Answer
$$-2+\frac{9}{2} \ln 3$$
Work Step by Step
\[
\iint_{D} x d A=\int_{1}^{3} \int_{0}^{\ln x} x d y d x
\]
$D$ has been interpreted as a type I region (using Vertical cross-sections).
We can also Interpret $D$ as a type II region (using horizontal cross-sections).
So:
\[
\iint_{D} x d A=\int_{0}^{\ln 3} \int_{e^{y}}^{3} x d x d y
\]
\[
\begin{array}{l}
=\int_{0}^{\ln 3}\left[\frac{x^{2}}{2}\right]_{x=e^{y}}^{x=3} d y \\
=\frac{1}{2} \int_{0}^{\ln 3} 9-e^{2 y} d y \\
=\frac{1}{2}\left[-\frac{1}{2} e^{2 y}+ 9 y\right]_{0}^{\ln 3} \\
=\frac{1}{2}\left[-\frac{1}{2} e^{2 \ln 3}+9 \ln 3\right]-\frac{1}{2}\left[0-\frac{1}{2}\right] \\
=\frac{1}{2}\left[9 \ln 3-\frac{9}{2}+\frac{1}{2}\right] \\
=-2+\frac{9}{2} \ln 3
\end{array}
\]
