Answer
$$V = \pi \left( {{e^2} + 6 + e - {e^{ - 2}} - {e^{ - 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = {e^{x/2}} + {e^{ - x/2}},{\text{ }}y = 0,{\text{ }}x = - 1,{\text{ }}x = 2 \cr
& V = \pi \int_a^b {{{\left[ {R\left( x \right)} \right]}^2}} dx \cr
& {\text{Let }}R\left( x \right) = {e^{x/2}} + {e^{ - x/2}} \cr
& {\text{So}},{\text{ the volume of the solid of revolution is}} \cr
& V = \pi \int_{ - 1}^2 {{{\left( {{e^{x/2}} + {e^{ - x/2}}} \right)}^2}} dx \cr
& V = \pi \int_{ - 1}^2 {\left( {{e^x} + 2{e^0} + {e^{ - x}}} \right)} dx \cr
& {\text{Integrate}} \cr
& V = \pi \left[ {{e^x} + 2x - {e^{ - x}}} \right]_{ - 1}^2 \cr
& V = \pi \left[ {{e^2} + 2\left( 2 \right) - {e^{ - 2}}} \right] - \pi \left[ {{e^{ - 1}} + 2\left( { - 1} \right) - {e^{ - \left( { - 1} \right)}}} \right] \cr
& V = \pi \left[ {{e^2} + 4 - {e^{ - 2}}} \right] - \pi \left[ {{e^{ - 1}} - 2 - e} \right] \cr
& V = \pi \left( {{e^2} + 4 - {e^{ - 2}} - {e^{ - 1}} + 2 + e} \right) \cr
& V = \pi \left( {{e^2} + 6 + e - {e^{ - 2}} - {e^{ - 1}}} \right) \cr} $$
