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