Answer
$$\dfrac{1}{12}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^1_0 \int^{1-x^2}_0 \int^{4-x^2-y}_3 (x) \space dz \space dy \space dx =\int^1_0 \int^{1-x^2}_0 (x) (1-x^2-y) \space dy \space dx \\= \int^1_0 x[(1-x^2)^2-\frac{1}{2}(1-x^2)] \space dx\\= \dfrac{1}{2} \times \int^1_0 x (1-x^2)^2 \space dx \\=[\dfrac{-1}{12} \times (1-x^2)^3]^1_0 \\=\dfrac{1}{12}$$