Double Integral through a Change of Coordinates (the Jacobian)

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  • čas přidán 21. 04. 2020

Komentáře • 32

  • @steveying1305
    @steveying1305 Před 3 měsíci +2

    This is amazing, the best explanation of transformation!!!!!!

  • @tanishkumar6682
    @tanishkumar6682 Před 2 lety +4

    Thankyou, so much for such a good explanation
    i was not able to understand this integration technique through any video on you tube except your video.You made this concept crystal clear for me.

  • @mr_mouli_10
    @mr_mouli_10 Před 8 měsíci

    keep going on upload the videos lots lo love from INDIA thank u sir for such a nice explanation

  • @Pitr0x
    @Pitr0x Před rokem

    thanks, its help a lot

  • @gauravshahi2308
    @gauravshahi2308 Před 2 lety +2

    Thank u sooooooo much

  • @onlineskillsbyimranbilal
    @onlineskillsbyimranbilal Před 2 lety +2

    Explained well

  • @KBMNVLpNdLumkstz
    @KBMNVLpNdLumkstz Před rokem +2

    thank you very helpful

  • @truman0725
    @truman0725 Před rokem +1

    Great video! Clear speaking and nice handwriting, thank you!

  • @jonteyh3388
    @jonteyh3388 Před rokem

    @Andrew Bulba Thanks!

  • @lyzongaming01
    @lyzongaming01 Před 3 měsíci

    Love from India🤝 respect to you sir folding this in such less time with ease

  • @thenoobchess
    @thenoobchess Před měsícem

    The determinant would be -1/2. You forgot you write the negative sign. But thank you for your great explanation

    • @Nyaterikk
      @Nyaterikk Před měsícem

      he took the absolute value of the determinant

  • @aidamahmoudi2627
    @aidamahmoudi2627 Před 29 dny

    thank u so much for this video. it was very helpful! :)
    btw anyone noticed that small hair on the paper?! lol

  • @nwakilenwabueze5850
    @nwakilenwabueze5850 Před 3 měsíci

    Rahhhh thanksss

  • @HashemAljifri515
    @HashemAljifri515 Před rokem

    If the determinant was negative, should I make it positive when evaluating double integral?, cuz i have a homework that contains two answers with the same boundaries but the determinant is negative

    • @hassanawad2131
      @hassanawad2131 Před rokem +4

      The rule implies that you put the determinant in absolute value

  • @HashemAljifri515
    @HashemAljifri515 Před rokem +1

    Why do we need to integrate using the jacobian coordinates? It seems more complex than cylindrical coordinates!

    • @andrewbulawa3273
      @andrewbulawa3273  Před rokem +2

      It's the same thing! You use the Jacobian whenever you change coordinates. When we switch from Euclidean coordinates to polar or cylindrical coordinates, dxdy changes to r drdθ. The r in front of drdθ comes from the Jacobian. You may not have been told that when you first learned it, but that's what it is.

    • @HashemAljifri515
      @HashemAljifri515 Před rokem +1

      @@andrewbulawa3273 Oh I get it now! So I can use any of these methods to find a volume of the solid including polar coordinates?

  • @focf1981
    @focf1981 Před 4 lety +2

    Lol, there "u" are. Classic. Too bad you weren't using "r" as one of the variables.

  • @CrumbleRumble-kw3zf
    @CrumbleRumble-kw3zf Před 2 měsíci

    Isn't the primitive of 0 a constant ?

    • @CrumbleRumble-kw3zf
      @CrumbleRumble-kw3zf Před 2 měsíci

      Thank you for the video

    • @andrewbulawa3273
      @andrewbulawa3273  Před 2 měsíci +1

      Indeed it is. But then when you evaluate the constant function, say f(v) = c, at -5 and -9 and subtract, you get f(-5)-f(-9) = c - c = 0.

  • @Artaxerxes.
    @Artaxerxes. Před 2 lety +4

    Shouldn't the Jacobian be the transpose of what you wrote. Doesn't make a difference when calculating det though. Maybe I'm mistaken

    • @hikarifathan5143
      @hikarifathan5143 Před rokem +3

      I'm thinking the same too

    • @andrewbulawa3273
      @andrewbulawa3273  Před rokem +3

      Yes. You are correct. ..and what I was calling the Jacobian, I should have called the Jacobian Determinant.

  • @badrmoubile9219
    @badrmoubile9219 Před rokem

    I toughed the hair was on my screen hahaha

  • @MrJohanh80
    @MrJohanh80 Před 9 měsíci

    So the area of the parallelogram is equal to 0? 😮

    • @danieljorgeescobari6532
      @danieljorgeescobari6532 Před 8 měsíci +2

      The x-y plane crosses the region R in a way that the positive volume is the same es the negative...add them and you get 0

    • @andrewbulawa3273
      @andrewbulawa3273  Před 8 měsíci +6

      We're not computing area here. To compute area of a region, we would integrate the function f(x,y)=1. Here the function is f(x,y)=x-y, which gives something else.