Answer
$4\pi $
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=2\int^2_{-2} \int^{\sqrt{4-x^2}/2}_0 \int^{x+2}_0 \space dz \space dy \space dx \\=2\int^2_{-2} \int^{\sqrt{4-x^2}}_0 (x+2) \space dy \space dx \\=\int^2_{-2} (x+2)\sqrt{4-x^2} \space dx \\= \int^2_{-2}2 \sqrt{4-x^2}dx+\int^2_{-2}x\sqrt{4-x^2} \space dx \\=[x( 4-x^2)^{1/2}+4 \sin^{-1}(\dfrac{x}{2})]^{2}_{-2}+[-\dfrac{1}{3} (4-x^2)^{3/2}]^{2}_{-2} \\=(4) \times \dfrac{\pi}{2}-4 \times (-\dfrac{\pi}{2}) \\ = 4\pi $$
