Flux through surfaces | MIT 18.02SC Multivariable Calculus, Fall 2010

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  • เผยแพร่เมื่อ 2 ม.ค. 2011
  • Flux through surfaces
    Instructor: Christine Breiner
    View the complete course: ocw.mit.edu/18-02SCF10
    License: Creative Commons BY-NC-SA
    More information at ocw.mit.edu/terms
    More courses at ocw.mit.edu

ความคิดเห็น • 51

  • @juanpreciado4081
    @juanpreciado4081 7 ปีที่แล้ว +18

    Why didn't you teach them using the parametrization and taking the cross product of partial derivatives of parametrization to get the normal vector? This dividing by "a" seems to only work sometimes.

    • @chriskitley9875
      @chriskitley9875 6 ปีที่แล้ว +1

      Juan Preciado I think this is resuscitation, not a lecture.

    • @FaizaanDatoo
      @FaizaanDatoo 4 ปีที่แล้ว

      @@chriskitley9875 Recitation*

    • @robertoberidojr.435
      @robertoberidojr.435 3 ปีที่แล้ว

      It confused me also

    • @robertoberidojr.435
      @robertoberidojr.435 3 ปีที่แล้ว

      I was use to getting the cross product of the partial derivatives

  • @santiagoalpuy983
    @santiagoalpuy983 11 ปีที่แล้ว +2

    Thank you for teaching in such a clever way. Keep up the good work. Greetings from Uruguay.

  • @kevingomez4096
    @kevingomez4096 9 ปีที่แล้ว

    She's fantastic. Thank you.

  • @SGonzalezINTJ
    @SGonzalezINTJ 12 ปีที่แล้ว

    I love you! Thank you!

  • @dheerajtiwari7534
    @dheerajtiwari7534 2 ปีที่แล้ว

    Lot of thanks from india👍

  • @takahirokobayashi1385
    @takahirokobayashi1385 7 ปีที่แล้ว

    Awesome lecture

  • @zmh786
    @zmh786 13 ปีที่แล้ว +15

    how did you work out the normals?

    • @ShaunGan
      @ShaunGan 6 ปีที่แล้ว +1

      did you find out the answer to this, especially for the curved side

    • @PhysicsMath
      @PhysicsMath 4 ปีที่แล้ว

      You can take gradient of that field and devide magnitude of that gradient

    • @TacoShmaucko
      @TacoShmaucko 4 ปีที่แล้ว

      the normal vector is like the parametric equations of ;if you plug that into a graphing calculator using parametric functions you get a full circle, but here we only have from 0-pi/2, which is controlled by the boundary of integration of d-theta. so the professor could have crossed the with the the F = < z ,x , y > and gotten the same. Her (a*d-theta*dz) is her jacobian for her cylindrical coordinates ( r , theta , z). Know t is theta in my example, I'm not writing cos(theta).

    • @robertoberidojr.435
      @robertoberidojr.435 3 ปีที่แล้ว

      @@JG-zn6od amazing. Is that really it?

    • @robertoberidojr.435
      @robertoberidojr.435 3 ปีที่แล้ว

      @@JG-zn6od please teach me, is the gradient of the surface different from taking the partial derivatives Ang getting its cross product?

  • @xxxcoolboyxxx
    @xxxcoolboyxxx 13 ปีที่แล้ว

    @MrVandeju Because.. you have to convert to spherical coordinate system.
    The formula for converting Rectangular --> Spherical is that z = row cos(phi), and here we know row = a since a is the radius.

  • @Rkmath-gq5nq
    @Rkmath-gq5nq 5 ปีที่แล้ว

    Thanku mam for doing your work very well

  • @esraasalamaq
    @esraasalamaq 4 ปีที่แล้ว

    شكرا😍

  • @giorgiogamo130370
    @giorgiogamo130370 9 ปีที่แล้ว +1

    Simple and excellent explanation, thanks for share!

  • @taylershaw2034
    @taylershaw2034 5 ปีที่แล้ว +8

    the normal calculation explanation confused me so much I stopped watching

  • @joshuaaa3337
    @joshuaaa3337 4 ปีที่แล้ว

    hey smartie. your explanation is fantastic, but i wanna ask you something. When the flux is outward, we have to take into account the top and the bottom of the cylinder, but why did you ignore them? Some people may think that they will cancel each other. Actually, this thought is wrong. The bottom will have zero flux, but the top will have some flux.

