Answer
$$\dfrac{\pi}{2}$$
Work Step by Step
$$Volume=V=2 \int_0^{\pi/2} \int_{-\cos y}^{0} \int_{0}^{-2x} \ dz \ dx \ dy \\= \int_0^{\pi/2} \int_{-\cos y}^{0} (-2x) \times (2) \ dx \ dy \\=2 \int_0^{\pi/2} \cos^2 y \ dy \\=2 [\dfrac{y}{2} +\dfrac{1}{4} \times \sin 2y]_0^{\pi/2} \\=\dfrac{\pi}{2}$$
