See diagram.we use washers to get the area and we will be integrating with respect to xouter radius is the distance between y = e^x and y = -1
![r_{\text{out} } = e^x ) = e^x + 1](/tex.php?f=r_{\text{out} } = e^x ) = e^x + 1)
inner radius of washer is the distance between y = 1 and y = -1
![r_{\text{in} } = 1 - (-1) = 2](/tex.php?f=r_{\text{in} } = 1 - (-1) = 2)
the area of one cross-section of a washer is
![\begin{aligned} a(x) & = \pi(r_{\text{out} } )^2 - \pi(r_{\text{in} } )^2 \\ & = \pi(e^x + 1 )^2 - \pi(2)^2 \\ & = \pi \big[(e^x + 1 )^2 - 4 \big] \end{aligned}](/tex.php?f=\begin{aligned} a(x) & = \pi(r_{\text{out} } )^2 - \pi(r_{\text{in} } )^2 \\ & = \pi(e^x + 1 )^2 - \pi(2)^2 \\ & = \pi \big[(e^x + 1 )^2 - 4 \big] \end{aligned})
solid lies between x = 0 and x = 2. the volume is
![\begin{aligned} v & = \int_0^2 a(x) dx \\ & = \int_0^2 \pi \big[(e^x + 1 )^2 - 4 \big] dx \end{aligned}](/tex.php?f=\begin{aligned} v & = \int_0^2 a(x) dx \\ & = \int_0^2 \pi \big[(e^x + 1 )^2 - 4 \big] dx \end{aligned})
if you need this expanded, then
![\begin{aligned} v & = \int_0^2 \pi \left(e^{2x} + 2e^x + 1 - 4 \right) dx \\ & = \int_0^2 \pi \left(e^{2x} + 2e^x -3 \right) dx \end{aligned}](/tex.php?f=\begin{aligned} v & = \int_0^2 \pi \left(e^{2x} + 2e^x + 1 - 4 \right) dx \\ & = \int_0^2 \pi \left(e^{2x} + 2e^x -3 \right) dx \end{aligned})
the question doesn't say to evaluate, so we're done.