Answer
$4$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ denotes the unit vector.
and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The flux through a surface can be defined only when the surface is orientable.
Here, we have $\iint_S x z dS =\iiint_{D}x(4-2x-2y) \times \sqrt {(-2)^2+(-2)^2+1^2} dA$
or, $=3 \times \int_{0}^{2} \int_0^{2- x} 4x-2x^2 -2xy dy dx$
or, $= 3 \times \int_{0}^{2} (4x-2x^2) (2-x) -x (2-x)^2 dx$
or, $=3 \times \int_{0}^{2} x(2-x)^2 dx$
or, $=3 \times \int_0^2 2x^2-x^3 dx$
Thus, we have $\iint_S x z dS=3[\dfrac{2x^3}{3}-\dfrac{x^4}{4}]_0^2=4$
