfaces can _contain_ lines (and points as well!), but can we say that the _area_ of the surface is _made of_ those lines? (or, even worse, from these points?) :P And one cannot get something from just adding up nothings, even infinitely many. This is not where all those dimensions come from.
***** The mere fact that I disagree with the majority, doesn't yet make me wrong. And the fact that something is commonly used, doesn't mean yet that it is correct. Even the fact that it seems to "work", doesn't mean yet that it really works this way, because it can also work "by accident", from different reasons. (From false assumptions you can derive everything, be it true or false.) As I said, infinitely thin shells are surfaces, they don't have volume. So no matter how many of them you "add up", the sum will still be 0. Notice that when you're calculating integrals, you're not really doing the summation directly. Usually, you're not even doing the limit directly. You just apply known formulas. These formulas were known before calculus has been "invented" by Newton, Leibniz & co., because they've been known from antiquity, and they've been derived without any notions of infinitesimals, limits, 0-thickness shells etc. The Greeks have got them from the Egyptians and from the Arabs, the Arabs got them from the Hindus, etc. Unfortunately, just the formulas, without their derivations. So Newton & co. needed to "reinvent" these derivations to be able to pretend that they invented calculus all by themselves :P They had to "made up a theory for the formulas", and they did it wrong. But the deception worked so well, that it works to this day, since people are being taught these techniques in schools and being taught to not question it (since questioning usually lead nowhere anyway, because tachers don't know the answers too :P ).
As far as I understand it, it's not that you're adding 'volumes' with "0-volume objects" (even though such a volume is just 0). But the idea is explained either with infinitesimals or limits with which you are looking at the volumes as the depth of the strips are approaching 0 or are extremely close to 0 (but not exactly 0). An infinitesimal is something that is extremely extremely close to 0 but is not exactly 0 and so anything with a depth that is infinitesimally small still has a depth (so to speak) but it's just that it's very close to 0. For example, when differentiating the principle we're always taught is using limits. And so we look at what happens to the gradient of a function at a particular point. Therefore, we take two points on a tangent line at that point. The idea of the limit is just looking at what happens when the vertical distance and horizontal distance approach 0. This doesn't mean you're doing the calculation 0/0 but rather doing calculations that get more and more accurate the smaller the distances get. And anyway, it seems more to me that we just use this idea to derive other formulae and not actually on it's own to calculate stuff.
The fact is that what he just said is another way to explain summation of infinitely small objects to estimate the volume of one object by cross sections or cylindrical shells. ECT. Many methods however all he did was rephrase the process if you don't understand it's because you don't want to be wrong and you are just being rebellious. Zz. To argue this point is just wrong because by definition the process is logical.
We can take volum of revolution of curve between (x,y)represent by equation of y^2=r^2-x^2 and then take Integration to give you half volume of the sphear
More on the last few sentences. The volume of the cone is 1/3*pi*r^3. Caliveris states that shapes with the same cross section have the same volume. Therefore, with similar volume, it is possible to use the subtraction above. Otherwise, just integrate pi*(r^2-x^2)) from r to -r. This is the integral equation for volume (pi*r^2). This works since integrals are the some of all the cross-sectional disks and the equation of a circle is all the points create the radius.
@Kosekans Okay. We'll integrate unity over a region in 3D space, E. E is defined as a sphere of radius R centered about the origin. Integrating it will give us the mass of a sphere with radius R and uniform density of 1 throughout, also known as the volume of said sphere. We'll need to set up a triple integral in order to solve this problem. We'll be working with spherical coordinates for simplicity, since we have a spherical region.
@vampiracy Our limits of integration are φ=0 to φ=π, and θ=0 to θ=2π. Integrating sinφ with respect to φ, from φ=0 to φ=π, gives us 2. Integrating 2 with respect to θ, from θ=0 to θ=2π, gives us 4π. Multiplying this with the (R^2) we took out gives us a surface area of (4π)(R^2).
+mathematicsonline I enjoy math. I have a math minor from college I've never used and I watch Numberphile often. I also really like your derivations. I am curious about one thing. Way back in high school (or maybe early college) I thought up a question I still haven't seen answered anywhere. If one were to cut a hole through a sphere and supposing the hole had a perfectly *square* (not round) cross section, how would one calculate the volume removed (or remaining)? Assume for simplicity that the central axis of the hole coincides with a diameter of the sphere. I'm sure this could be done with some relatively complicated calculus ( relative, at least, to the complexity of the derivations presented here). However, I'm hoping for a more intuitive approach, provided that such is even possible.
A very good question. Yes, of course it is possible to find the volume with some good advanced calculus. I do not know the answer at the moment but if I figure it out, I will let you know. Volume of Sphere: 4/3 pi r^3 Volume of Rectangular Prism: 2rs^2 4/3 pi r^3 - 2rs^2 < ACTUAL remaining volume I would need to find the volume of a rectangular prism whose ends are part of the sphere's surface area I do not know how to call that shape but then you would subtract that from 4/3 pi r^3. This might be a problem too hard for me. I don't know how functions behave in 3-D. Good luck.
The challenge of this proof is that is requires “believing” that surface area = 4 x pi x r^2. My daughter (in 8th grade) just shared this proof and I explained why the proof inherently had an underlying assumption that she couldn’t prove. Calculus and a volume integration is a far better way to prove this.
Great explanation except for where it gets to, "The amazing fact is the surface area of the sphere is equal to exactly 4 times the area of the circle." Why 4 times? Why not 3 times or 5 times or any other number of times? That wasn't explained. It was skipped over. I would like to have heard and seen the animated explanations of why the surface area of the sphere is only 4 times the area of the circle.
Waiting for explanations like these for a long long time....................Thanks Best videos ever!! Wish this concept could be applied to other areas of study!! Pure Genius Thanks so much.
My discovery according to reseach and experment. The volume of the sphere is - circle area times diameter or (pie radius squared) X diameter = volume, And the sphere surface area is - circumference times diameter or (2 pie radius) X diameter = surface area, Some equations: volume over 1/2 surface area = radius, surface area over diameter = circumference, volume over 2 radius cubed = pie, volume over 1/2 radius = surface area, surface area times 1/2 radius = volume, surface area over circumference = diameter or surface area /2 pie radius - diameter, volume over circle area = diameter or volume/pie radius squared = diameter, Enjoy your computation by using my formulas thank you!!!
T6hanks this was helpful although I had to watch the video twice to understand to anyone who is still confused after watching it the first time hit the replay button and go to 3:00 exactly and he will explain everything that may have confused you again in simpler terms
I am so, so so so so pissed that I eventually had to look this up. I wanted so badly to figure this out by myself... Tried a bunch of things out in my head but there were always imperfections because I wasn't using pyramids. Beautifully explained and an amazing graphical presentation. Thank you.
me too, truly great video though. i think the discrepancy occurred because we assume all the bases of the pyramids are the same, so we don't label them individually. so the nth base value is an unknown concept until you physically see it. of course it could've been a different process for you but that's where i had my "ooooohhh" moment
Same thoughts, (found a mathematical proof for the surface formula, tho - rly hard ^^ ) but i kinda worked out the "curved surface" thing: Lets just simplify it and say that a "sircle" is a figure with "infinite" sides. It's not a number. And the amount of pyramids are also infinite. But there is nothing like a perfect sircle, as when you zoom and zoom you'll find irregular patterns in the atoms, etc.. So the sircle isn't round, and the base of each pyramid represents each irregularity
Thank You for making math formulas vivid and clear
4 роки тому
Very good. But the reasoning within the three-dimensional plane with the use of the integral is more interesting, not to mention that it does not directly need other formulas, just integrate.
How to find Pi ? Draw a circle of radius 0.5. Draw another radius at 90 degrees. Join 2 ends. We get chord 0.707106781...Half of it is 0.35355339...Take 7 radii equal to 3.5 and subtract 0.35355339.. and we get 3.146446609...It is Circumference and also Pi (14 - root2)/4.. Truth is a roaring fire. --- SJR.
the formula depends on the given. If the given is diameter,you could use V= (3.14)(diameter cubed)and divided by 6.but if it's given is radius it will be V=4/3 (Pi)(radius cubed)
Thankyou . Nicely animated. I liked the drole ... area is B1 plus B2 plus B3 ! Archimedes and projecting a sphere area onto a cylinder can prove the formula for surface area is 4 Pi r squared . You used square based pyramid bases which visually fit nicely, but curiously this process also works out the same formula using thousands of radius high CONES instead of thousands of radius high square based pyramids. Both base shapes and any errors are irrelevant because you suddenly swop the real result of the integration for the previously worked out area formula. NEAT .This is definitely early calculus, did the ancients do it this way ?
Superb work guyzz, finally found the type of explanation that i dreamt of but never had experienced , thank you soo much for making it mathematicsonline
The volume of a sphere is the integral of its surface- area with respect to R. The surface- area of a sphere is the derivative of the volume with respect to R.
@vampiracy . Our limits of integration are ρ=0 to ρ=R, φ=0 to φ=π, and θ=0 to θ=2π, and our differential is [(ρ^2)sinφ dρ dφ dθ]. Integrating (ρ^2)sinφ with respect to ρ, from ρ=0 to ρ=R, gives us (1/3)(R^3)sinφ. Integrating (1/3)(R^3)sinφ with respect to φ, from φ=0 to φ=π, gives us (2/3)(R^3). Integrating (2/3)(R^3) with respect to θ, from θ=0 to θ=2π, gives us (4/3)(π)(R^3). This gives us the volume of the sphere to be (4/3)(π)(R^3). There's your proof! :D
I was so confused when i was making my own formulas up for a test and the volume formula came out to be 4/3pir³ instead of just pir³ like i theorized, and i was so intrigued as to where that 4/3rds came from
@mishraonline Okay. We'll integrate over a disk D of radius R. Our integrand will be infinitely small pieces of area, which are sqrt([F_φ]^2+[F_θ]^2+1), and our differential will be [(R^2)sinφ dφ dθ]. Since (R^2) is a constant, we can take is to the front of the integral.
"(found a mathematical proof for the surface formula, tho - rly hard ^^ )", I would recommend a look at Archimedes proof for this. He figured out how to prove that the surface area of a sphere is equal to that of a cylinder of the same diameter and height, combined with trig it produces a simple and beautiful proof.
If you think surface=4pi*r^2 is wierd I'd say If radius is half of D (dimension), square area=D*D*4. Circle=r*r*3.14. Cube is 4 squares =D*D*4*4. Sphere=r*r*4*3.14. To me I think an area of circle should be just as much smaller than a square as sphere is than cube.
Imagine the circle to be made up of infinitesimal shells. The volume of each shell is dV=4*pi*r^2*dr. Integral(0 to r) dV = integral(0 to r) 4*pi*r^2*dr = 4/3*pi*r^3.
