I need the calculate the following integral:
$\iint_D x^2y\log(y) + 2x^2y \;\textrm{d}x\;\textrm{d}y$, where $D$ is the region bounded by $e^x \leq y \leq e^{2x}, \; 1 \leq x^2y \leq 2$
What is being especially troublesome for me is figuring out how to deal with specifying the bounds of integration, given the information about $D$.
How should I go about specifying the bounds of integration?
Following @Dan’s comment, a reasonable way to approach this kind of problem is to look for a coordinate transformation that straightens out the boundary of the region and turns it into a square or triangle.
Trying $u=x^2y$ looks promising since its level curves form two edges of $D$ and it appears as a factor in the integrand. From the given inequalities, we have $x\gt0$ and $y\gt1$, so we can rewrite the first one as $1\le{\log y\over x}\le2$, which suggests trying $v={\log y\over x}$. Solving for $x$ and $y$ looks unpleasant, so we’ll defer that for now.
The Jacobian of this map is $$J=\pmatrix{2xy & x^2 \\ -{\log y\over x^2} & \frac1{xy}}$$ which is defined everywhere in $D$ and has determinant $\log y+2\gt2$, so is nonsingular on $D$. The Inverse Function Theorem tells us that the inverse map exists and has Jacobian $J^{-1}$, so the “scale factor” for this change of variables is $\det(J^{-1})=\frac1{\det J}=\frac1{\log y+2}$. The rest of the computation should be pretty simple.