Answer
$241 \pi$
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, $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The surface integral over $S_1$:
$\iint_S x^2+y^2+z^2 dS= \iint_{D} (9 \sin^2 u+9 \cos^2 u+v^2) (3) dA=(3) \iint_{D}9+v^2 dA= (3) \times \int_{0}^{2} \int_0^{2 \pi} 9+v^2 du dv=124 \pi$
The surface integral over $S_2$
$\iint_S x^2+y^2+z^2 dS= \iint_{D} (v^2 \sin^2 u+v^2 \cos^2 u+2^2) v dA=\iint_{D}v^3+4v dA= \int_{0}^{3} \int_0^{2 \pi} v^3+4v du dv=\dfrac{153 \pi}{2}$
The surface integral over $S_3$
$\iint_S x^2+y^2+z^2 dS= \iint_{D} (v^2 \sin^2 u+v^2 \cos^2 u+(0)^2) v dA=\iint_{D}v^3 du dv=2 \pi \int_{0}^{3} v^3 dv dv=\dfrac{81 \pi}{2}$
Thus, we have $\iint_{S} xz dS=\iint_{S_1} xz dS+\iint_{S_2} xz dS+\iint_{S_3} xz dS=124 \pi+\dfrac{153 \pi}{2}+\dfrac{81 \pi}{2}=241 \pi$
