2

Let $R$ be the region bounded by $x+y=1, x=0, y=0.$ Show $$\iint_R \cos\frac{x-y}{x+y}\, dx\,dy=\frac{\sin1}{2}.$$

So I've let $u=x-y$, $x+y=v$.

Graphically, the domain is a triangle with the slope as the $y=1-x$ line with vertices $(0,1)$ and $(1,0)$.

It seems pretty obvious that the domain for $v$ is $0$ to $1$.

For $u$, I've imagined a bunch of $y=x-u$ lines whereby $u$ is the variable. So in my mind as $u$ varies we get a stack of lines with gradient $1.$ The domain for $u$ will be when these lines intersect $(0,1)$ and $(1,0)$. Graphically, it seems like $u$ will vary from $1$ to $-1$.

Calculating the inverse Jacobian:

$$J= \begin{vmatrix} 1 & -1\\ 1 & 1 \end{vmatrix}=2 $$

$$\frac{1}{J}=\frac{1}{2}.$$

But when I start plugging in the substitution the problem starts:

$$\iint_R \cos\frac{x-y}{x+y}\, dx\,dy$$ $$\frac{1}{2}\int^1_0\int^1_{-1}\cos\frac{u}{v}\,du\,dv$$ $$\frac{1}{2}\int^1_0 2v\sin\frac{1}{v}\,dv.$$

After this I'm stuck. Am I on the right track?

amWhy
  • 210,739
koifish
  • 3,282

2 Answers2

3

You are not using the correct bounds for the integral as correctly mentioned in the other answer.

The region of interest is $$R=\{(x,y): x\ge 0, y\ge 0, x+y\le 1\}$$

A different change of variables, namely $$(x,y)\to (u,v)\text{ such that } u=\frac{x-y}{x+y}, v=x+y$$

makes the new ranges independent of each other. Since the region $R$ is transformed to some $$D=\{(u,v): -1\le u\le 1, 0\le v\le 1\}$$

The absolute value of the Jacobian is $v/2$, so that the integral is

\begin{align} I=&\iint_D \cos u \left(\frac{v}{2}\right)\, du\, dv \\&=\frac{1}{2}\int_{-1}^{1} \cos u\,du\int_0^1 v\,dv \end{align}

StubbornAtom
  • 17,932
2

Your way for solving the problem is right but you may lose the proper range of $u$ and $v$. Try to draw the triangle region and then find the corresponding vertices according to the transformation $u=x-y, v=x+y$. I am sure you will find another triangle with vertices $(0,1),(-1,-1),(1,1)$. So the right range for the integrals would be $$0\leq v\leq1, -v\leq u\leq v$$

Mikasa
  • 67,942
  • Thanks for your answer! But why isn't $u$ ranging from $-1$ to $1$ possible? Even from your inequality, since v ranges only from $0$ to $1$, u must also range from $-1$ to $1$? Am I wrong in saying that or is there some conceptual misunderstanding? – koifish Sep 06 '18 at 15:33
  • You're welcome. Plz do what i pointed above. Your converted region would be as written. This is a common way for doing such these problems. :-) – Mikasa Sep 06 '18 at 15:58