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math_proof

  • 2 years ago

change in variables in double integrals

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  1. math_proof
    • 2 years ago
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    \[\int\limits_{}^{}\int\limits_{}^{}e ^{xy}dA\]

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

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

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

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

  6. zepdrix
    • 2 years ago
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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.

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

  8. zepdrix
    • 2 years ago
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    hmm

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

  10. zepdrix
    • 2 years ago
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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

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

  12. zepdrix
    • 2 years ago
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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

  13. zepdrix
    • 2 years ago
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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

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

  15. zepdrix
    • 2 years ago
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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

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

  17. zepdrix
    • 2 years ago
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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. \[\huge \left|\begin{matrix}x_u & x_v \\ y_u & y_v\end{matrix}\right|\]

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

  19. zepdrix
    • 2 years ago
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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=.

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

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

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

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

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

  25. zepdrix
    • 2 years ago
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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

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

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

  28. zepdrix
    • 2 years ago
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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

  29. zepdrix
    • 2 years ago
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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 :)

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

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

  32. math_proof
    • 2 years ago
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    so much homework

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

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

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

  36. zepdrix
    • 2 years ago
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    ugh yah D:

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