Determine the flux of $u=(xz,yz,z^3)$ out of the unit sphere $x^2+y^2+z^2=1$
$\textbf{Solution}$: I have no idea how this solution can be wrong but apparently it is, By Stoke's theorem, instead of computing the surface integral, we may evaluate $$\iiint\limits_V(\nabla \cdot \mathbf{F}) dV = \iiint\limits_{x^2+y^2+z^2\leq 1}(3z^2+2z)dV $$ Now, by symmetry, we know that $\iiint\limits_{x^2+y^2+z^2\leq 1} z dV=0$ and that $\iiint\limits_{x^2+y^2+z^2\leq 1} z^2 dV=\iiint\limits_{x^2+y^2+z^2\leq 1} y^2 dV=\iiint\limits_{x^2+y^2+z^2\leq 1} x^2 dV$ and hence that $\iiint\limits_{x^2+y^2+z^2\leq 1} 3z^2 dV=\iiint\limits_{x^2+y^2+z^2\leq 1} x^2+y^2+z^2 dV=\frac{4}{3}\pi$ since $x^2+y^2+z^2=1$. Hence my answer would be $\frac{4}{3}\pi$ but the stated answer is $\frac{4}{5}\pi$. Am I correct or is the given answer incorrect?
The answer has been given in the comments though we will elaborate and fill in some details here. We would like to compute the flux $F$, given the flow field $\textbf{u} = (xz, yz, z^3)$, through the surface of the unit sphere $1 = x^2 + y^2 + z^2$. We integrate the local flux $d F$ through a surface element $dS$ over the surface of the sphere:
$$ F = \int d F = \int_S \textbf{u} \cdot \textbf{n} dS $$
Applying Gauss' theorem we get:
$$ F = \int_S \textbf{u} \cdot \textbf{n} dS = \int_V \nabla \cdot \textbf{u} dV = \int_V 2z + 3z^2 dV $$
This integral is readily computed in spherical coordinates:
$$ F = \int_V 2r \cos \phi + 3 r^2 \cos^2 \phi dV $$ $$ = \int_V \left( 2r \cos \phi + 3 r^2 \cos^2 \phi\right) r^2 \sin \phi d \phi d \Theta dr $$
$$ = \int_V 2r^3 \cos \phi \sin \phi d \phi d \Theta dr + \int_V 3r^4 \cos^2 \phi \sin \phi d \phi d \Theta dr $$
$$ = F_1 + F_2 $$
Computing $F_1$ we find it equal to zero as you stated and mentioned in the comments:
$$ F_1 = \int_V 2r^3 \cos \phi \sin \phi d \phi d \Theta dr $$
$$ = \int_0^{2 \pi} d \Theta \int_0^1 2r^3 dr \int_0^{\pi} \cos \phi \sin \phi d \phi $$
$$ = \pi \int_0^{\pi} \cos \phi \sin \phi d \phi = \pi \int_0^{\pi} \frac{\sin 2 \phi}{2} d \phi = 0$$
Computing $F_2$ we get:
$$ F_2 = \int_V 3r^4 \cos^2 \phi \sin \phi d \phi d \Theta dr $$
$$ = \int_0^1 3r^4 dr \int_0^{2 \pi} d \Theta \int_0^{\pi} \cos^2 \phi \sin \phi d \phi $$
$$ = \frac{6}{5} \pi \int_0^{\pi} \cos^2 \sin \phi d \phi $$
$$ = -\frac{6}{5} \pi \int_0^{\pi} \cos^2 d(\cos \phi) $$
$$ = -\frac{2}{5} \pi \left[ \cos^3 \phi \right]_0^{\pi} = \frac{4}{5} \pi $$
And so:
$$ F = F_1 + F_2 = \frac{4}{5} \pi$$