Answer
$\dfrac{13 \pi}{20}$
Work Step by Step
Flux through $S_2$ $=\int_{0}^{\pi/2}\int_0^{2 \pi} \int_{0}^{1} \rho^2 \times (\rho^2 \sin \phi) \times d \rho d\phi d \theta$
or, $=\int_{0}^{\pi/2}\int_0^{2 \pi} \int_{0}^{1} \rho^4 \sin \phi \times d \rho d\phi d \theta$
or, $=\int_{0}^{\pi/2} \sin \phi d\pi \times \int_0^{2 \pi} d \theta \times \int_0^1 \rho^4 d\rho$
or, $=\dfrac{2\pi}{5}$
Flux through $S_1$ $=-\int_{0}^{\pi/2}\int_{0}^{1} r^2 \sin^2 \theta r
dr d \theta$
or, $=-\int_{0}^{\pi/2}\sin^2 \theta d\theta \times [\int_{0}^{1} r^3 dr]$
or, $=(-0.5)\times [\theta-\dfrac{\sin 2\theta}{2}]_0^{2 \pi}$
or, $=-\dfrac{\pi}{4}$
Hence, the flux through S is given by$=\dfrac{2\pi}{5}-\dfrac{\pi}{4}=\dfrac{13 \pi}{20}$
