Answer
$$V = \pi \left( {1 - \ln 2} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Using the Shell method, the Volume of the solid can be}} \cr
& {\text{calculated by:}} \cr
& V\int_a^b {2\pi x\left( {f\left( x \right) - g\left( x \right)} \right)dx} \cr
& {\text{With }}a = 0,\,\,\,\,b = \ln 2,\,\,\,\,f\left( x \right) = {e^{ - x}}{\text{ and }}g\left( x \right) = 0 \cr
& {\text{Then,}} \cr
& V = \int_0^{\ln 2} {2\pi x\left( {{e^{ - x}} - 0} \right)dx} \cr
& V = 2\pi \int_0^{\ln 2} {x{e^{ - x}}dx} \cr
& {\text{Integrating by parts}} \cr
& \,\,\,u = x,\,\,\,\,dv = {e^{ - x}} \cr
& \,\,du = dx,\,\,\,\,\,v = - {e^{ - x}} \cr
& V = 2\pi \left[ { - x{e^{ - x}}} \right]_0^{\ln 2} - 2\pi \int_0^{\ln 2} {\left( { - {e^{ - x}}} \right)dx} \cr
& V = 2\pi \left( { - \ln 2{e^{ - \ln 2}}} \right) + 2\pi \int_0^{\ln 2} {{e^{ - x}}dx} \cr
& V = 2\pi \left( { - \ln 2\left( {\frac{1}{2}} \right)} \right) - 2\pi \left[ {{e^{ - x}}} \right]_0^{\ln 2} \cr
& V = - \pi \ln 2 - 2\pi \left[ {{e^{ - \ln 2}} - {e^0}} \right] \cr
& V = - \pi \ln 2 + 2\pi \left[ {\frac{1}{2}} \right] \cr
& V = - \pi \ln 2 + \pi \cr
& V = \pi \left( {1 - \ln 2} \right) \cr} $$
