Answer
$$V = \frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}}{\text{, }}y = {e^{ - x}},{\text{ }}y = 0,{\text{ }}x = 0{\text{ and }}x = p > 0 \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}\left[ {0,\infty } \right) \cr
& {\text{Revolving the region about the }}x{\text{ - axis}}{\text{, using the washer }} \cr
& {\text{method}}{\text{.}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {e^{ - x}} \geqslant 0{\text{ on the interval }}\left[ {0,\infty } \right) \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_0^\infty {\pi \left[ {{{\left( {{e^{ - x}}} \right)}^2} - {{\left( 0 \right)}^2}} \right]} dx \cr
& V = \pi \int_0^\infty {{e^{ - 2x}}} dx \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - 2x}}} dx \cr
& {\text{Integrating}} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{2}{e^{ - 2x}}} \right]_0^b \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{2}{e^{ - 2b}} + \frac{1}{2}{e^{ - 2\left( 0 \right)}}} \right] \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2} - \frac{1}{2}{e^{ - 2b}}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& V = \pi \left[ {\frac{1}{2} - \frac{1}{2}\left( 0 \right)} \right] \cr
& V = \frac{\pi }{2} \cr} $$
