Answer
$2(\pi+e-e^{-1})$
Work Step by Step
$R= [-1,1] \times [0,\pi]$
$V=\int_{R}\int f(x,y)dA$
$=\int^{\pi}_{0}\int^{1}_{-1} (1+e^{x} siny)dxdy$
$=\int^{\pi}_{0} [x+e^{x}siny]^{1}_{-1}dy$
$=\int^{\pi}_{0} [1+esiny-(-1+e^{-1}siny)]dy$
$=\int^{\pi}_{0}[2+siny(e-e^{-1})]dy$
$=[2y-cosy(e-e^{-1})]^{\pi}_{0}$
$=2\pi -cos\pi(e-e^{-1}) -(0-cos0(e-e^{-1}))$
$=2\pi + (e-e^{-1}) + (e-e^{-1})$
$=2\pi +2(e-e^{-1})$
$=2(\pi+e-e^{-1})$
