Answer
$18$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^3_0 \int^{\sqrt{9-x^2}}_0 \int^{\sqrt{9-x^2}}_0 dz dy dx = \int^3_0 \int^{\sqrt{9-x^2}}_0 \sqrt{9-x^2}\\= \int^3_0 (9-x^2) \space dy \space dx \\=\int^3_0 (9-x^2)\space dx \\=[9x-\dfrac{x^3}{3}]^3_0 \\=18 $$
