Let $D$ be the region in the first quadrant of the $x,y$-plane, bounded by the curves $y=x^2$, $y=x^2+1$, $x+y=1$, and $x+y=2$.
I want to find a change of variables for $\iint_D x\sin(y-x^2) \, dA$.
Since $y-x^2=0$, $y-x^2=1$, my inclination is to let $u=y-x^2$, and since $x+y=1$ and $x+y=2$ I want to let $v=x+y$. However, when solving for $x$ in terms of $u$ and $v$ I get a horrid looking thing.
Could someone provide me with a better way of approaching this problem? Thank you!
Here's half an answer. Maybe you went down this path already:
If you let $u=x^2, v=x+y$ some things are less horrid. First, the Jacobian is $\frac{1}{2x}$ which cancels the first $x$ in the integrand. The pre-image $D^*$ of the domain is bounded by $v=1, v=2, v=u+\sqrt{u},$ and $v=u+\sqrt{u}+1.$
At first glance $D^*$ tempts you to integrate in the order $du \; dv.$ But you run into the same horridness solving those equations for $u$. But happily, the 4 points of intersections are not that bad and (here's the good part) the middle two have the same $u$ coordinate $a=(3-\sqrt{5})/2$. So you can break $D^*$ into two regions.
First region is $u=0$ to $a$, $v=1$ to $u+\sqrt{u} +1$.
Second region is $u = a$ to $1$, $v=u+\sqrt{u}$ to $2$.
Everything works swimmingly until the end when there are two integrals:
$$\int_0^a \frac{1}{2} \cos(1-\sqrt{u}-u) \; du$$
and
$$\int_a^1 \frac{1}{2} \cos(2-\sqrt{u}-u) \; du.$$
I was hoping for a clever substitution that would change one of them into the other and then they'd cancel out. I've had no luck with that so far.