Answer
$$\dfrac{16}{3}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=8\int^1_0 \int^{\sqrt{1-x^2}}_0 \int^{\sqrt{1-x^2}}_0 \space dz \space dy \space dx \\= 8\int^1_0 \int^{\sqrt{1-x^2}}_0 \sqrt{1-x^2} \space dy \space dx \\= 8 \times \int^1_0 (1-x^2)dx \\=\dfrac{16}{3}$$
