Answer
$\dfrac{16}{3}$
Work Step by Step
Consider $V=\int_{0}^2 \int_{0}^{4-2x} \int_{0}^{4-2x-y} dz dy dx= \int_{0}^2 \int_{0}^{4-2x} [z]_{0}^{4-2x-y} dy dx$
or, $=\int_{0}^2 \int_{0}^{4-2x}(4-2x-y) dy dx$
or, $= \int_{0}^{2}(4y-2xy-\dfrac{y^2}{2}]_0^{4-2x} dx$
or, $=\int^{0}_{2} 4(4-2x)-2x(4-2x)-\dfrac{(4-2x)^2}{2} dx$
or, $=\int_0^2 2x^2-8x+8 dx$
or, $=[(2/3) x^3-4x^2+8x]_0^2$
or, $=\dfrac{16}{3}$