Good video but i would have liked you mentioning that the number of pyramids needs to go up to an infinite number in order for the sum of the bases being equal to the surface area of the sphere. Also the explanation of how the the surface area comes together was, well.. not really explanatory haha. But if you don't care about details, this video is really nice. Also the animations! 👌
4πr^2 video explanation ua-cam.com/video/6EzQEdBX_30/v-deo.html
At first the sphere isn't rounded because it is made of a few pyramids. The trick is to increase the number of pyramids while maintaining the same radius therefore the bases of the pyramids become smaller. By increasing the number of pyramids and the bases are getting smaller, so the collective shape of the pyramids gets rounder and it resembles more the shape of a sphere. If we have an infinite amount of pyramids then the volume of the pyramids eventually becomes equal to the volume of a perfect sphere.
Thanks for the surface explanation
Asslam o alikum. Thank you for the video. Have a good day.
I love how you concisely and accurately explained the concept of an integral without ever mentioning calculus.
It's a really nice application of Archimedes' method for finding the area of a circle
A R T
Ok nerd
You seriously in a nutshell said: "a quick explanation why the surface area of a sphere is 4pir^2 is because of the amazing fact that surface area of a sphere is actually 4pir^2"
that explanation requires another video, right?
ua-cam.com/video/GNcFjFmqEc8/v-deo.html
Lol v:
Here the explanation about surface area of sphere ua-cam.com/video/6EzQEdBX_30/v-deo.html
VidyVid 👀
Exactly my thought. You don't explain one thing by throwing another question into the room.
1:20 This is the most beautiful thing I've even seen in a math channel.
agreed
Me too
me too
SUMYIU li
Check out 3Blue1Brown
If we have an infinite amount of pyramids then the volume of the pyramids eventually becomes equal to the volume of a perfect sphere.
And I do have a video response explaining the surface area of a sphere, thanks for watching.
where are tangent theta ,please tell me about it🔬🔬🔬🔬🔬🔬🔬🔬🙏
why cant we integrate
Anime Movies and series This is explaining what the integration is doing. most people have no idea what integration is, and even if you do, just showing an integral isn't helpful or intuitive
Great video. But even if we have a lot of "pyramids", they are never real pyramids because there is always a very small round part at the base. Where is the sum of these tiny round parts in the formula ?
@@emmanuelkuzniak4534 I think that's why calculus eventually gets involved, to add up all those little pieces (pretty much integration, but not precisely)
Volume and Surface of sphere with Integrals:
Koords of a sphere:
x=R*sin(a)*cos(b) a=[0,2*Pi] b=[-Pi/2,Pi/2]
y=R*cos(a)*cos(b)
z=R*sin(b)
derivatives of a:
x=R*cos(a)*cos(b)
y=-R*sin(a)*cos(b)
z=0
derivatives of b:
x=-R*sin(a)*sin(b)
y=-R*cos(a)*sin(b)
z=R*cos(b)
crossprodukt from derivatives of a and derivatives of b:
x=-R²*sin(a)*cos²(b)
y=-R²*cos(a)*cos²(b)
z=-R²*cos²(a)*cos(b)*sin(b)-R²*sin²(a)*cos(b)*sin(b)
=-R²*cos(b)*sin(b)*(sin²(a)+cos²(a))
=-R²*cos(b)*sin(b)
norm n of this vector:
n²=x²+y²+z²
=R^4*sin²(a)*cos4(b)+R^4*cos²(a)*cos4(b)+R^4*cos²(b)*sin²(b)
=R^4*cos²(b)*(sin²(a)*cos²(b)+cos²(a)*cos²(b)+sin²(b))
=R^4*cos²(b)*((sin²(a)+cos²(a))*cos²(b)+sin²(b))
=R^4*cos²(b)*(cos²(b)+sin²(b))
=R^4*cos²(b)
n=R²*cos(b)
calculation of surface area A and volume V of the sphere with integral:
A=int(int(n,a=0..2*Pi),b=-Pi/2..Pi/2)
=int(R²*cos(b)*2*Pi,b=-Pi/2..Pi/2)
=R²*sin(Pi/2)*2*Pi-R²*sin(-Pi/2)*2*Pi
=2*Pi*R²*(sin(Pi/2)-sin(-Pi/2))
=2*Pi*R²*(1-(-1))
A=4*Pi*R²
V=int(A,R=0..r)
=4/3*Pi*r^3-4/3*Pi*0^3
V=4/3*Pi*r^3
Yea, i'am bored and have no gf
wtf.... wow
you could also just say volume is the integral of the surface area since he jut blatantly uses the surface area formula here
Philipp Petsch Have u copied or written it yourself??? OMG !!! I was just shocked. I think you could be the boss of mathematics hmmm... 👍👍👍👍😊😊😊😊😊😊
But I haven't understood even a bit(Lol)😕😕😕😕😕😕😕😕
Christina DC thank you for the compliment. :) Didn't copy it, but when you know how integrals work, then it's not that hard. Had to learn this staff last year and so i know it. But please don't be sad, at first, for me it was hard to unterstand too. It takes some time and explanation. I only gave formula here. I can try my best to explain some more of what i did here if you want, but I can't guarantee that it helps, because for me it's not easy to explain something in english. It's not my main language and google translate sucks sometimes. :D
Philipp Petsch did you seriously just calculated how to calculate the area of a sphere with trigonometry in a UA-cam comment section? Man you must be really bored
Day XX of quarantine, learning how to find the volume of a sphere
You will be einstein by the end of it
@@user-oi7sf1jm3h lol
We all gotta keep busy!
