Answer
$\dfrac{16}{3}$
Work Step by Step
Here, we have $V=\int_{0}^2 \int_{0}^{4-2x} \int_{0}^{4-2x-y} dz dy dx= \int_{0}^2 \int_{0}^{4-2x} [z]_{0}^{4-2x-y} dy dx$
and $\int_{0}^2 \int_{0}^{4-2x}4-2x-y dy dx= \int_{0}^{2}(4y-2xy-\dfrac{y^2}{2}]_0^{4-2x} dx$
$\implies \int^{0}_{2} [4(4-2x)-2x(4-2x)-\dfrac{1}{2}(4-2x)^2] dx$
$\implies \int_0^2 2x^2-8x+8 dx$
Hence, $V=[\dfrac{2 x^3}{3}-4x^2+8x]_0^2=[\dfrac{2 (2-0)^3}{3}-4(2-0)^2+8(2-0)]=\dfrac{16}{3}$
