Answer
$$A = 2\pi $$
Work Step by Step
$$\eqalign{
& y = {e^{ - x}},{\text{ }}x \geqslant 0,{\text{ revolved about the }}y{\text{ - axis}} \cr
& {\text{The volume is given by}} \cr
& A = 2\pi \int_a^b {xf\left( x \right)} dx \cr
& A = 2\pi \int_0^{ + \infty } {x{e^{ - x}}} dx \cr
& {\text{Using the definition of improper integrals}} \cr
& A = 2\pi \mathop {\lim }\limits_{b \to + \infty } \int_0^b {x{e^{ - x}}} dx \cr
& {\text{Integrating by parts}} \cr
& A = 2\pi \mathop {\lim }\limits_{b \to + \infty } \left[ {x{e^{ - x}} - {e^{ - x}}} \right]_0^b \cr
& A = 2\pi \mathop {\lim }\limits_{b \to + \infty } \left[ {\left( {b{e^{ - b}} - {e^{ - b}}} \right) - \left( {0{e^{ - 0}} - {e^{ - 0}}} \right)} \right] \cr
& A = 2\pi \mathop {\lim }\limits_{b \to + \infty } \left[ {\left( {b{e^{ - b}} - {e^{ - b}}} \right) + 1} \right] \cr
& {\text{Evaluate the limit}} \cr
& A = 2\pi \left[ {\left( { + \infty {e^{ - \infty }} - {e^{ - \infty }}} \right) + 1} \right] \cr
& A = 2\pi \left[ {0 + 1} \right] \cr
& A = 2\pi \cr} $$
