Answer
$$V = \pi {\left( {\sqrt e - 1} \right)^2}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {\ln \left( {{x^2}} \right)} ,{\text{ }}y = \sqrt {\ln x} ,\;{\text{ }}y = 1 \cr
& \sqrt {\ln \left( {{x^2}} \right)} = 1 \cr
& \ln \left( {{x^2}} \right) = 1 \cr
& 2\ln x = 1 \cr
& \ln x = \frac{1}{2} \cr
& x = {e^{1/2}} \cr
& x = \sqrt e \cr
& {\text{and}} \cr
& \sqrt {\ln x} = 1 \cr
& x = e \cr
& {\text{Using the washer method about the }}x{\text{ - axis}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{From the graph of the region shown below}} \cr
& \sqrt {\ln \left( {{x^2}} \right)} \geqslant \sqrt {\ln x} {\text{ on the interval }}\left[ {1,\sqrt e } \right]{\text{ }} \cr
& {\text{and}} \cr
& {\text{1}} \geqslant \sqrt {\ln x{\text{ }}} {\text{on the interval }}\left[ {\sqrt e ,e} \right]{\text{ }} \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_1^{\sqrt e } {\pi \left[ {{{\left( {\sqrt {\ln \left( {{x^2}} \right)} } \right)}^2} - {{\left( {\sqrt {\ln x} } \right)}^2}} \right]} dx + \int_{\sqrt e }^e {\pi \left[ {1 - {{\left( {\sqrt {\ln x} } \right)}^2}} \right]} dx \cr
& V = \pi \int_1^{\sqrt e } {\left( {\ln \left( {{x^2}} \right) - \ln x} \right)} dx + \pi \int_{\sqrt e }^e {\left( {1 - \ln x} \right)} dx \cr
& V = \pi \int_1^{\sqrt e } {\left( {2\ln x - \ln x} \right)} dx + \pi \int_{\sqrt e }^e {\left( {1 - \ln x} \right)} dx \cr
& V = \pi \int_1^{\sqrt e } {\ln x} dx + \pi \int_{\sqrt e }^e {\left( {1 - \ln x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {x\ln x - x} \right]_1^{\sqrt e } + \pi \left[ {x - x\ln x + x} \right]_{\sqrt e }^e \cr
& V = \pi \left[ {x\ln x - x} \right]_1^{\sqrt e } + \pi \left[ {2x - x\ln x} \right]_{\sqrt e }^e \cr
& V = \pi \left[ {\sqrt e \ln \sqrt e - \sqrt e } \right] - \pi \left[ {\left( 1 \right)\ln \left( 1 \right) - 1} \right] + \pi \left[ {2e - e\ln e} \right] \cr
& - \pi \left[ {2\sqrt e - \sqrt e \ln \sqrt e } \right] \cr
& {\text{Simplifying}} \cr
& V = \pi \left[ {\frac{1}{2}\sqrt e - \sqrt e } \right] - \pi \left[ { - 1} \right] + \pi \left[ e \right] - \pi \left[ {2\sqrt e - \frac{1}{2}\sqrt e } \right] \cr
& V = \pi \left[ {\frac{1}{2}\sqrt e - \sqrt e - 2\sqrt e + \frac{1}{2}\sqrt e + 1 + e} \right] \cr
& V = \pi \left( {e - 2\sqrt e + 1} \right) \cr
& V = \pi {\left( {\sqrt e - 1} \right)^2} \cr
& {\text{Using the shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& y = \sqrt {\ln \left( {{x^2}} \right)} \to x = {e^{{y^2}/2}} \cr
& {\text{ }}y = \sqrt {\ln x} \to x = {e^{{y^2}}} \cr
& {e^{{y^2}}} \geqslant {e^{{y^2}/2}}{\text{ on the interval }}\left[ {0,1} \right]{\text{ }} \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^1 {2\pi y\left( {{e^{{y^2}}} - {e^{{y^2}/2}}} \right)} dy \cr
& V = 2\pi \int_0^1 {\left( {y{e^{{y^2}}} - y{e^{{y^2}/2}}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{1}{2}{e^{{y^2}}} - {e^{{y^2}/2}}} \right]_0^1 \cr
& V = 2\pi \left[ {\frac{1}{2}{e^{{{\left( 1 \right)}^2}}} - {e^{{{\left( 1 \right)}^2}/2}}} \right] - 2\pi \left[ {\frac{1}{2}{e^{{{\left( 0 \right)}^2}}} - {e^{{{\left( 0 \right)}^2}/2}}} \right] \cr
& V = 2\pi \left[ {\frac{1}{2}e - {e^{1/2}}} \right] - 2\pi \left[ {\frac{1}{2} - 1} \right] \cr
& V = 2\pi \left[ {\frac{1}{2}e - \sqrt e } \right] - 2\pi \left( { - \frac{1}{2}} \right) \cr
& V = 2\pi \left[ {\frac{1}{2}e - \sqrt e } \right] + \pi \cr
& V = \pi \left( {2\left[ {\frac{1}{2}e - \sqrt e } \right] + 1} \right) \cr
& V = \pi \left( {e - 2\sqrt e + 1} \right) \cr
& V = \pi {\left( {\sqrt e - 1} \right)^2} \cr
& {\text{Shell method is easier to apply}} \cr} $$
