Change of Variables

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  • Опубліковано 3 бер 2020
  • The Jacobian
    Check out my Double Integrals Playlist: • Changing order of inte...
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КОМЕНТАРІ • 54

  • @shandyverdyo7688
    @shandyverdyo7688 4 роки тому +7

    Me : *trying to sleep.
    UA-cam : hey, we have a new video for you.
    Me : Ok, here i am.

    • @gnikola2013
      @gnikola2013 4 роки тому +2

      You don't need sleep, you need calculus

  • @michaelzumpano7318
    @michaelzumpano7318 4 роки тому

    Beautifully done!

  • @TheMauror22
    @TheMauror22 4 роки тому +1

    Awesome video! I think it would be great if you do a video proving the change of variables theorem. I love when you do a video proving a cool theorem. Please do more of that!

  • @emman100
    @emman100 4 роки тому +3

    I never understood why we had to do the Jacobian until now. Also nice suit.

    • @tomatrix7525
      @tomatrix7525 3 роки тому +1

      Emmanuel Azadze Yep, he presented it really well. Jacobian basically calculates the area change. So we change it to a nicer shape and account for area change via jacobian.

  • @rohithnarra9026
    @rohithnarra9026 3 роки тому

    Dr. Peyam you are the GOAT! My multivariable calculus and advanced calculus professors were unable to explain why the change of variables formula involves a Jacobian determinant and what that means. You are the best and keep doing what you are doing as you are f****ng amazing at it!

    • @drpeyam
      @drpeyam  3 роки тому

      Thanks so much!!!

  • @tgx3529
    @tgx3529 4 роки тому

    It was nice. But only 0

  • @frozenmoon998
    @frozenmoon998 4 роки тому +1

    I was awaiting to hear "Where is my BpRp" but you were close :).

  • @dhunt6618
    @dhunt6618 4 роки тому

    Man, so glad you found that lost 16/sqrt(3) slice of pi! I bet it was delicious!

  • @avinashpandey8933
    @avinashpandey8933 4 роки тому +1

    From india lots of love

  • @adityaujjwalmain5943
    @adityaujjwalmain5943 4 роки тому +2

    Niceeeee

  • @manfredwitzany2233
    @manfredwitzany2233 4 роки тому +1

    Realy good work. How would a physist solve this problem? He would directly formulate the problem in adequate coordinates, namely rotate the coordinate system and use elliptical cooridantes. It is intersting to see, the way to solve is nearly the same. Only matrix diagonalisation is not necessery using this way.

  • @tomatrix7525
    @tomatrix7525 3 роки тому

    Which video is it that you were talking about at the start to show the linear algbra approach to calculating our change of variable

  • @ugursoydan8187
    @ugursoydan8187 3 роки тому

    have you got a proof for the jacobian matrix technique? Why are we doing a matrix calculation?

  • @cameronspalding9792
    @cameronspalding9792 4 роки тому

    If you integrate a function over a subset of the complex numbers: is that not similar to computing an integral of a subset of R^2

  • @hamanahamana3799
    @hamanahamana3799 4 роки тому +2

    Good video! I have a couple questions though.
    1. at 8:48, why do we need the absolute value sign? I know you said "to keep it positive," but why do we need to keep it positive?
    2. I'm confused about how you went from dxdy/dudv to the Jacobian. Could someone explain it to me?

    • @drpeyam
      @drpeyam  4 роки тому +1

      1) Basically the Jacobian measures a positive change in the area or volume, and you need that because if f is a positive function, then the integral after change of vars should be positive

    • @drpeyam
      @drpeyam  4 роки тому

      2) dxdy/dudv is just notation, it doesn’t mean anything. The Jacobian is the correct formula to use, and again because the det measures a change in area/volume

    • @hamanahamana3799
      @hamanahamana3799 4 роки тому

      Awesome. Thank you!

  • @mingmiao364
    @mingmiao364 4 роки тому +1

    Awesome presentation! Would you be interested in making a video on differential forms? Cheers man

  • @cheny0165
    @cheny0165 3 роки тому

    Hi Dr Peyam, I'm confused about why r ranges from 0 to 1. If we fix a θ, r is always =1 since it's a unit circle, surely it's never 0. But again we would be integrating from 1 to 1 which is 0 which raises another problem...Thanks!

