Answer
$\dfrac{\sqrt {21}}{3}$
Work Step by Step
The flux through a surface can be defined only when the surface is orientable.
We know that $\iint_S F \cdot dS=\iint_S F \cdot n dS$
Here, $n$ denotes the unit vector.
Since, $\iint_S x dS =\iiint_{D}x \sqrt {(-4)^2+(2)^2+1^2} dA=\sqrt{21} \iint_D dA$
$=\sqrt{21} \int_{0}^{1} \int_{2x-2}^0 x dy dx$
$=\sqrt{21} \int_{0}^{1} [xy]_{2x-2}^0 x dx$
$=- \sqrt{21} \int_{0}^{1} 2x^2-2x dx$
$=- \sqrt{21} [(2x^3/3)-x^2]_{0}^{1}$
$=-\sqrt {21} [\dfrac{2x^3}{3}-x^2]_0^1$
$=\dfrac{\sqrt {21}}{3}$