Exercise the mind!
Me too !
This is what real mathematics is. We were so unliky that our teachers never demonstrated these concepts while were were learning at college. I love maths from my childhood, but our teachers were unable to show us the beauty of mathematics, and I lost the chance to build a career based on engineering subjects, and ended up becoming management graduate. However, there is a saying that you are never too late to learn something...so now I am learning these amazing techniques and teaching students these beautiful techniques of mathematics...
I was really ashamed inside that I didn't know why the formula of sphere's volume is that. I thought they taught but I forgot how they derived it but it turns out, they did not. They just made us memories the formula. This is so dope thank you.
0:26
this version of interstellar brought to you by mathematicsonline
Im 14 and my maths teacher wanted usto make a presentation about this and this video really helped me understand the steps. thanks man x
Same but I'm 10
Same but I'm 3
I'm a fetus
I am a sperm
I am the parents of myself
I love your animation. Surely, students will understand how the formula is obtained. I wish you would make similar videos for other formulas.
I don't know a lot of english myself and i understood your lesson 1000000000000x times better than any other video in my language, 1000^999999 times thanks!
A simpler method is as follows...
Draw your sphere, centre (0,0).
Allow sphere radius to be r.
Select a value of x to the right of (0,0).
Erect a perpendicular (perp) of height y.
Rotate that perp about the x axis to form a disc.
Allow that disc to have width dx.
The incremental volume of that disc is its area A = pi*y^2 multiplied by its width dx....
dV = pi*y^2*dx
The perp height y is related to x by the classical equation of a circle...
y^2 + x^2 = r^2
make y the subject...
y^2 = r^2 - x^2
It will follow that...
dV = pi*(r^2 - x^2).dx
To determine the full volume of the sphere, integrate that last equation -r to +r...
V = integral of pi*(r^2 - x^2).dx between -r and +r
V = pi*( r^2*x - x^3/3 )
Insert the limits.... -r and +r
V = pi*( r^3 - r^3/3 - (-r^3 + r^3/3) ) = pi*( 2*r^3 -(2/3)*r^3 )
V = pi*r^3*(2 - 2/3) = pi*r^3*(6/3 - 2/3) = (4/3)*pi*r^3
V = (4/3)*pi*r^3
TroyaE117 You are right but your method relies on differential calculus. The method and reasoning shown in this video is that which was used by the ancient greeks long before Newton and Liebnitz.
So did Archimedes or Pythagoras invent this idea?
+Edward Suleski
Wikipedia - history of calculus credits Eudoxus who predates Archimedes by 100 years. Its called the method of exhaustion.
Everyone has heard of Archimedes and Pythagoras so they get all the credit while poor old Eudoxus is flagged as a spelling mistake.
Not fair eh?
+TroyaE117 AS jack hodges pointed out, your solution requires calculus. To most people, calculus is *not* simpler. Therefore, your solution, though correct is not simpler than the one presented in this video. No offense intended. Just pointing out what I would have thought to have already been obvious.
+Chris Sekely Chris, If you think the method in this video is simpler, I suggest you brush up on your calculus.You will find it very interesting, transparent, and very much faster.Best...Troya
Lol "and the amazing fact is, the surface area of a sphere is equal to four of these circles!"
Yeah but how ?
The best explanation can be shown with integrals
This one its better, its more intuitive
we are stupid
I think that this is just a (nice) algebraic trick because you can't divide a sphere into equal sized square-based pyramids.
This is why I failed geometry in high school. I watched the video twice and it still makes no sense to me.
I never had a problem with algebra, but geometry just makes my mind shut down.
amazing work man but you didnt explain why the surface area of the sphere is 4 of the circle area. but thank you any way for your efforts.
+ricky bobby That's not the topic of the video.
There's another video where he explains that
Elementary Foundation Stone Of Differential Geometry
Given a 120 centimeter Diameter multiplied by three, the length of the circle will be 360 centimeters.
Given 360 degrees to the circles length, each degree will be one centimeter in length.
And not one or any number of mathematicians, geometer's or geniuses can alter or change the fact that;
A circles length is exactly three times that of its diameter length.
Area of a 120 centimeter diameter Circle
3 X r2 (Three times the radius squared)
120 cm x 120 cm; Square = 14. 400 sq cm
1 right angle = 120 cm x 120 cm ÷ 2 = 60 cm radius
60 cm x 60 cm = 3,600 sq cm
3,600 sq cm x 3 = 10, 800 sq cm to the area of the Circle
Which is three quarters of the area to the 14, 400 sq cm square.
In sum
A Circles Area is always three quarters of the area gained by squaring its diameter length.
A circles length is always three quarters of the perimeter length gained by squaring its diameter length.
Twelve Steps From The Cube, To The Sphere (Volume & Surface Area)
Calculating the surface area and volume of a 6 centimeter diameter sphere, obtained from a 6 centimeter cube.
1. Measure the (a) cubes height to obtain its Diameter Line, which in this case is 6 centimeter’.
2. Multiply 6 cm x 6 cm to obtain the square area of one face of the cube; and also add them together to obtain the length of perimeter to the square face = Length 24 cm, Square area 36 sq cm.
