Відео

Nice try. Would be much easier to use Cavalieri's principle, pyramid with its reflection. If you are really interested in this task, ask yourself "what exactly three here"? You will be surprised how far this question goes and how many interesting things opens. Here is a clue for you for the first step: 3 here is the number of dimensions. For flat triangle you will gave 2 and for 4 dimensional cone you will have 4. But the way you find it will lead you to many discoveries in math.

Absulutely love this video

Wrapping my mind around shapes just being equivalent to a rectangular prism with side lengths that have multiplication in them and pi was such a trip.

Nicely explained

Very nice proof.

I like how he performed integration without mentioning it through its notation once.

Please excuse dumb question. At 6:00, "according to the right hand side if n is larger all these terms will vanish". Why would they vanish exactly?

Great explanation. 👍

the summation of squared terms can be generalized in a easier way with a little geometry... you can find it here... ua-cam.com/video/aXbT37IlyZQ/v-deo.html

Protocalculus!

I really hope that I will be able to understand this level of mathematics one day !

Me gustó demasiado el vídeo, las animaciones y la explicación, te felicito. Una pregunta, tú hiciste las animaciones de los cilindros infinitos en el cono? si es así, en dónde los realizaste? quedaron geniales, saludos

Hablas muy deprisa y las imágenes aparecen y desaparecen con mucha rapidez: no da tiempo de asimilar todo lo que cuentas. No obstante, tu trabajo de edición y maquetación tiene mucho mérito. ¿Por qué lo has grabado en inglés? Ya hay demasiadas cosas editadas en ese idioma. Agradezco el español.

Nunca me habian explicado las derivadas mejor, q arte

Sum of consecutive squares again! Why do figurate numbers appear everywhere!!!! And BTW I love figurate numbers - think they are cool.

Great!!!

Thank you too Joan!! Great work on the video, congratulations!!! 👏👏

1+ sinx = sin²x + cos²x + 2·sinˣ/₂·cosˣ/₂ 1-sinθ = sin²θ + cos²θ- 2·sinᶿ/₂·cosᶿ/₂

Consider the region R delimited by the curve y = C− x^2 and the x axis. Use integration to find the value of C > 0 so that the volume of the solid obtained by rotating R about the x axis is 64√2π/15 Note: Plane sections known to the x axis are circles. Would you help me?

999th like and 100th comment somehow Also thanks

He said: "no integrals"... And then he used what is integral in disguise.

Very detailed and informative. Thanks.

This was a super video!! #1 subs provider -> P r o m o S M!!

LIMIT is a strong & mutual bridge between algebra to calculus.....you have proven this concept with excellence.....keep it up👍

Simply marvellous👌👌👌👌👌

What I like is a visual proof for a piramide exists. That 3 piramids fit in a cube ( heigh* base). Thus 1/3 of base *h and then using a function that maps every slice to a cone/elliptic cone or weird complex piramide is by observation just a constant that maps very slice exactly to A (piramid) to A (any complex cone/pirmide) should logically lead to : Volume = A_piramid*h/3 to Volume = A_complex_cone/piramid*h/3, because for every slice we define the function to simply map the areas, this is a continous constant function.

Spotted an error at 7:11 , the 3rd row should be : the same, but the a_1*3 instead of a_1*2. Since n=3 there.

