What is the value of the $\iint \ln(y)\,dx\,dy$ $$ \int_1^e \int_{\ln y}^1 \ln y \, dx \, dy $$
I know I neeed to change the limits of integration and I have drawn a picture of the original region but I am confused how I find the new limits of integration. Please help.
 A: First let's make sure we agree on the region, since that's the hardest part. Work from the inside out: $x$ is trapped between two functions of $y$, then $y$ is trapped between two numbers. The functions of $y$ are $x = \ln(y)$, so we draw the curve $y = e^x$; we need to stay to the right of that, and the constant line $x = 1$, so we draw the vertical line $x = 1$, which we need to stay to the left of. Now we only want the part where $y$ is between $1$ and $e$. By looking at the picture, we see that it's the "almost triangle" region above the line $y = 1$, bounded on the top and left by the curve, and on the right by the vertical line.
To switch the order of integration, you need to redescribe the region as follows. Work from the inside out again, but now the "inside" will be $dy$ and the outside $dx$. First, $y$ is trapped between two functions of $x$. Then, $x$ is trapped between two numbers. The integral becomes
$$
\int_0^1 \int_1^{e^x} \ln(y)\ dy\ dx.
$$
This you can do, using that $y\ln(y) - y$ is an antiderivative of $\ln(y)$. 
You get
\begin{align*}
\int_0^1 \int_1^{e^x} \ln(y)\ dy\ dx &= \int_0^1 (y\ln(y) - y) \bigg{|}^{e^x}_1 \ dx\\
&= \int_0^1 xe^x - e^x + 1 \ dx\\
&= xe^x - e^x - e^x + x \bigg|^1_0 \\
&= (e - e - e + 1) -(0 - 1 - 1 + 0)\\
&= 3-e
\end{align*}
