## math_proof Group Title change in variables in double integrals one year ago one year ago

1. math_proof Group Title

$\int\limits_{}^{}\int\limits_{}^{}e ^{xy}dA$

2. math_proof Group Title

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

3. math_proof Group Title

|dw:1353376012156:dw|

4. math_proof Group Title

|dw:1353376473982:dw|

5. math_proof Group Title

can this be new region of integration?

6. zepdrix Group Title

$\large u=xy \quad v=\frac{y}{x}$ 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 Group Title

how did you get that u and v?

8. zepdrix Group Title

hmm

9. math_proof Group Title

you just pick any that you want ?

10. zepdrix Group Title

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 Group Title

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

12. zepdrix Group Title

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 Group Title

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 Group Title

i don't ;/

15. zepdrix Group Title

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

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17. zepdrix Group Title

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 Group Title

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

19. zepdrix Group Title

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 Group Title

$\large x=\frac{u}{y} \qquad y=vx$

21. zepdrix Group Title

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

22. zepdrix Group Title

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

23. zepdrix Group Title

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

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wow thats gonna be messy

25. zepdrix Group Title

$\huge x_u = \frac{1}{2v \sqrt{\frac{u}{v}}}$ Yah they get pretty bad :( They clean up nicely though when you take the determinate :D

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

27. zepdrix Group Title

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 Group Title

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 Group Title

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 Group Title

lol alright thanks a lot for ur help

31. zepdrix Group Title

$\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$

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so much homework

33. zepdrix Group Title

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

34. zepdrix Group Title

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

35. math_proof Group Title

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

36. zepdrix Group Title

ugh yah D: