multivariable calculus divergence theorem help

I am stuck on a problem:

Use the Divergence Theorem to evaluate $\iint \mathbf{F} \cdot d\mathbf{S}$, where $$\mathbf{F}(x,y,z)=z^2x\mathbf{i}+(\frac{1}{3}y^3+\tan(z))\mathbf{j}+(x^2z+y^2)\mathbf{k}$$ and $S$ is the top half of the sphere $x^2+y^2+z^2=1$.

[Hint: Note that $S$ is not a closed surface. First compute integrals over $S_1$ and $S_2$, where $S_1$ is the disk $x^2+y^2\le 1$, oriented downward, and $S_2 = S \cup S_1$.]

my working process:

$\iint \mathbf{F} \cdot d\mathbf{S} = \iiint\limits_E div\mathbf{F} \cdot d\mathbf{V} = \iiint\limits_E (x^2+y^2+z^2) d\mathbf{V}$.

for $S_2$, parametrise $x=r\sin(\phi)\cos(\theta)$, $y=r\sin(\phi)\sin(\theta)$, $z=r\cos(\theta)$ and $0\le r \le 1,\;\; 0\le \theta \le 2\pi,\;\; 0\le \phi \le \pi.$

=$\iiint\limits_{S_2} (r^2\cos^2(\theta)+r^2\sin^2(\phi)) dV+\iint\limits_{S_1} \mathbf{F} \cdot d\mathbf{S}$

=$\frac{14\pi}{15}$+$\iint\limits_{S_1} \mathbf{F} \cdot d\mathbf{S}$

(the correct answer is $\frac{13\pi}{20}$)

could you point out the mistake? Thank you very much!!!

Applying divergence theorem, you get flux over closed surface $S_2$.

In spherical coordinates, $x^2 + y^2 + z^2 = \rho^2$

where $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi. z = \rho \cos\phi$

As it is part of the sphere above $z = 0, \phi \leq \pi/2$.

So the volume integral is,

$ \displaystyle \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^4 \sin\phi ~d\rho ~d\phi ~d\theta = \frac{2\pi}{5}$

Now the unit normal vector to the disk surface $S1: x^2 + y^2 \leq 1, z = 0 ~$ is $~\hat n = (0, 0, -1)$

$\vec F \cdot \hat n = - y^2$

Using polar coordinates, the surface integral over disk surface is,

$ \displaystyle \int_0^{2\pi} \int_0^1 - r^3\sin^2\theta ~ dr ~ d\theta = - \frac{\pi}{4}$

So flux over spherical surface $S$ is,

$ \displaystyle \frac{2\pi}{5} - \left(- \frac{\pi}{4}\right) = \frac{13\pi}{20}$