3. Multiply the square area, by the length of diameter line to obtain the cubic capacity = 216 cubic cm.
4. Divide the cubic capacity by 4, to obtain one quarter of the cubic capacity of the cube = 54 cubic cm.
5. Multiply the one quarter cubic capacity by 3. to obtain the cubic capacity of the Cylinder = 162 cubic cm.
6. Multiply the area of one face of the cube by 6, to obtain the cubes surface area = 216 square cm.
7. Divide the cubes surface area by 4, to obtain one quarter of the cubes surface area = 54 square cm.
8. Multiply the one quarter surface area of the cube by 3, to obtain the three quarter surface area of the Cylinder = 162 square cm.
CYLINDER TO SPHERE
9. Divide the Cylinders cubic capacity by 4, to obtain one quarter of the cubic capacity of the Cylinder = 40 & a half cubic cm.
10. Multiply the one quarter cubic capacity by 3, to obtain the three quarter cubic capacity of the Sphere = 121 & a half cubic cm, to the volume of the Sphere.
11. Divide the Cylinders surface are by 4, to obtain one quarter of the surface area of the Cylinder = 40 & a half square cm.
12. Multiply the one quarter surface area by 3 to obtain the three quarter surface area of the Sphere = 121 & a half square cm, to the surface area of the Sphere
Confirmation by Weight
Given that the 6 Centimeter Diameter Line Sphere was obtained from a Wooden Cube weighing 160 grams, prior to it being turned on a wood lathe into the shape of a sphere
The Cylinder of the Cube would weigh 120 grams
The waste wood shavings would weigh 40 grams
Given that the Cylinder weighed 120 grams
The waste wood shavings would weigh 30 grams.
Note: And ironically you can also obtain this same result by volume, using Archimedes Principle.
www.fromthecircletothesphere.net
he did that in annotated video
Yes he did... Watch it again man, and all the way through.
2年ぶりに見つけた、、！
This video is easy to understand!!
Thanks☺️
Very nicely presented! Visuals are quite awesome, illustrative, and creative. Spheres are quite fascinating, even eerie, which is why they are sometimes called SphEeries. Similarly, circles are often called Quirkles, because they are quirky! Of course, Calculus provides more rigor in solving this vital volume problem, but the visuals here have plenty of helpful VIGOR to help people get to more RIGOR! Thanks for all your dedication to edication (oops, education)!
Great visual proof sir! I like the ways you derive the formulas by simple methods that anyone can understand. Good job
It's very helpful and nice. May I ask you, which Programm do you use to do the graphics?
I totally agree with your method, which is exactly what I was saying, the calculus (integration) method is the most complete and thorough.
The limit of the length of one side of the base is 0, eliminating the curved surface effect. We agree ! :-)
Wow, this is as great explanation. Thanks!
I like knowing where the formula comes instead of just blindly accepting it.
Came across this video while researching for an assignment, and wow, your channel's very interesting!It seems like you explain the steps behind the formulas,and I really need that to solve math problems.
I love proofs.
surface area of sphere cut at h height from crown
Same!
Very nice
Nicole Melendez Pena 👍
V lmao
I'm mesmerized. I was looking for this kind of explanation from my childhood. Thank you so much.
A VERY GOOD EXPLANATION FOR THE DERIVATION OF THE FORMULA OF VOLUME OF A SPHERE.IT IS THE BEST METHOD A BEGINNER CAN UNDERSTAND THE DERIVATION OF VOLUME OF A SPHERE FORMULA. EXTREMELY USEFUL FOR STUDENTS.
Excellent! Visuals, explanation, pacing, structure of the presentation - I can think about spheres differently now thank you so much 🎉👍🍻
Circle some equations: circumference over 2 pie = radius,
area over 1/2 circumference = radius,
area over radius squared = pie.
circumference over diameter = pie,
area over 1/2 radius = circumference
circumference times 1/2 radius = area, Thank you Enjoy your computation by using the formulas.
Just wanted to say: this was an EXCELLENT explanation, and superbly animated. You've done a really great thing here--thank you for making this!
Just learned more in a few minutes than I did in all of high school math. If only teachers could have explained it like this and not to like 30 other kids at the same time
@vampiracy
Just as a note: sqrt([F_φ]^2+[F_θ]^2+1)=1. This is because we're talking about a sphere of radius R. This is represented in spherical coordinates as ρ=F(φ,θ)=R. Since R is a constant, taking it's derivative with respect to any variable will yield 0, so [F_φ] and [F_θ] are both 0. When we plug this into our surface area differential, we get sqrt(0^2+0^2+1) or sqrt(1), which equals 1.
Understanding how this thing works is way better than memorizing all the formulas in the Geometry
The hemisspehe volume is - circle base area times radius or (pie radius squared) X radius,
The hemisphere surface area is - circle base circumference times radius or (2 pie radius) X radius,
Thank you!!!
Bravo! Beautiful high school level explanation of how to derive the spherical volume formula from the surface area formula.
Glad you liked it
I agree with your assertion about surface area proof is 'lacking' however this is the best non calculus solution I have seen for deriving a spheres volume.
"Moreover, the bases of the pyramids are not flat, but have a curved surface",
That doesn't matter as the sum of all bases are substituted with 4Pir^2 which is curved, he has essentially done a limit by setting their sum equal to this.
Calculating these things with calculus is commonplace even though the formula has been around much longer
bro all other people use integrals and i see this
my savior
Volume sphere = V = Pi * 4/3 * r^3;
Surface of sphere S = Pi * dV/dr = Pi * 4 * R^2
Circumference of disc through center C = Pi * dS/dr = Pi * 2 * r
Problem is that when you differentiate, the constant Pi is lost.
