Answer
$$V = \frac{{49}}{2}\pi $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}}{\text{, }}y = \ln x,{\text{ }}y = \ln {x^2},{\text{ }}x = 0{\text{ and }}x = \ln 4 \cr
& y = \ln x,{\text{ }} \to x = {e^y} \cr
& y = \ln {x^2}{\text{ }} \to x = {e^{y/2}} \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}y = 0{\text{ to }}y = \ln 8 \cr
& {\text{Revolving the region about the }}y{\text{ - axis}}{\text{, using the washer }} \cr
& {\text{method}}{\text{.}} \cr
& V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - q{{\left( y \right)}^2}} \right]} dy \cr
& {e^y} \geqslant {e^{y/2}}{\text{ on the interval }}\left[ {0,\ln 8} \right] \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_0^{\ln 8} {\pi \left[ {{{\left( {{e^y}} \right)}^2} - {{\left( {{e^{y/2}}} \right)}^2}} \right]} dy \cr
& V = \int_0^{\ln 8} {\pi \left( {{e^{2y}} - {e^y}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\frac{1}{2}{e^{2y}} - {e^y}} \right]_0^{\ln 8} \cr
& V = \pi \left[ {\frac{1}{2}{e^{2\left( {\ln 8} \right)}} - {e^{\ln 8}}} \right] - \pi \left[ {\frac{1}{2}{e^{2\left( 0 \right)}} - {e^0}} \right] \cr
& V = \pi \left( {32 - 8} \right) - \pi \left( { - \frac{1}{2}} \right) \cr
& V = \frac{{49}}{2}\pi \cr} $$
