Answer
$V = \pi \int_1^3 {{{\ln }^2}x} dx$
Work Step by Step
$$\eqalign{
& y = \ln x,{\text{ }}y = 0,{\text{ }}x = 3;{\text{ about the }}x{\text{ - axis}} \cr
& {\text{Graph of the region shown below }}\left( {{\text{Use Geogebra}}} \right) \cr
& \cr
& {\text{When we slice through the point }}x,{\text{ we get a disk with }} \cr
& {\text{radius}}\ln x.{\text{ The area of this cross section is }}A = \pi {r^2}, \cr
& {\text{Let }}r = \ln x,{\text{ then}} \cr
& A\left( x \right) = \pi \underbrace {{{\left( {\ln x} \right)}^2}}_{{\text{radius}}} \cr
& {\text{The solid lies on the interval }}1 \leqslant x \leqslant 3,{\text{ so using the volume}} \cr
& {\text{definition }}\left( {{\text{page 448}}} \right) \cr
& V = \int_a^b {A\left( x \right)} dx \cr
& {\text{The volume is given by}} \cr
& V = \int_1^3 {\pi {{\left( {\ln x} \right)}^2}} dx \cr
& V = \pi \int_1^3 {{{\ln }^2}x} dx \cr} $$
