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math_proof Group Title

change in variables in double integrals

  • 2 years ago
  • 2 years ago

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  1. math_proof Group Title
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    \[\int\limits_{}^{}\int\limits_{}^{}e ^{xy}dA\]

    • 2 years ago
  2. math_proof Group Title
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    where R is bounded by hyperbolas xy=1 and xy=4 and lines y/x=1 and y/x=3

    • 2 years ago
  3. math_proof Group Title
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    |dw:1353376012156:dw|

    • 2 years ago
  4. math_proof Group Title
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    |dw:1353376473982:dw|

    • 2 years ago
  5. math_proof Group Title
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    can this be new region of integration?

    • 2 years ago
  6. zepdrix Group Title
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    \[\large u=xy \quad v=\frac{y}{x}\] Making this substitution? Hmmm yah I think that will work out. Just need to get your substitutions in terms of x= and y= and then take the jacobian from there.

    • 2 years ago
  7. math_proof Group Title
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    how did you get that u and v?

    • 2 years ago
  8. zepdrix Group Title
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    hmm

    • 2 years ago
  9. math_proof Group Title
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    you just pick any that you want ?

    • 2 years ago
  10. zepdrix Group Title
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    I chose u based on the integrand, it simplifies the exponential to something like e^u, which makes it a little easier to deal with. The v was based on the limits of integration, I'm not quite sure if i picked v correctly though :D i still need to check hehe

    • 2 years ago
  11. math_proof Group Title
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    so use \[\int\limits_{1}^{3}\int\limits_{1}^{4}e ^{u}dudv\]

    • 2 years ago
  12. zepdrix Group Title
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    Wellllll, if you change variables you have to multiply by the Jacobian as part of your change of differentials. It's a bit messy in this one also :\ hmm

    • 2 years ago
  13. zepdrix Group Title
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    Do you happen to know what the right answer is suppose to be? XD I want to know if I'm on the right track here lol

    • 2 years ago
  14. math_proof Group Title
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    i don't ;/

    • 2 years ago
  15. zepdrix Group Title
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    Ok well I'll show you how I did it at least XD No promises that it's correct though lolol

    • 2 years ago
  16. math_proof Group Title
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    oki thanks for your time

    • 2 years ago
  17. zepdrix Group Title
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    Hmm so the Jacobian looks like this, it's the determinate of this matrix. When we change from dy dx to the other differentials, we have to multiply by this factor. \[\huge \left|\begin{matrix}x_u & x_v \\ y_u & y_v\end{matrix}\right|\]

    • 2 years ago
  18. zepdrix Group Title
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    In order to do so, we need our substitutions written in the form x= and y=

    • 2 years ago
  19. zepdrix Group Title
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    Starting with this substitution, \[\large u=xy \quad v=\frac{y}{x}\]We'll move some stuff around to get x= and y=.

    • 2 years ago
  20. zepdrix Group Title
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    \[\large x=\frac{u}{y} \qquad y=vx\]

    • 2 years ago
  21. zepdrix Group Title
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    \[\large x=\frac{u}{(vx)} \qquad y=v\left(\frac{u}{y}\right)\]

    • 2 years ago
  22. zepdrix Group Title
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    \[\large x=\sqrt{\frac{u}{v}} \qquad y=\sqrt{uv}\]

    • 2 years ago
  23. zepdrix Group Title
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    From there, we can apply the Jacobian. Taking the partials and placing them into the matrix.

    • 2 years ago
  24. math_proof Group Title
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    wow thats gonna be messy

    • 2 years ago
  25. zepdrix Group Title
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    \[\huge x_u = \frac{1}{2v \sqrt{\frac{u}{v}}}\] Yah they get pretty bad :( They clean up nicely though when you take the determinate :D

    • 2 years ago
  26. math_proof Group Title
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    but the integrate will be the one that i wrote times the jocubian?

    • 2 years ago
  27. zepdrix Group Title
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    Yah, it'll be in this form \[\large \int\limits \int\limits e^{xy} dA \rightarrow \qquad \int\limits \int\limits e^u \left|\begin{matrix}x_u & x_v \\ y_u & y_v\end{matrix}\right| du dv\]

    • 2 years ago
  28. zepdrix Group Title
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    After calculating the Jacobian, I came up with\[\frac{1+v^2}{4v}\]I may have made mistakes somewhere though :D so make sure you try to do it!! I think it'll take too much time to show all the steps to get it though :3 Hopefully you have somewhat of an understanding of that thingy ma jigger :D

    • 2 years ago
  29. zepdrix Group Title
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    Keep in mind that you don't have to do the quotient or product rule on any of them, since they're only partials (one of the variables is held constant). Only messy chain rules :)

    • 2 years ago
  30. math_proof Group Title
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    lol alright thanks a lot for ur help

    • 2 years ago
  31. zepdrix Group Title
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    \[\huge \int\limits_{u=1}^4 \int\limits_{v=1}^3e^u \frac{1+v^2}{4v}dv \;du\]\[=\huge \int\limits\limits_{u=1}^4 e^u \; du \int\limits\limits_{v=1}^3\frac{1+v^2}{4v} \; dv\]

    • 2 years ago
  32. math_proof Group Title
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    so much homework

    • 2 years ago
  33. zepdrix Group Title
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    So I'm coming up with somethign like that D: pshh i dunno..

    • 2 years ago
  34. zepdrix Group Title
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    You have evil maf teacher too? :3 assigns bajillion problems? lol

    • 2 years ago
  35. math_proof Group Title
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    hmmm may be not billion but that kind of problems that take 2hrs to sold one like this one:p

    • 2 years ago
  36. zepdrix Group Title
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    ugh yah D:

    • 2 years ago
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