Answer
2
Work Step by Step
$R=[0,2] \times [0,\frac{\pi}{4}]$
$V=\int\int_{R} xsec^{2}ydA$
$V= \int^{\frac{\pi}{4}}_{0} [\frac{x^{2}}{2}sec^{2}y]^{2}_{0} dy$
$=\int^{\frac{\pi}{4}}_{0} 2sec^{2} ydy$
$=[2tany]^{\frac{\pi}{4}}_{0}$ (since $\frac{d}{dy}tany=sec^{2}y$)
$=2[1-0]$
$=2$