Solving a double integral by changing variables

Question

I got a double integral exercise which is shown as below:

$$\int_0^1\int_0^{1-x}\exp\left(\frac y{x + y}\right)\,\mathrm dx\,\mathrm dy$$

I have a hint that I should put $u = x + y$ and $v = y$ to solve this question but I still find it hard to solve.

Answer 1

$$\int_0^1\int_0^{1-x}\exp\left(\frac y{x + y}\right)\,\mathrm dx\,\mathrm dy$$

$$=\int_0^1\int_0^{1-y}\exp\left(\frac y{x + y}\right)\,\mathrm dx\,\mathrm dy$$

so $|J|=1$ sine $v=y$ and $u=x+y$

$0<y<1$ and $x<1-y$

$0<y<1$ and $x+y<1$

$0<v<1$ and $v<u<1$ ( since $y\leq x+y$ )

\begin{eqnarray} J=\left| \begin{array}{cc} \frac{d }{du}x & \frac{d}{dv}x \\ \frac{d}{du}y & \frac{d}{dv} y \\ \end{array} \right| =\left| \begin{array}{cc} 1 & -1 \\ 0 & 1 \\ \end{array} \right| \end{eqnarray}

$$=\int_0^1\int_v^{1}\exp\left(\frac v{u}\right)\,\mathrm du\,\mathrm dv$$

$$=\int_0^1\int_0^{u}\exp\left(\frac v{u}\right)\,\mathrm dv\,\mathrm du$$