# How to find limits of integration in changed coordinates

Given integral = $\iint_R \frac{x-y}{(x+y)^3}dxdy$ :$R=[0,1]\times[0,1]$

Let $u = x+y, v= x-y \implies x = \frac{u+v}{2}, y = \frac{u-v}{2}$

Now x limits are from 0 to 1 and y limits are from 0 to 1. I am not able to understand how to find limits of integration in terms of u, v..

Pls enlighten me.

We need to consider the original domain in x-y plane which is a square $[0,1]\times[0,1]$ and the consider on that plane the lines

• $u = x+y$
• $v= x-y$

to see that the range for the two new variables is

• $0\le u\le 2$
• $-1\le v\le 1$

but note that in this case the two variables are not independent thus we need to fix the limits of variation for a first variable and the find the range for the second that is for example

• $0\le u\le 1$
• $-u\le v\le u$

and

• $1\le u\le 2$
• $u-2\le v\le 2-u$
• I did not understand 2nd part. I know $0\leq u \leq 1 , -1 \leq v \leq 1$ But in changed coordinates i think it is not rectangle with limits u = [0,2], v = [-1,1]. So can u pls elaborate how did u get limits of u,v in 2nd part. I am not able to get what it represents in changed coordinates. I recently started multivariable calculus. Kindly elaborate – Magneto Jun 17 '18 at 5:50
• @anirudhb As you can see in the figure we can take the variation for $u$ from $0 to 2$ in the domain and at each values for $u$ corresponds a range for $v$ which depends upon $u$. For example for $u=0 \implies v=0$ and for $u=2 \implies v=0$. In general to find the limits we need to divide the range for $u$ in 2 parts. For the for $0\le u \le 1\implies -u\le v\le u$ and for $1\le u \le 2\implies u-2\le v \le 2-u$. – user Jun 17 '18 at 5:55
• @anirudhb Recall that the original domain is the square and for $u=0$ we are describing a single point of the domain that is the origin $(0,0)$. For $u=1$ we are describing the segment from the vertex $(1,0)$ to $(0,1)$ and so on. – user Jun 17 '18 at 6:14
• Yes it is correct, I also did in that way! – user Jun 17 '18 at 6:55
• I can’t suggest others methods. – user Jun 17 '18 at 6:56