Help getting new bounds for a change of variable I have the domain $D: x^2-y^2=1, x^2-y^2=4,y=0, y=\frac{x}{2}$ where $x\geq 0$.
I have to calculate this double integral:
$$\iint_D \left(1-\left(\frac{y}{x}\right)^4\right)e^{x^2-y^2} dxdy $$
So my first idea was, to use these new variables:
$$u=\frac{y}{x}, v=x^2-y^2; 
$$
$$J=\frac{1}{2(u^2-1) } $$
So this would give me :
$$\iint (1-u^4)e^v\frac{1}{2(u^2-1)}dudv = -\frac{1}{2}\iint (1+u^2)e^v dudv$$
But I have a problem getting the bounds for $u$. The other one is easy, $v\in [1,4]$, and I got $u=\frac{1}{u}$. Is it safe to assume: $u\in[0,1]$? I am unsure about this one.
Did I do the above steps correctly? Any help/insight would be appreciated.
 A: I like to do change of variables in three steps, in no particular order. You have correctly determined good candidates for the substitutions, namely $u=y/x, v= x^2-y^2$.


*

*Change the region.

Our region $D$ satisfies $1\le x^2-y^2\le 4$ and $0\le y\le x/2$. Here's a picture.

The  $\color{green}{\text{ left and right }}$ boundaries are indeed just $v=1$ and $v=4$; now let's try to get the $\color{red}{\text{ up and down }}$ boundaries. The bottom one is the $x$-axis, i.e. $y=0$, or $u=0$. The top one isn't bad either: it's $y=x/2$, or $y/x=1/2$, which is just $u=1/2$. So our new region in $u,v$ is $[0,1/2]\times [1,4]$.


*Compute the Jacobian determinant.

We have
$$
J = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}  \end{pmatrix} = \begin{pmatrix}  -\frac{y}{x^2} & \frac{1}{x} \\
 2 x & -2 y \\ \end{pmatrix}
$$Then we have
$$
|\det (J) |= \left|\frac{2 y^2}{x^2}-2\right| = |2(u^2-1)| = 2(1-u^2)
$$Note the use of absolute values here. At this point, before we examine the function, we can quickly check to see we're on the right track by computing the area of the region pre- and post-transform:
$$
\iint _D 1\,dx dy = \int _0^{1/\sqrt{3}}  \int_{\sqrt{y^2+1}}^{\sqrt{y^2+4}}1\,dxdy+\int _{1/\sqrt{3}}^{2/\sqrt{3}}  \int_{2y}^{\sqrt{y^2+4}}1\,dxdy = \frac{3 \log (3)}{4} ;
$$
$$
\int _0 ^{1/2} \int _1^4 \frac{1}{2(1-u^2)}\,dvdu = \frac{3\log(3)}{4}
$$Again, this is not strictly necessary, but it makes us feel reassured we're on the right track.


*Change the integrand.

Now that we're done with where and how, we can focus on what. This is often the easiest step, and indeed the integrand is quickly seen to be $(1-u^4) e^v$. (Note that I have not yet divided by the Jacobian determinant.)

Now the integration is straightforward.
$$
\iint _D \left(1-\left(\frac{y}{x}\right)^4\right)e^{x^2-y^2}\,dx dy
$$
$$
= \int _0 ^{1/2} \int _1^4 (1-u^4) e^v\cdot \frac{1}{2(1-u^2)}\,dvdu
$$
$$
= \frac{1}{2}\int _0 ^{1/2} \int _1^4 (1+u^2) e^v\,dvdu
$$
$$
= \frac{13}{48}(e^4-e)
$$
