Answer
$$\dfrac{20}{3}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=\int^4_0 \int^{\sqrt{4-x}}_0 \int^{2-y}_0 \space dz \space dy \space dx\\=\int^4_0 \int^{\sqrt{4-x}}_0 (2-y) \space dy \space dx\\=\int^4_[2(4-x)^{1/2}-(\dfrac{4-x}{2})] \space dx\\=[-\dfrac{4}{3} \times (4-x)^{3/2}+\dfrac{1}{4} \times (4-x^2)]^4_0 \\=\dfrac{4}{3}(4)^{3/2}-\dfrac{1}{4}(16)\\=\dfrac{20}{3}$$
