Answer
$2$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$\int^2_0 \int^{2-x}_0 \int^{4-2x-2y}_{(2-x-y)/2} \space dz \space dy \space dx =\int^2_0 \int^{2-x}_0 (3-\dfrac{3x}{2}-\dfrac{3y}{2}) \space dy \space dx $
or, $=\int^2_0 [3 \times (1-\dfrac{x}{2})(2-x)-\dfrac{3}{4} \times (2-x)^2] \space dx $
or, $=\int^2_0 [6-6x+\dfrac{3x^2}{2}-\dfrac{3(2-x)}{4}] \space dx $
or, $= [6x-3x^2+\dfrac{x^3}{2}+\dfrac{(2-x)^3}{4}]^2_0$
or, $=2$