And when you perform the operations, you have to add Pi.
Notice the topology: volume, surface and circumference. Surface area of sphere-> circumference of the disc is intuitive.
But volume of sphere -> surface of sphere is less intuitive, at least a priori
Strikes me as a bizarre approach to the question but it just popped into my mind.
To respond to Daren Soobrayen: Good observation. This explanation might help: When I studied calculus in undergraduate school, they taught us to add together an infinite number of squares to determine the area of an irregularly shaped object. So imagine a flat surface such as a butterfly shape drawn on a piece of paper. You want to determine the area of that shape. If you draw an infinite number of squares and/or rectangles on that shape, then add together all of those areas (which are easily determined by using the formulas for a square and/or a rectangle), you can then compute the area of the butterfly shape. So looking at the sphere, imagine an infinite number of pyramids each with a base the size of a pin point; or imagine pyramids the size of a human hair. Then add all of the bases (rectangles) together, and you come up with the surface area of the sphere. Just make the pyramids' bases so small that they fill in every nook and cranny of the curved surface area of the sphere. I hope conceptualizing it like this helps out. If you make the bases (and the pyramids) small enough, they'll fill in even a curved surface.
Brilliant !....
Why haven't I got a teacher like you, when I was at school ?.....
You are fantastic....
Thank you so much for that magic video...
:-)
From Brussels, with Love....
I loved the explanation for the surface area formula. First time I see a proof by amazement.
You could also do it by cutting the sphere into infinitely many shells (hollow spheres.) Each of which has a surface area equal to 4(pi)(radius of the shell)^2. Then the volume of any shell would be equal to the surface area of it times dr (the infinitely small width of the shell.) then the total volume of the sphere would be the summation of the volumes of these shells. This would be equal to the integral of the surface area equation with respect to r from 0 to r. This integral is equal to (4/3)(pi)r^3, the volume of a sphere. I just appreciate the fact that volume of the sphere is equal to the integral of its surface area
+wv1seahawks Infinitely thin shells have no volume. You cannot get a volume with adding 0-volume objects, no matter how many you will add up (even infinitely many).
This is a common misconception that haunts Mathematics from centuries: that volumes are made of surfaces, surfaces are made of lines, and lines are made of points. This is utter bullshit, because their dimensions don't match. Points don't have length. Lines don't have breadth. Surfaces don't have volume. (Remember Euclid?) True, lines _contain_ points, but it doesn't mean that their _length_ is made of the "infinitesimal lengths" of those points, because again: *points don't have length*. True, surfaces can _contain_ lines (and points as well!), but can we say that the _area_ of the surface is _made of_ those lines? (or, even worse, from these points?) :P
And one cannot get something from just adding up nothings, even infinitely many. This is not where all those dimensions come from.
+Bon Bon no it is common to use shells to find volume. you are wrong here m8.
***** The mere fact that I disagree with the majority, doesn't yet make me wrong. And the fact that something is commonly used, doesn't mean yet that it is correct. Even the fact that it seems to "work", doesn't mean yet that it really works this way, because it can also work "by accident", from different reasons. (From false assumptions you can derive everything, be it true or false.)
As I said, infinitely thin shells are surfaces, they don't have volume. So no matter how many of them you "add up", the sum will still be 0.
Notice that when you're calculating integrals, you're not really doing the summation directly. Usually, you're not even doing the limit directly. You just apply known formulas.
These formulas were known before calculus has been "invented" by Newton, Leibniz & co., because they've been known from antiquity, and they've been derived without any notions of infinitesimals, limits, 0-thickness shells etc. The Greeks have got them from the Egyptians and from the Arabs, the Arabs got them from the Hindus, etc. Unfortunately, just the formulas, without their derivations. So Newton & co. needed to "reinvent" these derivations to be able to pretend that they invented calculus all by themselves :P They had to "made up a theory for the formulas", and they did it wrong. But the deception worked so well, that it works to this day, since people are being taught these techniques in schools and being taught to not question it (since questioning usually lead nowhere anyway, because tachers don't know the answers too :P ).
As far as I understand it, it's not that you're adding 'volumes' with "0-volume objects" (even though such a volume is just 0). But the idea is explained either with infinitesimals or limits with which you are looking at the volumes as the depth of the strips are approaching 0 or are extremely close to 0 (but not exactly 0). An infinitesimal is something that is extremely extremely close to 0 but is not exactly 0 and so anything with a depth that is infinitesimally small still has a depth (so to speak) but it's just that it's very close to 0.
For example, when differentiating the principle we're always taught is using limits. And so we look at what happens to the gradient of a function at a particular point. Therefore, we take two points on a tangent line at that point. The idea of the limit is just looking at what happens when the vertical distance and horizontal distance approach 0. This doesn't mean you're doing the calculation 0/0 but rather doing calculations that get more and more accurate the smaller the distances get.
And anyway, it seems more to me that we just use this idea to derive other formulae and not actually on it's own to calculate stuff.
The fact is that what he just said is another way to explain summation of infinitely small objects to estimate the volume of one object by cross sections or cylindrical shells. ECT. Many methods however all he did was rephrase the process if you don't understand it's because you don't want to be wrong and you are just being rebellious. Zz. To argue this point is just wrong because by definition the process is logical.
