Answer
$$128 \pi $$
Work Step by Step
$ g(x,y,z) = x^2+y^2+z^2 =25 \implies \nabla g =2 xi+2y j +2z k$
$\implies |\nabla g|=\sqrt {(2x)^2+(2y)^2+(2z)^2 }=\sqrt { 4 (x^2+y^2+z^2) }=\sqrt { 4 (25)}=10$
Now, $\ Normal \ Vector, n =\dfrac{\nabla g }{|\nabla g|}=\dfrac{xi+yj+zk}{5} \implies F \cdot n =\dfrac{x^2}{5} z+\dfrac{y^2}{5} z +\dfrac{z}{5} \\ d \sigma=\dfrac{10}{2z} \ d A$
We need to set up the integral for the above flux area as follows: $F_1=\iint_{S} F \cdot n \ d S=\iint_{R} (\dfrac{x^2}{5} z+\dfrac{y^2}{5} z +\dfrac{z}{5}) ( \dfrac{5}{z} ) \ dA = \int_{0}^{2 \pi} \int_{0}^{4} (r^2+1) r \ dr \ d \theta =\int_{0}^{2 \pi} 72 \ d \theta = 144 \pi$
We need to set up the integral for bottom flux area as follows: $F_2=\iint_{\ bottom} F \cdot n \ d \sigma=\iint_{R} (-1) \ dA = (-1) \times $ area of the circular region $=-16 \pi$
Total flux, $F=F_1+F_2=144 \pi+(-16 \pi) =128 \pi $
