Let $R$ be the triangle with vertices in $(0,0)$, $(0,1)$, $(1,0)$. Find: $\iint_R \exp((y-x)/( y+x))\, dx\,dy.$

Question

I have a problem with this:

Let $R$ be the triangle with vertices in $(0,0)$, $(0,1)$, $(1,0)$.

Find: $$\iint_R \exp\left({y-x\over y+x}\right)\, dx\,dy.$$ I tried to use the substitution $u=y+x$ and $v=y-x$, getting Jacobian determinant $\mathbf J={1\over 2}$, but I don't know how to find the the new integration limits.

Answer 1

prior to your change of coordinates your limits were $x = 0, y = 0, x+y = 1$

If $u = x+y, v = x-y$

$x = \frac 12 u + \frac 12 v\\ y = \frac 12 u - \frac 12 v$

and substitute

$u = v\\ u = -v\\ u = 1$

Answer 2

Your initial area of integration is given by

$$R = \{(x,y) \in \mathbb{R}^2 : 0 \le x \le 1, 0 \le y \le 1-x\}$$

Let $u = y+x$, $v = y-x$. We have

$$ 0 \le y+x \le 1 \implies u \in [0,1]$$ $$ \left|v\right| = \left|y-x\right| \le |y| + |x| = y+x = u \implies v \in [-u,u]$$

so the new area of integration is $$S = \{(u,v) \in \mathbb{R}^2 : 0 \le u \le 1, -u \le v \le u\}$$

Therefore, your integral is

\begin{align} \iint_R \exp \left(\frac{y-x}{y+x}\right)\,dx\,dy &= \frac12\iint_{S} \exp\left(\frac{v}{u}\right)\,du\,dv \\ &=\frac12 \int_0^1 \int_{-u}^u \exp\left(\frac{v}{u}\right)\,dv\,du\\ &= \frac12 \int_0^1 u\left(e - \frac1e\right)\,du\\ &= \frac14 \left(e - \frac1e\right)\\ &= \frac{e^2-1}{4e} \end{align}