We can take volum of revolution of curve between (x,y)represent by equation of y^2=r^2-x^2 and then take Integration to give you half volume of the sphear
More on the last few sentences. The volume of the cone is 1/3*pi*r^3. Caliveris states that shapes with the same cross section have the same volume. Therefore, with similar volume, it is possible to use the subtraction above.
Otherwise, just integrate pi*(r^2-x^2)) from r to -r. This is the integral equation for volume (pi*r^2). This works since integrals are the some of all the cross-sectional disks and the equation of a circle is all the points create the radius.
for a class 9th student, ur videos for surafce areas and volumes are perfect.....amazing videos
@Kosekans
Okay. We'll integrate unity over a region in 3D space, E. E is defined as a sphere of radius R centered about the origin. Integrating it will give us the mass of a sphere with radius R and uniform density of 1 throughout, also known as the volume of said sphere. We'll need to set up a triple integral in order to solve this problem. We'll be working with spherical coordinates for simplicity, since we have a spherical region.
@vampiracy
Our limits of integration are φ=0 to φ=π, and θ=0 to θ=2π. Integrating sinφ with respect to φ, from φ=0 to φ=π, gives us 2. Integrating 2 with respect to θ, from θ=0 to θ=2π, gives us 4π. Multiplying this with the (R^2) we took out gives us a surface area of (4π)(R^2).
+mathematicsonline I enjoy math. I have a math minor from college I've never used and I watch Numberphile often. I also really like your derivations. I am curious about one thing. Way back in high school (or maybe early college) I thought up a question I still haven't seen answered anywhere. If one were to cut a hole through a sphere and supposing the hole had a perfectly *square* (not round) cross section, how would one calculate the volume removed (or remaining)? Assume for simplicity that the central axis of the hole coincides with a diameter of the sphere. I'm sure this could be done with some relatively complicated calculus ( relative, at least, to the complexity of the derivations presented here). However, I'm hoping for a more intuitive approach, provided that such is even possible.
A very good question. Yes, of course it is possible to find the volume with some good advanced calculus.
I do not know the answer at the moment but if I figure it out, I will let you know.
Volume of Sphere: 4/3 pi r^3
Volume of Rectangular Prism: 2rs^2
4/3 pi r^3 - 2rs^2 < ACTUAL remaining volume
I would need to find the volume of a rectangular prism whose ends are part of the sphere's surface area
I do not know how to call that shape but then you would subtract that from 4/3 pi r^3.
This might be a problem too hard for me. I don't know how functions behave in 3-D.
Good luck.
omg, the dissection but so easy to understand jeezzzzz
The challenge of this proof is that is requires “believing” that surface area = 4 x pi x r^2. My daughter (in 8th grade) just shared this proof and I explained why the proof inherently had an underlying assumption that she couldn’t prove. Calculus and a volume integration is a far better way to prove this.
Clear, easy to understand, excellent presentation...very cool :-) Thank you!!!
An actually good video my ma'am made our class watch
Great explanation except for where it gets to, "The amazing fact is the surface area of the sphere is equal to exactly 4 times the area of the circle." Why 4 times? Why not 3 times or 5 times or any other number of times? That wasn't explained. It was skipped over. I would like to have heard and seen the animated explanations of why the surface area of the sphere is only 4 times the area of the circle.
it's been a year since your comment... meh
that's not the aim of the video
should he explain how to get pi with archimede's method?
How wonderful idea!! If we can think of this idea, we don't need to do any (inaccurate) experiment. Thank you for sharing.
Waiting for explanations like these for a long long time....................Thanks Best videos ever!! Wish this concept could be applied
to other areas of study!! Pure Genius Thanks so much.
My discovery according to reseach and experment. The volume of the sphere is - circle area times diameter or (pie radius squared) X diameter = volume,
And the sphere surface area is - circumference times diameter or (2 pie radius) X diameter = surface area,
Some equations: volume over 1/2 surface area = radius,
surface area over diameter = circumference,
volume over 2 radius cubed = pie,
volume over 1/2 radius = surface area,
surface area times 1/2 radius = volume,
surface area over circumference = diameter or surface area /2 pie radius - diameter,
volume over circle area = diameter or volume/pie radius squared = diameter, Enjoy your computation by using my formulas thank you!!!
It was amazing and very easy to understand.............Truthfully speaking it is very very interesting . hats off.................
T6hanks this was helpful although I had to watch the video twice to understand to anyone who is still confused after watching it the first time hit the replay button and go to 3:00 exactly and he will explain everything that may have confused you again in simpler terms
Thanks for this great work. With which programm did you make this animation?
I am so, so so so so pissed that I eventually had to look this up. I wanted so badly to figure this out by myself... Tried a bunch of things out in my head but there were always imperfections because I wasn't using pyramids. Beautifully explained and an amazing graphical presentation. Thank you.
me too, truly great video though. i think the discrepancy occurred because we assume all the bases of the pyramids are the same, so we don't label them individually. so the nth base value is an unknown concept until you physically see it. of course it could've been a different process for you but that's where i had my "ooooohhh" moment
This helped a lot! As a 6th grader I was able to understand it thank you!!!
Same thoughts, (found a mathematical proof for the surface formula, tho - rly hard ^^ ) but i kinda worked out the "curved surface" thing:
Lets just simplify it and say that a "sircle" is a figure with "infinite" sides. It's not a number. And the amount of pyramids are also infinite. But there is nothing like a perfect sircle, as when you zoom and zoom you'll find irregular patterns in the atoms, etc.. So the sircle isn't round, and the base of each pyramid represents each irregularity
Thank You for making math formulas vivid and clear
Very good. But the reasoning within the three-dimensional plane with the use of the integral is more interesting, not to mention that it does not directly need other formulas, just integrate.
