Answer
$$\dfrac{128}{15}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$\int^2_0 \int^{4-x^2}_0 \int^{4-x^2-y}_0 \space dz \space dy \space dx = \int^2_0 \int^{4-x^2}_0 (4-x^2-y) \space dy \space dx \\=\int^2_0 [(4-x^2)^2-\frac{1}{2}(4-x^2)^2] \space dx \\=\dfrac{1}{2} \times \int^2_0 (4-x^2)^2 \space dx \\=\int^2_0 (8-4x^2+\dfrac{x^4}{2})\space
dx \\=\dfrac{128}{15}$$
