How to evaluate this surface integral $ \iint_T e^{(y-x)/(y+x)} dA$ where $T$ is the triangular region with vertices $(0,0)$, $(1,0)$ and $(0,1)$?

Question

I am given the following problem:

Use a change of variables to evaluate $\int \int_T e^{(y-x)/(y+x)} dA$ where $T$ is the triangular region with vertices $(0,0)$, $(1,0)$ and $(0,1)$.

Here is what I have tried. I transformed the surface integral into the following double integral:

$$ \int_0^1 \int_0^{1-x} e^{(y-x)/(y+x)}\ dy\ dx $$

Now I'm trying to find the antiderivative of $e^{}$ w.r.t. $y$ using $u$-substitution. However, it's not working out for me. Setting $u = (y-x)/(y+x)$ I get

$$ du = \frac{2x}{(y+x)^2} dy\\ dy = \frac{(y+x)^2}{2x} du $$

The fraction $\frac{(y+x)^2}{2x}$ is not expressible in terms of $u$, so I can't perform $u$-substitution. What am I doing wrong?

Answer 1

In order to evaluate that integral, you need a change of variables:

$$u= x -y,\quad v = x + y.$$

With that, you find $x$ and $y$:

$$y = \dfrac{u+v}{2}, \quad x = \dfrac{u-v}{2}.$$

Then, the Jacobian determinant:

$$\left \lvert \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v} \dfrac{\partial y}{\partial u} \right \rvert = 1/2.$$

Now, we need the limits for $u$ and $v$. For that, you need to draw the figure in the $uv$-plane:

So the limits are:

$$0 \le v \le 1,\quad -v \le u \le v.$$

The integral is:

$$\int_{0}^1 \! \int_{-v}^{v} \exp(u / v)\, \dfrac{1}{2} \,du \, dv = \dfrac{e^2-1}{4 e}.$$

Answer 2

Write the integrand as $$e^{\left(1-\frac{2}{1+y/x}\right)}.$$ Then perform the transformation $u=y/x.$