Deriving the Formula for the Volume of a Sphere | Calculus Application

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  • Опубликовано: 3 янв 2025

Комментарии • 16

  • @HAGARCIA
    @HAGARCIA 2 дня назад +1

    Muito bom, mestre! O senhor integra com base em coordenadas cilíndricas.

  • @rakeshupadhyay9082
    @rakeshupadhyay9082 4 дня назад +3

    Thanks for the basics

  • @englishforfunandcompetitio248
    @englishforfunandcompetitio248 4 дня назад +3

    There ain't sny redundant repeating. All he said and did was necessary. This video is meant for students and not for experts. To teach a student in an effective way, all this is necessary.
    Keep on doing the same way dude. I found it effective and NOT boring at all.
    Warm regards,
    Prof: Ahmad.

    • @learnmathbydoing
      @learnmathbydoing  4 дня назад

      Thank you so much for your kind words, Professor Ahmad! 😊

  • @DrakeLarson-js9px
    @DrakeLarson-js9px День назад +1

    Interesting video! (and English voice)

  • @pikabroiboy5093
    @pikabroiboy5093 6 дней назад +1

    You just taught me basic calculus and something I've been wondering about geometry, thank you

  • @maxime9636
    @maxime9636 2 дня назад +1

    🙏🏻👍🏻👍🏻👍🏻❤️❤️❤️

  • @rebeccachauhan9003
    @rebeccachauhan9003 5 дней назад

    This derivation was very hard when I personally tried to do it. Excellent work

  • @rickrys2729
    @rickrys2729 6 дней назад +1

    Thanks - something I could follow with basic integral calculus.

  • @bloodywingrich8844
    @bloodywingrich8844 5 дней назад +1

    So cool!

  • @KipIngram
    @KipIngram 5 дней назад +1

    Uh... it comes from circumference = pi*diameter. It's nothing but the simplest possible calculus. Let r be the radius of an arbitrary circle and let c be the circumference. You can envision your sphere as a stack of rings, with changing radius. Consider a hemisphere and let angle a measure the angle between horizontal and the radius R of a sphere you're building. The radius of a circle reached at angle a is just R*cos(a) The width an arc strip for that circle is just R*da. So the area of that strip is
    A = 2*pi*R*cos(a)*R*da
    A = 2*pi*R^2*cos(a)*da
    Now integrate that from 0 to pi/2. The integral of cos() is sin(), so we have
    Hemisphere area = 2*pi*R^2*sin(a), evaluate (0 to pi/2)
    The result is 2*pi*R^2. Double that to get the whole sphere - the surface area of a sphere is 4*pi*R^2.
    That's step 1. Now do a similar process adding up the volume of spherical subshells of an entire sphere. We just need to integrate from r=0 to r=R, the final sphere radius:
    Sphere volume = integral of 4*pi*r^2 dr from 0 ro R.
    The result, OF COURSE is (4/3)*pi*R^3.
    So there you go. Once you accept that the ratio of a circle's circumference to its diameter is a constant regardless of the size of the circle, and name that ratio pi, the rest is just simple math. There is no mystery here. And accepting the circumference /diameter thing isn't hard either, because THEY'RE BOTH LENGTHS. They're going to scale in exactly the same way.

    • @learnmathbydoing
      @learnmathbydoing  5 дней назад

      Thanks for sharing your detailed breakdown of the formula derivation. It's always interesting to see different ways of approaching a problem.

  • @longbeachboy57
    @longbeachboy57 5 дней назад +1

    When I saw this the first time, I kind of realized what 'integration' was for...

  • @shrikantlimaye9213
    @shrikantlimaye9213 2 часа назад +1

    भास्कराचार्य यांनी हीच पद्धत वापरून 'स्फिअर व्हॉल्यूम'' काढला,ईसवी सन १११०

  • @BerndLercher
    @BerndLercher 4 дня назад +2

    Come to the point. Redundant repeating is boring.