Answer
$$V = \frac{{255}}{{32}}\pi $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}}{\text{, }}y = {e^{ - x}},{\text{ }}y = {e^x},{\text{ }}x = 0{\text{ and }}x = \ln 4 \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}\underbrace {\left[ {0,\ln 4} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Revolving the region about the }}x{\text{ - axis}}{\text{, using the washer }} \cr
& {\text{method}}{\text{.}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {e^x} \geqslant {e^{ - x}}{\text{ on the interval }}\left[ {0,\ln 4} \right] \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_0^{\ln 4} {\pi \left[ {{{\left( {{e^x}} \right)}^2} - {{\left( {{e^{ - x}}} \right)}^2}} \right]} dx \cr
& V = \pi \int_0^{\ln 4} {\left( {{e^{2x}} - {e^{ - 2x}}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\frac{1}{2}{e^{2x}} - \frac{1}{2}{e^{ - 2x}}} \right]_0^{\ln 4} \cr
& V = \frac{\pi }{2}\left[ {{e^{2x}} - {e^{ - 2x}}} \right]_0^{\ln 4} \cr
& V = \frac{\pi }{2}\left[ {{e^{2\left( {\ln 4} \right)}} - {e^{ - 2\left( {\ln 4} \right)}}} \right] - \frac{\pi }{2}\left[ {{e^{2\left( 0 \right)}} - {e^{ - 2\left( 0 \right)}}} \right] \cr
& V = \frac{\pi }{2}\left[ {16 - \frac{1}{{16}}} \right] - \frac{\pi }{2}\left[ 0 \right] \cr
& V = \frac{{255}}{{32}}\pi \cr} $$
