Double Integral $\iint e^\frac{x+y}{x-y} \,dx \,dy$ solution

Question

In the Multivariable Calculus Class I'm taking we were tasked to solve $$\iint_{R} e^\frac{x+y}{x-y} \,dx \,dy\,,$$ where $R$ is the trapezoidal region with vertices $(1,0)$, $(2,0)$, $(0,-1)$ and $(0,-2)$ as part of our optional Problem Set. While I intuitively knew how to solve the other problems of the Set, I didn't with this one.

Since integrating $e^\frac{x+y}{x-y}$ isn't that easy, I tried finding another way of solving this and found the method of Jacobians online (we haven't learned that yet). I figured that we could maybe set $u=x+y$ and $v=x-y$ based on the exemplary problem I saw online. Then I would arrive at $$x = \frac12(u+v)$$ $$y=\frac12(u-v)$$

This gives the Jacobian $$J(u,v)=\begin{vmatrix} \frac12 & \frac12 \\ \frac12 & -\frac12 \end{vmatrix}=-\frac12.$$ Where do I go from here to find the region $D$ in the plane of $(u,v)$ which corresponds to $R$.

My question therefore:

1. How do I find $D$ ?

and

2. Is there another way to solve this without knowledge of Jacobians? I would assume that the chance of us being given this problem as a brutal introduction to Jacobian, kind of like "You all didn't know how to solve this and are very confused now? Let's show you a method how to solve it!" is rather low.

Answer 1

You have the right idea. Now let's proceed from where you stopped.

$u = x + y, v = x-y, x = \frac{u+v}{2}, y = \frac{u-v}{2}, |J| = \frac{1}{2}$

Now please note the original region is bound between $4$ lines given as $x-y = 1, x-y = 2, x = 0, y = 0$

So the bound of $v$ is simply $1 \leq v \leq 2$. Now,

$x = 0, y = 0$ lead to bound of $u$ as $-v \leq u \leq v$.

And it results in a straightforward integral,

$\displaystyle \int_1^2 \int_{-v}^{v} e^{(u/v)} |J| \ du \ dv = \frac{1}{2} \big(e - \frac{1}{e}\big) \int_1^2 v \ dv = \frac{3(e^2-1)}{4e}$

Answer 2

Note that $x=0$, $y=0$, $x-y=1$ and $x-y=2$ are the boundaries for the integration region, which, in polar coordinates, translates into

$$\theta \in (-\frac\pi2,0),\>\>\>\>\> r\in \left(\frac1{ \cos\theta - \sin\theta }, \frac2{\cos\theta - \sin\theta }\right)$$ Then, the integral is

\begin{align} \iint_{R} e^\frac{x+y}{x-y} dx dy & = \int_{-\frac\pi2}^0 \int_{\frac1{ \cos\theta - \sin\theta }}^{\frac2{ \cos\theta - \sin\theta }}e^{\tan(\frac\pi4+\theta)}\>rdrd\theta \\ & = \frac34 \int_{-\frac\pi2}^0 e^{\tan(\frac\pi4+\theta)} \sec^2(\frac\pi4+\theta)d\theta = \frac34(e-e^{-1})=\frac32 \sinh(1) \end{align}

Answer 3

Another simple change of variables is $(x,y)\mapsto (u,v)$ such that $$u=\frac{x+y}{x-y}\quad,\quad v=x-y$$

Then $uv=x+y$ and $v=x-y$ implies $x=\frac{v(u+1)}{2}$ and $y=\frac{v(u-1)}{2}$.

The region $R$ is defined by $$R=\left\{(x,y): x>0,y<0,1<x-y<2\right\}$$

This means $1<v<2$ and from $v(u+1)>0, v(u-1)<0$ we have $-1<u<1$.

So $R$ is transformed to some $D$ where

$$D=\left\{(u,v): -1<u<1,1<v<2\right\}$$

Absolute value of jacobian of transformation is $\frac{v}{2}$, so the integral becomes

\begin{align} \iint_R \exp\left\{\frac{x+y}{x-y}\right\}\,\mathrm dx\,\mathrm dy &=\frac12\iint_D v\, e^{u}\,\mathrm du\,\mathrm dv \\&=\frac12\int_1^2 v\,\mathrm dv\int_{-1}^1 e^u\,\mathrm du \\&=\frac34\left(e-\frac1e\right) \end{align}