Answer
$\approx 3.4895$
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, $\iint_S x^2y^2 z^2 dS =\iint_{D} x^2y^2z^2 \times \sqrt{1+(dz/dx)^2+ (dz/dy)^2} dA$
or, $\iint_S x^2y^2 z^2 dS =\iint_{D} x^2y^2z^2 \sqrt{1+16x^2+ 4y^2} dA$
or, $\iint_S x^2y^2 z^2 dS =\iint_{D} x^2y^2 \times (3-2x^2-y^2)^2 \times \sqrt{1+16x^2+ 4y^2} dA$
or, $\iint_S x^2y^2 z^2 dS =\int_{-\sqrt{3/2}}^{\sqrt{3/2}} x^2y^2 \times (3-2x^2-y^2)^2 \times \sqrt{1+16x^2+ 4y^2} dy dx$
Need to use calculating tool.
$ \iint_S x^2y^2 z^2 dS \approx 3.4895$
