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I want to do the following integral using the change variables $u = x+y$ and $ v= y/x$:

$$\int_U \frac{1}{x^2}\,{dU}$$

where $U$ is the region such that $1 \le x+y \le 2$ and $x \le y \le 2x$, and given that $x >0$.

I calculated the Jacobian $\displaystyle \begin{vmatrix} \dfrac{\partial (u, v)}{\partial (x, y)} \end{vmatrix} = \begin{vmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{vmatrix} = \bigg|\dfrac{x+y}{x^2}\bigg| = \frac{x+y}{x^2}$

But the trouble is I can't write $\dfrac{1}{x^2}$ in terms of $u$ and $v$. How do I do that?

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3 Answers 3

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You need to solve $u=x+y$ and $v=y/x$ for $x$ by eliminating $y$. For example, $y=vx$ and thus $u=x+vx$, so $x=u/(v+1)$. Of course you get the same result using the equations the other way around: $y=u-x$ and thus $v=(u-x)/x$, so $x=u/(v+1)$.

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  • $\begingroup$ Thanks. I've now calculated the integral but I've a sign error due to (I think) the fact that I said $x \le y \le 2x$ $\iff 2 \le y/x \le 1$ (my logic: inequalities flip as we divide by $x$) so $v$ goes from $2$ to $1$. Is this wrong? It gives me $-\log{2}$ while the answer is $\log{2}$. $\endgroup$
    – user129566
    Apr 14, 2016 at 1:56
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    $\begingroup$ @user129566: In your comment, it should be $1\le y/x\le2$. Inequalities change direction when you divide by negative numbers, not when you divide by positive numbers. $\endgroup$
    – joriki
    Apr 14, 2016 at 1:58
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Now what you need to do is to represent x and y in terms of u and v.

Since you have two equations: $x+y=u$ and $\frac{y}{x}=v$, you can write $y=xv$ and plug this into the first equation. You'll get $x+xv=u$, rewrite this as $x=\frac{u}{1+v}$.

Then you substitute every x and y in the integral with u and v, and try to solve this integral in terms of u and v.

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$$y= u-x$$ $$v=(u-x)/x$$ $$xv=u-x$$ $${x^2}v=ux-{x^2}$$ $${x^2}(v+1)=ux$$ $${x^4}{(v+1)^2}={u^2}{x^2}$$ $${x^2}{(v+1)^2}={u^2}$$Any good?

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