Answer
$\sqrt{2}$
Work Step by Step
V=$\int\int f(x,y)dA$
=$\int^{\frac{\pi}{2}}_{0} \int^{\frac{\pi}{4}}_{0}2 \sin x \cos y dydx$
=$\int^{\frac{\pi}{2}}_{0}[2sinx siny]^\frac{\pi}{4}_0dx$
=$\int^{\frac{\pi}{2}}_{0}(\sqrt{2} \sin x)dx$
$
=[-\sqrt{2} \cos x]^\frac{\pi}{2}_0$
=$\sqrt{2}$
