The double integral $\int_{1}^{2}\int_{x}^{2x}f(x,y)dxdy$ under the tranformation $x=u-uv $ and $y=uv$ is__________________

I have calculated jacobian as u but I am not able to find out the limits of integration for u and v .

u and v in terms of x and y are $u=x+y$ and $v=\frac{y}{x+y}$.

x goes from 1 to 2 and y goes from x to 2x .

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    $\begingroup$ Is the integral supposed to be $\int_1^2 \int_x^{2x} f(x,y)\,dy\,dx$ ? $\endgroup$ – StubbornAtom Jan 19 at 12:40

The region of integration is the set of points $(x,y)$ such that $ 1\le x\le 2$ and $x \le y \le 2x$. Notice that under the change of variables, the bottom line $y=x$ can be written as $uv = u - uv$, or $v = 1/2$. Along the top line $y=2x$, we have $uv/2 = u - uv$, or $v= 2/3$. More generally, along any of the lines $y=tx$ for $t\in[1,2]$, we have $uv/t = u - uv$, or $(1+1/t)v = 1$ i.e. $v = \frac{t}{t+1} = 1-\frac1{t+1}$. We also need to describe the vertical lines $x=1,2$. For fixed $u$ and $x$, $v = 1- x/u$. A graph of the level sets of $u,v$:

enter image description here

Thus the region is $$ 2 \le u \le 6,\quad \max(1/2, 1-2/u) \le v \le \min(2/3, 1-1/u).$$ Alternatively, by describing the lines $x=x_0$ as $u=x_0/(1-v)$, we obtain $$ \frac12 \le v \le \frac23 , \quad \frac1{1-v} \le u \le \frac2{1-v}.$$

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    $\begingroup$ One could add, that it's probably best to integrate first over $u$ i.e. $$ \int_{\frac{1}{2}}^{\frac{2}{3}} {\rm d}v \int_{\frac{1}{1-v}}^{\frac{2}{1-v}} {\rm d}u \, \left| \frac{\partial (x,y)}{\partial (u,v)} \right| \, f\left(x(u,v),y(u,v)\right) \, . $$ $\endgroup$ – Diger Jan 20 at 14:38
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    $\begingroup$ @Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks) $\endgroup$ – Calvin Khor Jan 20 at 14:54

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