  • @mihaitensor
    @mihaitensor 12 ปีที่แล้ว +2

    good explanation

  • @marie-angesayegh5480
    @marie-angesayegh5480 8 ปีที่แล้ว +1

    why is there no z in the normal vector

    • @IMACORPTECH
      @IMACORPTECH 7 ปีที่แล้ว +4

      Hi, it means, does not depend of z, n can be anywhere along z

    • @Rayquesto
      @Rayquesto 7 ปีที่แล้ว +2

      Normal points perpendicular to the surface. There's not really a surface in the x-y plane. So, no z component for normal vector.

  • @bernarddoherty4014
    @bernarddoherty4014 5 ปีที่แล้ว

    You say the normal, but actually you mean the unit normal because you are dividing but 'a'. Am I correct?

  • @bryankerr6725
    @bryankerr6725 5 ปีที่แล้ว

    isn't the first one =0, the divergence of F is 0... surface integral is equal to divF of the volume integral

    • @varunmallya5852
      @varunmallya5852 4 ปีที่แล้ว +1

      Gauss theorem is invalid for unclosed surfaces

  • @MauroVictor
    @MauroVictor 8 ปีที่แล้ว +2

    Very clear and objective. Thank you very much

  • @jijobuje
    @jijobuje 4 ปีที่แล้ว

    I need a video that this topic is fully explained like teaching a little child. So i can solve even the hardest problem of this topic. But in every video or even my own face to face lectures, teachers don't teach the every aspect of the topic so i am struggling and confusing while trying to solve my textbooks questions.

  • @imegatrone
    @imegatrone 12 ปีที่แล้ว

    I Really Like The Video Flux through surfaces From Your

  • @mdmobashshir432
    @mdmobashshir432 8 ปีที่แล้ว +1

    5:45 that zcosϴ should be zacosϴ. you guys should put an annotation as this have been watched 31k times. LUV from INDIA.

    • @_arteetmarte
      @_arteetmarte 8 ปีที่แล้ว

      No, she did it right... (Remember it's the unit vector, so you have to divide zacosϴ by a, the norm.)

    • @mdmobashshir432
      @mdmobashshir432 8 ปีที่แล้ว

      When #MOBI says she wrong. It doesn't matter she is MIT or Oxford, She is damn wrong. No excuse whatsoever will make her right.

    • @MeagaRain84
      @MeagaRain84 7 ปีที่แล้ว +4

      ur stupid ... she did it right

    • @mdmobashshir432
      @mdmobashshir432 7 ปีที่แล้ว

      MeagaRain84 ok

    • @ronnies.3440
      @ronnies.3440 7 ปีที่แล้ว

      You're* stupid..

  • @TheImpressionable
    @TheImpressionable 11 ปีที่แล้ว

    :D:D:D

  • @EltonSilvaCruz
    @EltonSilvaCruz 11 ปีที่แล้ว

    :D

  • @monu_7712
    @monu_7712 4 ปีที่แล้ว +1

    flux across Surface
    ∬ F.ñ ds

  • @financewithsom485
    @financewithsom485 6 ปีที่แล้ว

    That's mit

  • @jr7sa
    @jr7sa 7 ปีที่แล้ว +1

    Jumped tons of steps. Bad

    • @ManilaSayang
      @ManilaSayang 6 ปีที่แล้ว

      This is a class in MIT. And you are from a shit-hole country, so you can't cope up.

    • @newtubedavid
      @newtubedavid 5 ปีที่แล้ว +5

      You can find this material in any multivariable calculus textbook. It isn't some revolutionary material out of MIT. If your argument is that he cannot understand this material based on his geographical location, then it's only more pitiful that you don't have that excuse and are still here watching the same video because you obviously have trouble with the same concept. Since you yourself have pointed out your issue to be a lack of intelligence seeing that you are from "the proper country" to understand this and still don't...

  • @asmcriminaL
    @asmcriminaL 8 ปีที่แล้ว +4

    I do not like the generalized stuff. e.g. using a and h.

    • @lucasm4299
      @lucasm4299 6 ปีที่แล้ว +2

      asmcriminaL
      You should get used to it. They’re just like numbers

  • @Zeppelinpuppy
    @Zeppelinpuppy 9 ปีที่แล้ว +31

    Terrible explanation