change in variables in double integrals

- anonymous

change in variables in double integrals

- jamiebookeater

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- anonymous

\[\int\limits_{}^{}\int\limits_{}^{}e ^{xy}dA\]

- anonymous

where R is bounded by hyperbolas xy=1 and xy=4 and lines y/x=1 and y/x=3

- anonymous

|dw:1353376012156:dw|

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## More answers

- anonymous

|dw:1353376473982:dw|

- anonymous

can this be new region of integration?

- 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

how did you get that u and v?

- zepdrix

hmm

- anonymous

you just pick any that you want ?

- 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

so use \[\int\limits_{1}^{3}\int\limits_{1}^{4}e ^{u}dudv\]

- 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

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

i don't ;/

- zepdrix

Ok well I'll show you how I did it at least XD No promises that it's correct though lolol

- anonymous

oki thanks for your time

- 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

In order to do so, we need our substitutions written in the form x= and y=

- 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

\[\large x=\frac{u}{y} \qquad y=vx\]

- zepdrix

\[\large x=\frac{u}{(vx)} \qquad y=v\left(\frac{u}{y}\right)\]

- zepdrix

\[\large x=\sqrt{\frac{u}{v}} \qquad y=\sqrt{uv}\]

- zepdrix

From there, we can apply the Jacobian. Taking the partials and placing them into the matrix.

- anonymous

wow thats gonna be messy

- 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

but the integrate will be the one that i wrote times the jocubian?

- 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

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

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

lol alright thanks a lot for ur help

- 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

so much homework

- zepdrix

So I'm coming up with somethign like that D: pshh i dunno..

- zepdrix

You have evil maf teacher too? :3 assigns bajillion problems? lol

- anonymous

hmmm may be not billion but that kind of problems that take 2hrs to sold one like this one:p

- zepdrix

ugh yah D:

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