Answer
$12 \pi $
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=\int^{2}_{-2} \int^{\sqrt{4-x^2}}_{-\sqrt{4-x^2}} \space dz \space dy \space dx = \int^2_{-2} \int^{\sqrt{4-x^2}}_{-\sqrt{4-x^2}} (3-x) \space dy \space dx \\=2\int^2_{-2}(3-x)\sqrt{4-x^2} \space dx \\= 3\int^2_{-2}2\sqrt{4-x^2} dx-2\int^2_{-2}x\sqrt{4-x^2} \space dx \\=3[x\sqrt{4-x^2} +4 \sin^{-1}( \dfrac{x}{2})]^2_{-2} \\=12 \sin^{-2}(1)-12 \sin^{-1} (6) \\=12(-\dfrac{\pi}{2})-12(\dfrac{\pi}{2}) \\=12\pi $$
