Answer
$\dfrac{13 \pi}{20}$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}=\dfrac{\partial (z^2 x)}{\partial x}+\dfrac{\partial (\dfrac{1}{3} y^3+\tan z)}{\partial y}+\dfrac{\partial (x^2z+y^2)}{\partial z}=z^2+y^2+x^2$
Flux through $S_2$ $=\int_{0}^{\pi/2}\int_0^{2 \pi} \int_{0}^{1} \rho^2 (\rho^2 \sin \phi) d \rho d\phi d \theta=\int_{0}^{\pi/2}\int_0^{2 \pi} \int_{0}^{1} \rho^4 \sin \phi d \rho d\phi d \theta=\int_{0}^{\pi/2} \sin \phi d\pi \int_0^{2 \pi} d \theta \int_0^1 \rho^4 d\rho=\dfrac{2\pi}{5}$
Flux through $S_1$ $=-\int_{0}^{\pi/2}\int_{0}^{1} r^2 \sin^2 \theta r
dr d \theta=-\int_{0}^{\pi/2}\sin^2 \theta d\theta \int_{0}^{1} r^3 dr=(-1/2) [\theta-\dfrac{\sin 2\theta}{2}]_0^{2 \pi}=-\dfrac{\pi}{4}$
Flux through $S$ $=\dfrac{2\pi}{5}-\dfrac{\pi}{4}=\dfrac{13 \pi}{20}$
