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Let $D$ be the region bounded by $x=0$, $y=0$, $x+y=1$ and $x+y=4$. Evaluate $$\iint_D \frac{dx\,dy}{x+y}$$ by making the change of variables $x=u-uv$, $y=uv$

My attempt
I understand I must first find the domain. $$x=u-uv, y=uv$$ $$x=u-y$$ $$u=x+y$$ and $$v=\frac{u}{y}$$ This gives me the domain $$ (x,y) \rightarrow(u,v) $$ $$ (0,1) \rightarrow(1,1) $$ $$ (1,0) \rightarrow(1,0) $$$$ (0,4) \rightarrow(4,1) $$ $$ (4,0) \rightarrow(4,0) $$ After i graph these points on the new $(u,v)$ graph i get a box and so my limits are $$ 1 \le u \le 4$$ $$0 \le v \le 1 $$ and now I am having trouble finding the jacobian since u is a function of v.
Any help would be great.

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I believe the way you need to re-arrange this is

$$x \ = \ u - uv \ = \ u \cdot (1-v) \ , \ y = uv \ \Rightarrow \ x = u - y \ \Rightarrow \ u = x + y $$

$$\Rightarrow \ y = (x + y) \cdot v \ \Rightarrow \ v = \frac{y}{x+y} \ . $$

Since this transformation does not reverse the orientation of the boundary, your Jacobian ought to come out positive.

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  • $\begingroup$ But now for my Jacobian I have $\frac{x}{y^2} + \frac{1}{y}$ and when i now sub the u and v terms back in and find the integral of this Jacobian*$\frac{1}{u}$ I get the wrong answer? $\endgroup$ – amanda Jun 14 '13 at 20:35
  • $\begingroup$ Isn't the Jacobian the determinant of the partial derivatives of $ \ x \ $ and $ \ y \ $ with respect to $ \ u \ $ and $ \ v \ $ ? I have a Jacobian of $ \ u \ $ , rather than $ \ 1/u \ $ . $\endgroup$ – colormegone Jun 14 '13 at 20:46
  • $\begingroup$ You are absolutely right, I'm so sorry. i have been awake for about 30 hours straight now, I think it's best if I get some sleep and continue on tomorrow. Thankyou so much for your help, I really appreciate it. $\endgroup$ – amanda Jun 14 '13 at 20:55
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    $\begingroup$ Yes, please be sure to get enough sleep before you take a math exam; otherwise you'll mess up things you do know how to do... $\endgroup$ – colormegone Jun 14 '13 at 20:57
  • $\begingroup$ I will make sure of that, I still have two days to study so I still got time I guess. Thankyou once again :) $\endgroup$ – amanda Jun 14 '13 at 21:01

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