anonymous
  • anonymous
change in variables in double integrals
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{}^{}\int\limits_{}^{}e ^{xy}dA\]
anonymous
  • anonymous
where R is bounded by hyperbolas xy=1 and xy=4 and lines y/x=1 and y/x=3
anonymous
  • anonymous
|dw:1353376012156:dw|

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anonymous
  • anonymous
|dw:1353376473982:dw|
anonymous
  • anonymous
can this be new region of integration?
zepdrix
  • zepdrix
\[\large u=xy \quad v=\frac{y}{x}\] 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.
anonymous
  • anonymous
how did you get that u and v?
zepdrix
  • zepdrix
hmm
anonymous
  • anonymous
you just pick any that you want ?
zepdrix
  • zepdrix
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
anonymous
  • anonymous
so use \[\int\limits_{1}^{3}\int\limits_{1}^{4}e ^{u}dudv\]
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
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
anonymous
  • anonymous
i don't ;/
zepdrix
  • zepdrix
Ok well I'll show you how I did it at least XD No promises that it's correct though lolol
anonymous
  • anonymous
oki thanks for your time
zepdrix
  • zepdrix
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. \[\huge \left|\begin{matrix}x_u & x_v \\ y_u & y_v\end{matrix}\right|\]
zepdrix
  • zepdrix
In order to do so, we need our substitutions written in the form x= and y=
zepdrix
  • zepdrix
Starting with this substitution, \[\large u=xy \quad v=\frac{y}{x}\]We'll move some stuff around to get x= and y=.
zepdrix
  • zepdrix
\[\large x=\frac{u}{y} \qquad y=vx\]
zepdrix
  • zepdrix
\[\large x=\frac{u}{(vx)} \qquad y=v\left(\frac{u}{y}\right)\]
zepdrix
  • zepdrix
\[\large x=\sqrt{\frac{u}{v}} \qquad y=\sqrt{uv}\]
zepdrix
  • zepdrix
From there, we can apply the Jacobian. Taking the partials and placing them into the matrix.
anonymous
  • anonymous
wow thats gonna be messy
zepdrix
  • zepdrix
\[\huge x_u = \frac{1}{2v \sqrt{\frac{u}{v}}}\] Yah they get pretty bad :( They clean up nicely though when you take the determinate :D
anonymous
  • anonymous
but the integrate will be the one that i wrote times the jocubian?
zepdrix
  • zepdrix
Yah, it'll be in this form \[\large \int\limits \int\limits e^{xy} dA \rightarrow \qquad \int\limits \int\limits e^u \left|\begin{matrix}x_u & x_v \\ y_u & y_v\end{matrix}\right| du dv\]
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
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 :)
anonymous
  • anonymous
lol alright thanks a lot for ur help
zepdrix
  • zepdrix
\[\huge \int\limits_{u=1}^4 \int\limits_{v=1}^3e^u \frac{1+v^2}{4v}dv \;du\]\[=\huge \int\limits\limits_{u=1}^4 e^u \; du \int\limits\limits_{v=1}^3\frac{1+v^2}{4v} \; dv\]
anonymous
  • anonymous
so much homework
zepdrix
  • zepdrix
So I'm coming up with somethign like that D: pshh i dunno..
zepdrix
  • zepdrix
You have evil maf teacher too? :3 assigns bajillion problems? lol
anonymous
  • anonymous
hmmm may be not billion but that kind of problems that take 2hrs to sold one like this one:p
zepdrix
  • zepdrix
ugh yah D:

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