Answer
$\dfrac{\sqrt {21}}{3}$
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 $\iint_S x dS =\iiint_{D}x \sqrt {(-4)^2+(2)^2+1^2} dA$
or, $=\sqrt{21} \times \int_{0}^{1} \int_{2x-2}^0 x dy dx$
or, $=\sqrt{21} \times \int_{0}^{1} [xy]_{2x-2}^0 \times x dx$
or, $=- \sqrt{21} \times \int_{0}^{1} (2x^2-2x) dx$
or, $=- \sqrt{21} \times [(]\dfrac{2x^3}{3})-x^2]_{0}^{1}$
or, $\iint_S x dS=-\sqrt {21} [(\dfrac{2(1)^3}{3}-0)-[(1)^2-0]]_0^1=\dfrac{\sqrt {21}}{3}$
