## Thomas' Calculus 13th Edition

$128 \pi$
We have: $g(x,y,z) = x^2+y^2+z^2 =25$ Thus, $\nabla g =2 xi+2y j +2z k$ and $|\nabla g|=\sqrt {(2x)^2+(2y)^2+(2z)^2 }=\sqrt { 4 (x^2+y^2+z^2) }=\sqrt { 4 (25)}=10$ and $\ Normal \ Vector, n =\dfrac{\nabla g }{|\nabla g|}=\dfrac{xi+yj+zk}{5}$ $F \cdot n =\dfrac{x^2}{5} z+\dfrac{y^2}{5} z +\dfrac{z}{5}$ and $d \sigma=\dfrac{10}{2z} \ d A$ We set up the integral and solve the flux of $F$ as follows: For above flux area: $\iint_{S} F \cdot n \ d S=\iint_{R} (\dfrac{x^2}{5} z+\dfrac{y^2}{5} z +\dfrac{z}{5}) ( \dfrac{5}{z} ) \ dA = \int_{0}^{2 \pi} \int_{0}^{4} (r^2+1) r \ dr \ d \theta =\int_{0}^{2 \pi} 72 \ d \theta = 144 \pi$ For bottom flux area: $\iint_{\ bottom} F \cdot n \ d \sigma=\iint_{R} (-1) \ dA = (-1) \times$ Area of the circular region $=-16 \pi$ Now, the total flux is equal to $=144 \pi+(-16 \pi) =128 \pi$