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

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 .

• Is the integral supposed to be $\int_1^2 \int_x^{2x} f(x,y)\,dy\,dx$ ? – 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 $$u,v$$:
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 $$\frac12 \le v \le \frac23 , \quad \frac1{1-v} \le u \le \frac2{1-v}.$$
• One could add, that it's probably best to integrate first over $u$ i.e. $$\int_{\frac{1}{2}}^{\frac{2}{3}} {\rm d}v \int_{\frac{1}{1-v}}^{\frac{2}{1-v}} {\rm d}u \, \left| \frac{\partial (x,y)}{\partial (u,v)} \right| \, f\left(x(u,v),y(u,v)\right) \, .$$ – Diger Jan 20 at 14:38