    • @drpeyam
      @drpeyam  3 роки тому

      But we’re ranging over the unit disk not the unit circle, so r goes from 0 to 1, which represents the inside of the disk

  • @brendonreidvictor3491
    @brendonreidvictor3491 4 роки тому

    Dr Peyam just wondering why is it that when we make a single variable U-Sub (i.e.the single variable calculus analogue of this) we do not need to take the absolute value of the dx/du?

    • @drpeyam
      @drpeyam  4 роки тому +1

      Because the way they teach you in calculus is wrong

    • @drpeyam
      @drpeyam  4 роки тому +1

      The Jacobian ua-cam.com/video/SrYStw84T4o/v-deo.html

    • @nathanisbored
      @nathanisbored 4 роки тому +2

      You don't need abs value basically because the x-axis is the "region" being considered, and the x-axis is one dimensional. So it only really has one orientation. With multiple variables, the region being considered in often two-dimensional, representing an area. So imagine the region as a lake of water. You can either be above the surface of the water or below. In other words, if the region is upside down, you get a different value. This doesn't happen with a line, because a line can't really be upside down.

  • @morphomorph4923
    @morphomorph4923 4 роки тому

    Does this also work if u & v aren't linear combinations of x & y?

  • @dougr.2398
    @dougr.2398 4 роки тому

    What does the region look like? A good drawing would be nice

    • @drpeyam
      @drpeyam  4 роки тому +1

      Just a diagonal ellipse

    • @dougr.2398
      @dougr.2398 4 роки тому

      Dr Peyam yes, the xy factor tells us so

  • @vectorial6467
    @vectorial6467 Місяць тому

    Dr peyan i canr find the link of methods to find the convinient change of variable via diagonalization can you give it to me please?

    • @drpeyam
      @drpeyam  Місяць тому

      I think it’s called quadratic forms!

    • @vectorial6467
      @vectorial6467 Місяць тому

      @@drpeyam ah alright ! do you have any videos regarding this topic?

  • @nathanisbored
    @nathanisbored 4 роки тому

    Since polar coords is really a second change of variables in disguise, surely there must have been a more direct change of variables to choose at the beginning to skip that step.
    Also, I remember learning this vaguely, but I could have sworn I remembered the jacobian formula involving some kind of cross product. What’s that about?

    • @drpeyam
      @drpeyam  4 роки тому

      Cross product? Not sure

    • @hyperboloidofonesheet1036
      @hyperboloidofonesheet1036 4 роки тому

      Original equation x²-xy+y²=2 becomes r²(1-sin(Θ)cos(Θ))=2

    • @nathanisbored
      @nathanisbored 4 роки тому

      @@drpeyam found my old notes, i guess this is what i was thinking of maybe? i.imgur.com/yj8nmss.jpg

    • @nathanisbored
      @nathanisbored 4 роки тому

      @@hyperboloidofonesheet1036 i know that but, im curious what the systematic method would be (he mentioned diagonalizing a matrix, but that doesnt seem to give you the direct transformation)

    • @drpeyam
      @drpeyam  4 роки тому

      Ah yes, that’s for the surface integral of a function! A kind of Jacobian, I guess, although I usually don’t think of it as one

  • @jonathangrey6354
    @jonathangrey6354 4 роки тому

    *The Jacobian*

  • @Hirnlappen
    @Hirnlappen 4 роки тому

    Shouldn't the original region be defined by an inequality?

    • @drpeyam
      @drpeyam  4 роки тому +1

      Yeah, but it’s a small detail

  • @lindsaywaterman2010
    @lindsaywaterman2010 2 роки тому

    I did a substitution of u = x -y and v = x+y and got the double integral of (3/4u^2 + 1/4v^2)/2 du dv and got a result of 8/sqrt(3) times pi. My value is twice Dr Peyam's value. Why? Since from the integral of the ellipse of u and v, a = sqrt(8/3) and b = sqrt(8); area = pi times ab. I think that Dr Peyam should have used u^2 + v^2 =1 in the integral and then integrate r dr d(theta), then the area of the unit circle would be pi and the total integral would be 8/sqrt(3) times pi, which gives my answer.

    • @drpeyam
      @drpeyam  2 роки тому

      But that’s incorrect though

    • @drpeyam
      @drpeyam  2 роки тому

      u^2 + v^2 = 1 only on the boundary of the region, it’s not equal to 1 inside the region

    • @lindsaywaterman2010
      @lindsaywaterman2010 2 роки тому

      @@drpeyam I take your point. Then, will you kindly use my u, v substititutions and arrive at the correct answer? Thank you in anticipitation.