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