Why is dxdydz=ρ^2sinφdρdθdφ? (using Jacobian)

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  • Опубликовано: 22 ноя 2024

Комментарии • 20

  • @bprpcalculusbasics
    @bprpcalculusbasics  3 месяца назад +1

    The geometry way: ruclips.net/video/uL_yq733CTY/видео.html

  • @xinpingdonohoe3978
    @xinpingdonohoe3978 2 месяца назад +8

    It's easier to do it in two steps.
    First is (x,y,z)→(r,θ,z) with (x,y)→(r,θ).
    x=r cos(θ), y=r sin(θ), x²+y²=r²
    dxdy=r drdθ, dxdydz=r drdθdz
    Then do the exact same thing again for (r,θ,z)→(ρ,θ,φ) with (r,z)→(ρ,φ).
    z=ρ cos(φ), r=ρ sin(φ), z²+r²=ρ²
    dzdr=ρ dρdφ, dzdrdθ=ρ dρdφdθ
    Combine them to get
    dxdydz=rρ dρdφdθ
    r=ρ sin(φ)
    dxdydz=ρ²sin(φ) dρdφdθ

    • @thenew3dworldfan
      @thenew3dworldfan 2 месяца назад

      Have you heard about the negation of the epsilon delta definition? The statement is: There exists some positive
      ϵ such that no matter how small a positive δ you choose, there is always some x within the δ-neighborhood of c (but not equal to c) for which the distance between 𝑓(𝑥) and L is at least
      𝜖.

  • @Arycke
    @Arycke 2 месяца назад +7

    Hey, great video! Typo in the title though sir.
    You have dpdødp

    • @mohannad_139
      @mohannad_139 2 месяца назад +1

      @@Arycke
      Yes he wrote dρdθdρ while he is supposed to write dρdθdφ

    • @bprpcalculusbasics
      @bprpcalculusbasics  2 месяца назад +2

      Thanks! I just fixed it.

  • @jacokriek
    @jacokriek 2 месяца назад +19

    You waited a semester too long to make the calc 3 videos😅

    • @JayOnDaCob
      @JayOnDaCob 2 месяца назад

      Nah like a year, this class was by far my worst class in uni so far

    • @k0pstl939
      @k0pstl939 2 месяца назад

      I had to take it twice

  • @yacovaaa9042
    @yacovaaa9042 2 месяца назад

    what about dxdydzdw?

  • @bitoty9357
    @bitoty9357 2 месяца назад +2

    why does the jacobian works?

    • @Seb2006-y4x
      @Seb2006-y4x 2 месяца назад

      What kind of Greek letter is that the one representing sine angle?

    • @erroraftererror8329
      @erroraftererror8329 5 дней назад

      @@Seb2006-y4xTheta or phi?

    • @Seb2006-y4x
      @Seb2006-y4x 5 дней назад

      @@erroraftererror8329 i think phi

  • @saharhaimyaccov4977
    @saharhaimyaccov4977 2 месяца назад

    If use x=y=z

  • @cyrusyeung8096
    @cyrusyeung8096 5 месяцев назад +1

    Can you make a video explaining the concept of Jacobian marix? In particular, why dxdydz = ||J(ρ,φ,θ)||dρdφdθ.

    • @mohannad_139
      @mohannad_139 2 месяца назад +5

      How is your comment 2 months ago?

    • @kunwarbtw981
      @kunwarbtw981 2 месяца назад

      yeah lol

    • @be-mr3tj
      @be-mr3tj 2 месяца назад +2

      Well to answer your question (I'll simplify to 2D) shifting the coordinates to a new point (x+dx, y+dy) comes with a new area. For rectangular cartesian coordinates this is is a rectangle with area dxdy. But for polar and other coordinate systems, this new additional area may not be a rectangle but can be approximated as a trapezium. This is essentially a rectangle in polar coordinates, but the area depends on r. (At different points, a small change in r,theta has differently scaled areas) The jacobian accounts for this scaling and thus any stretching in area that may occur when changing coordinates. I recommend looking into this further geometrically.

  • @maxvangulik1988
    @maxvangulik1988 2 месяца назад

    now do it in reverse
    dpdQd@=|J(x,y,z)|dxdydz
    p=sqrt(x^2+y^2+z^2)
    Q=arctan(sqrt(x^2+y^2)/z)
    @=arctan(y/x)
    dp/dx=x/p
    dp/dy=y/p
    dp/dz=z/p
    dQ/dx=xz/rp^2
    dQ/dy=yz/rp^2
    dQ/dz=-r/p^2
    d@/dx=-y/r^2
    d@/dy=x/r^2
    d@/dz=0
    J(x,y,z)=|x/p y/p z/p; xz/rp^2 yz/rp^2 -r/p^2; -y/r^2 x/r^2 0|
    =x/p•|yz/rp^2 -r/p^2; x/r^2 0|
    -y/p•|xz/rp^2 -r/p^2; -y/r^2 0|
    +z/p•|xz/rp^2 yz/rp^2; -y/r^2 x/r^2|
    =x^2/rp^3+y^2/rp^3+z^2/rp^3
    =1/rp
    =1/sqrt(x^2+y^2)sqrt(x^2+y^2+z^2)
    dpdQd@=dxdydz/sqrt((x^2+y^2)(x^2+y^2+z^2))
    figures, but it was fun to work out :)