Answer
$(48 \sqrt 2-12)\pi$
Work Step by Step
As we know that $div F=\dfrac{\partial A}{\partial x}i+\dfrac{\partial B}{\partial y}j+\dfrac{\partial C}{\partial z}k$
Now, we have
$Flux =\iiint_{o} 15x^2+15y^2+15z^2 dA$
Also,
$Flux =\nabla \cdot F= \int_{0}^{2 \pi}\int_{0}^{\pi}\int_{0}^{\sqrt 2} (15 \rho^2) (\rho^2 \sin \phi) d\rho d \phi d\theta$
This implies that
$\int_{0}^{2 \pi}(24 \sqrt 2-6)d\theta =(48 \sqrt 2-12)\pi$