5:28 bruhhhhhhhhhhhhhhhhhhhhhhhh

Another example of My "Inverted Reverse Neutrality". Superpositon=Every point of space=180^3>0.1=1<3>[1,080]=0.2^3>1<3=D×3>1R=%×°÷^3>1=0.180^3>1-^0.2×1^3>1=(-360-)<[0.1^3]>!^3=[540]<0.1×1^3=0.540=270>(-0-)^3>1<2=[0.540]^2=!^3×1,080>1^2=1/4>1-[#]÷27-^3>1<(#)=9^3=0.1^3>1<27-1^3=0.27-^3>1<9^3=-!^3>(#)=9^3>1=[#]÷1^3>1×27-^3>1+@>180^3>1=[0.1]=2-^3>1<0.1^3=000...=E=Mc^3>1=[Relitivity] 540=100%×°-!^3×0.1<!>1.0=180^3>1=0.360>[=)×(=]<0.180=1-!^3>180^3>0.1=|^2=[1080]×%+°>/~¡_!+]=(-360-)<720+1^3>1<4^-1<5>9=999...>0<3×180^3>1.1=("Two")>[1<E=M>C-^2>1]=1^2=2/3>1^3=0.111...+...111.0^3>111...=3×D<4^3>180^3>0.1^3=0+1=Eו••>1=[-4-]<3×D>1^2=1/3>0.1^3/-1^3=0.1^2-1/3>1=0.0×1<0.1>1+0.0^3>3=999...+0×111...=10,-0-:-,^100=54^3>1<180^3>1÷27-^3>[#]=1<3×0.1^3>1×2=[1^2]=/%×°50>!=270^3>111...=Pyramid×1^2=25%×°>1^3=1<3=Cube×1^3>1<3=1.620>1^2+0.1^3×2-1^3=[◇]<1,080-!^3>1-2^3>1<2^3=540.-1^3>1<("Two")=[-2-]+^3>1.0=!×/-%+°=100>1=440%×°>0.1^3=("One")>[0.1^3]=1×0.540^3

Joan la tienes en español o siemprebhaces el guión en ingles?

This is really high quality, I am sure that in the future you will get loads of views and subscribers!!!

The cone is simply a stack of incrementally widening discs of radius r = Rh/H at any height h going from 0 to H with radius r going from 0 to R. Area of the disc = π r^2 = π (Rh/H)^2 and height = dh leading to the volume of each infinitesimally thin disc, dV = π (Rh/H)^2 dh = π R^2/H^2 (h^2dh). Integration can be performed between any 2 heights to get the volume of any frustum, but to get the volume of the entire cone go from h=0 to h=H. V = π R^2/H^2 (h^3/3) from h=0 to h=H = 1/3 π R^2 H

nice vedio :} what program u use to explain

Hawa pani. Astronomy.

This formula actually generalizes straightforwardly to any suspension: Any height h suspension of a 2D shape with area A, i.e., any 3D shape that has a flat 2D base with area A and goes linearly to a point at height h has volume Ah/3. In fact, any height h suspension of an (n-1)D shape with (n-1)D volume A, i.e., any nD shape that has an (n-1) base with (n-1)D volume A and goes linearly to a point at height h has nD volume Ah/n. Proof: Layer at height h-t has (n-1)D volume L_t = A(t/h)^(n-1), so the nD volume of the entire shape is int_0^h L_t dt = int_0^h A(t/h)^(n-1) dt = A/h^(n-1) int_0^h t^(n-1) dt = A/h^(n-1) h^n/n = Ah/n.

Very nice proof and excellent graphics! Thank you.

I didn't understand where this was going at first, the ? 'upside down' confused me, I have never seen it before .... but then I got it :-)

Unexpected factorial in the thumbnail!

The cool part is that the core of the proof applies to more than cones but to pyramids as well - the base doesn't even need to be regular so far as I can tell.

Amazing!

It is hr^2 but not 1/3 pi hr^2 You guys will need to be re-educated

Maths is consistent because, nevermind what human language we speak and where we start out from, we understand each other perfectly.

It can also be proved through Calculus by integrating the equation y = k x

The word "determines" is not pronounced as you think it is pronounced.

Hi, vera good vídeo!

What software you use to make this videos , please tell me

I think the volume of a cone is easily proved by rotating the centroid of the area of the triangle about the the y-axis. Volume = Area x 2π(Centroid) = (hr/2)(2π)(r/3).

👍 from 100 th subscriber

it is easier to use integration with solids of revolution

interesting detour for the sum squares!