Answer
$$12 \pi$$
Work Step by Step
$$ Volume =4 \int_0^{2} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{4-x^2} \ dz \ dy \ dx \\= \int_0^{2} \int_{0}^{\sqrt {4-x^2}} 4 \times (4-x^2) \ dy \ dx \\= \int_0^{2} (4-x^2)^{3/2} \ dx \times 4 \\=[x (4-x^2)^{3/2} +6x \sqrt {4-x^2}+24 \sin^{-1} (\dfrac{x}{2}) ]_0^{2} \\24 \sin^{-1}(1) \\=12 \pi$$