Spectacular demonstration. Math is beautiful if you understand it.
Using the pyramid amazed me ❤
How to find Pi ? Draw a circle of radius 0.5. Draw another radius at 90 degrees. Join 2 ends. We get chord 0.707106781...Half of it is 0.35355339...Take 7 radii equal to 3.5 and subtract 0.35355339.. and we get 3.146446609...It is Circumference and also Pi (14 - root2)/4.. Truth is a roaring fire. --- SJR.
the formula depends on the given. If the given is diameter,you could use V= (3.14)(diameter cubed)and divided by 6.but if it's given is radius it will be V=4/3 (Pi)(radius cubed)
Really appreciate your visualizing work, it makes me understand a lot better!
Thankyou . Nicely animated.
I liked the drole ...
area is B1 plus B2 plus B3 !
Archimedes and projecting a sphere area onto a cylinder can prove the formula for surface area is 4 Pi r squared . You used square based pyramid bases which visually fit nicely, but curiously this process also works out the same formula using thousands of radius high CONES instead of thousands of radius high square based pyramids. Both base shapes and any errors are irrelevant because you suddenly swop the real result of the integration for the previously worked out area formula. NEAT .This is definitely early calculus, did the ancients do it this way ?
Superb work guyzz, finally found the type of explanation that i dreamt of but never had experienced , thank you soo much for making it mathematicsonline
Thanks for this very lucid example
great explanation !!! all says the formula of volume of a sphere but nobody explains how it came 😕😕😕😕😕
great explanation thanks for uploading!!!!
3blue1brown has done that
Video was wonderful but u have to watch this more that 1 time to understand each part easily....but nice video..keep it up!
The volume of a sphere is the integral of its surface- area with respect to R.
The surface- area of a sphere is the derivative of the volume with respect to R.
Thank you fo explaining. You just earned another subscriber
I'm fairly certain that someone just integrated the area of the circle to get volume of sphere.
Not area of circle but the surface area of sphere
The integration of 4 pi r^2 is volume of sphere
@vampiracy
. Our limits of integration are ρ=0 to ρ=R, φ=0 to φ=π, and θ=0 to θ=2π, and our differential is [(ρ^2)sinφ dρ dφ dθ]. Integrating (ρ^2)sinφ with respect to ρ, from ρ=0 to ρ=R, gives us (1/3)(R^3)sinφ. Integrating (1/3)(R^3)sinφ with respect to φ, from φ=0 to φ=π, gives us (2/3)(R^3). Integrating (2/3)(R^3) with respect to θ, from θ=0 to θ=2π, gives us (4/3)(π)(R^3). This gives us the volume of the sphere to be (4/3)(π)(R^3). There's your proof! :D
Your visualisation is beautiful, keep it up!
wow wow wow.....n times wow......my heartiest wow to your process of explaining as well as the animation 👍👍👍👍
I was so confused when i was making my own formulas up for a test and the volume formula came out to be 4/3pir³ instead of just pir³ like i theorized, and i was so intrigued as to where that 4/3rds came from
@mishraonline
Okay. We'll integrate over a disk D of radius R. Our integrand will be infinitely small pieces of area, which are sqrt([F_φ]^2+[F_θ]^2+1), and our differential will be [(R^2)sinφ dφ dθ]. Since (R^2) is a constant, we can take is to the front of the integral.
"(found a mathematical proof for the surface formula, tho - rly hard ^^ )",
I would recommend a look at Archimedes proof for this. He figured out how to prove that the surface area of a sphere is equal to that of a cylinder of the same diameter and height, combined with trig it produces a simple and beautiful proof.
In conclusion...after watching this i forgot the perimeter of a triangle
Hahah
omg sameeee
😂😂😂
Love this channel. I learn each day something new.
A very brief & to the point explanation. Keep it up!!!
Most beautiful explanation of tge proof....
very very good and easy to understand. Thank you.
If you think surface=4pi*r^2 is wierd I'd say
If radius is half of D (dimension), square area=D*D*4. Circle=r*r*3.14.
Cube is 4 squares =D*D*4*4. Sphere=r*r*4*3.14.
To me I think an area of circle should be just as much smaller than a square as sphere is than cube.
This channel is just absolutely beautiful.
Thanks man
You are great
Imagine the circle to be made up of infinitesimal shells. The volume of each shell is dV=4*pi*r^2*dr.
Integral(0 to r) dV = integral(0 to r) 4*pi*r^2*dr = 4/3*pi*r^3.
I love your videos
Thanks for fabulous work
We could ascertain which we can not imagine alone.
So awesome...I miss school.
Good video but i would have liked you mentioning that the number of pyramids needs to go up to an infinite number in order for the sum of the bases being equal to the surface area of the sphere. Also the explanation of how the the surface area comes together was, well.. not really explanatory haha. But if you don't care about details, this video is really nice. Also the animations! 👌
... also the infinite number for the height to be equal the radius
Marvelous, Never seen like that...... good work Thanks for spread this knowledge world wide through world wide web. Excellent from my side.
Will you please share which software do you use to animate your videos?
I understood it . Nice video. Thanks for explaning . I actually wanted to know that how the formula derived .
Outstanding explanation and your graphics are very helpful. Well done!