1

calculate $\iint_D \cos (\frac{x-y}{x+y})$ while $D$ is the trapezoid $(0,1),(0,2),(2,0),(1,0)$

first I found the lines of the trapezoid , let $A=(0,1),B=(0,2),C=(2,0),D=(1,0)$

so $AC: y=-0.5x+1$ ,$AD: y=-x+1$ $BC: y+x=2$ and $BD: y=-2x+2$

and also change variables $u=x-y$ and $v=x+y$ from here the jacobian is $|J|=\frac{1}{2}$

my problem was with finding the bounds of the integral , from the lines of the trapezoid I found that $1 \leq v \leq 2$

and from $BC$ and $BD$ I got that $x-y=0$ after subtracting them .. but what about the other bound of $u$? I cannot find it and in the solution in the book it says $-v \leq u \leq v$ why is that ?

Thanks for any tips and help, calculating the integral and actually getting the final solution is not important I just want to know how to find the bounds of $u$

I found this but I did not understand from there why it is $-v \leq u \leq v$

Adamrk
  • 909
  • 1
    The trapezoid is $D={(x,y): 1\leq x+y\leq 2, x\geq 0, y\geq 0}$. Therefore, since $x=\frac{u+v}{2}$, $y=\frac{v-u}{2}$, the inequalities become $$\begin{cases} 1\leq v\leq 2\ \frac{u+v}{2}\geq 0\ \frac{v-u}{2}\geq 0\ \end{cases}\Leftrightarrow \begin{cases} 1\leq v\leq 2\ u\geq -v\ v\geq u\ \end{cases}.$$ – Robert Z Jul 26 '23 at 10:04

0 Answers0