Answer
$$\dfrac{2}{3}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=2\int^1_0 \int^0_{-\sqrt{1-x^2}} \int^{-y}_0 \space dz \space dy \space dx \\=-2 \times \int^1_0 \int^0_{-\sqrt{1-x^2}}y \space dy \space dx \\= \int^1_0 (1-x^2)dx \\=\dfrac{2}{3}$$
