Very nice video, following the exploratory style that we all love. There's another nice way to distinguish between the m = 3 and m = 4 candidates for the sum of squares. So in the sum S(n) = 1^2 + 2^2 + 3^2 + . . . + n^2, each term is less than or equal to n^2. Hence since there are n terms in the sum, S(n) n^3, because n^4 grows faster than n^3. Therefore we discard the m = 4 case and proceed with m = 3. Overall very well done, hope you keep making videos. :)
@@laweaphysics4289 You're very welcome :) Yeah, that's the wonderful thing; there's always more than one way to go about it, and there's a degree of personal taste involved sometimes.
Lol nice joke/tease. Just to make it clear for other viewers this is just a joke the argument is not BS (circular) cause its just using the heights of the objects instead of the volumes. Also yes true to the label we do avoid integrals but that does not necessary mean we avoid calculus. In this case looking at limits, indeterminate forms, and a lot of perseverance to go through lots of computational algebra is all that's needed. This proof is INTEGRAL FREE!
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.
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.
I like this approach, in fact is interesting how that relates an abstract mapping with the visual example of slicing and modifying the cone/pyramid shape. And I also think is a good example to illustrate what a function does!😉😉
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
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).
@@laweaphysics4289 Thank you so much for believing in me! Now I am preparing to my UK GCSE Maths and honestly, I have never been more motivated! These videos are pulling out a lot of the stops and they make it easier to reach the understanding needed to get things done and don't even thinking about giving up the whole mess... So... Thank you for giving your support and knowledge to the world! Peace!✌️
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
Hola Joel, muchas gracias por tu comentario😉😉😉. La animación de los cilindros la hice en Power Point, haciendo que aparezcan y desaparezcan los distintos grupos de cilindros (estos son relativamente transparentes para que se viese el cono de fondo)
Thanks! I mainly use Power Point to make the animations and any recording program is ok for the voice. You could also utilize audio programs as Audacity for fixing tones and noise.
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?
You could try by using cylindrical coordinates. First, note that the region could be defined in some plane xy that we’ll rotate. This is equivalent to define the region in terms of the distance from the origin (x axis when x>0) and a vertical coordinate (y) and the region is defined by the domain of integration 0
I combine multiple free software (cause I can’t pay expensive programs 😅😅) -I mainly use Power Point to create the geometrical objects and equations, so I can animate them. -I also employ Audacity if I want to precisely modify audio clip. -And iMovie, Movie maker or your device’s incorporated video editor for joining the video clips with audio.
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.
Yeah, that’s true. As long as you have a “generalized cone” with section S that increases linearly with the height of the “cone” you could use the same approach even if the shape of the section changes from the base to the top. Good point this, you made me feel interesting about a new topic to study. Thanks!!!
what I appreciate about this proof is that it answers why 1/3 shows up in both pyramid and cone volume formulae, the limit of (1^2 + 2^2 + 3^2 + ... n^2) / n^3 = 1/3 . I find that much more satisfying than rotating a line :) Indeed when I first saw that pop up, I thought of the relationship between sums of cubes and squares... but when I finally remembered it (sum of cubes = square of sums) it doesn't quite fit this case... I
Form a traingle on the x & y coordinates so that the centerline of cone is on x axis and the slanting side of the cone is line passing through (0,0). Now revolve the slanting side round x axis which will generate a volume Pie .x .[y (square) dy Now integrate it Pie.x.[y (cube)/3] applying limits Volume of cone= 1/3. pie. x.[y (cube)] =1/3.pie.x .[y( cube)__0] *1/3 pie. x.y ( cube)*
@@laweaphysics4289 I have committed a mistake in it but the approach is correct. *Mistakenly in formula there is cube instead of square.*. Through integration you can also derive the formulae for area of circle, Volume of sphere and cicumfrence of circle etc
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.
Of course, the point here is to: -Show a method for the people who don’t know how to integrate. -Use a large variety of mathematical techniques. Anyway, I hope you liked the video. 😉😉
Muy buen video. Una pregunta,¿ Cómo reordenas las sumas parciales para que te quede con la ecuación de un polinomio? Entendí el razonamiento para determinar el grado del polinomio y demás pero me falta justamente el asegurar que es un polinomio. Muchas gracias!!
Gracias Ángel. Si no me equivoco preguntas por cómo llego a escribir la forma general de la suma hasta n^2 en forma de polinomio. En este caso he supuesto que la suma, al ser una función de n “que se comporta normalmente”, se puede desarrollar en potencias. Si el desarrollo no es infinito te queda un polinomio, obtienes el comportamiento lineal que muestro en el vídeo, y a partir de ahí se puede seguir. La demostración más rigurosa, tal y como yo lo veo, sería en los siguientes pasos: -Planteamos todo como en el vídeo y con lo que te he explicado. Por tanto, la fórmula del polinomio de grado 3 es nuestra “candidata” a fórmula general. - Se supone que la fórmula es la correcta para la suma de los cuadrados y se demuestra formalmente por inducción matemática, con lo que se establece que realmente hemos encontrado la fórmula buena. Espero que la aclaración te ayude, y si necesitas cualquier cosa no dudes en preguntar. Un saludo!!😉😉😉
the summation of squared terms can be generalized in a easier way with a little geometry... you can find it here... ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-aXbT37IlyZQ.html
Exactly, the key is in the techniques used to solve it (graphing functions, linear equations and finding the grade of the polynomial), so I try to present them in a way you could apply them in other mathematical problems.
