Double Integral (Choosing order of integration) A question from [Calculus Early Transcendentals Fourth Edition by Jon Rogaski, 15.2, #31]
Compute the integral of $f(x, y)=(\ln y)^{-1}$ over the domain $D$ bounded by $y=e^x$ and $y=e^\sqrt x$.
The answer is $$\int_1^e \int_{\ln^2 y}^{\ln y} (\ln y)^{-1} \,dx\,dy$$
I do not understand how to derive the bounds for $dx$
 A: I think the limits should be $0 \leq x \leq 1$ and $e^x \leq y \leq e^\sqrt{x}$:

Note that the limits have nothing to do with the integrand you're integrating over.
A: As @DavidGStork notes, the region of integration is much easier to identify when integrating with respect to $y$ first. However, the contents of the integrand make it much easier to integrate with respect to $x$ first, and rearrange the bounds of integration to match.
First, we write the bounds of region $D$ as a system of inequalities. Observing the graph of $D$, we get
$$\begin{align} y&\geq e^x & y&\leq e^\sqrt{x}\end{align}$$
Solving these inequalities, we get a range for $x$.
$$\begin{align} y&\geq e^x & y&\leq e^\sqrt{x}\\
\ln(y)&\geq x & \ln(y)&\leq\sqrt x\\
& & \ln(y)^2&\leq x\end{align}$$
$$\ln(y)^2\leq x\leq\ln(y)$$
Solving for the values of $y$ for which the two sides of this range are equal, we get $y\in\{1,e\}$. Checking with our graph of $D$, we confirm that the range of $y$ is $1\leq y\leq e$.
Thus we have our integral:
$$\int_1^e\int_{\ln(y)^2}^{\ln(y)}\frac{1}{\ln(y)}\ dx\ dy$$
which solves rather easily as follows
$$\begin{align}\int_1^e\int_{\ln(y)^2}^{\ln(y)}\frac{1}{\ln(y)}\ dx\ dy &=\int_1^e(1-\ln(y))\ dy\\
&=\left[2y-y\ln(y))\right]^{y=e}_{y=1}\\
&=e-2\end{align}$$
