How to Find the integral $\int_{S}−(xy^2)~dydz + (2x ^2 y)~dzdx − (zy^2)~dxdy$ where is the portion of the sphere $x^2 + y^2 + z^2 = 1$ 
Find the integral 
  $$
\int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$ 
  where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=\frac{1}{2}$. Choose the direction of the normal to be outside the sphere.

Can i get some help? I got that the flux is 
$$
\huge \phi = \huge 0.
$$ 
and I am not sure if thats right.
\begin{split}
\vec F &= (-xy^2,2x^2y,-zy^2)\\
\vec n_\mathrm{sphere} &= (2x,2y,2z)\\
\vec{F}\cdot{}\vec{n} = 2y^2(x^2 - z^2) &=^{{~(z^2 = 1-x^2-y^2)}}~4x^2y +2y^3 -2y\\
\int\int 4x^2y +2y^3 - 2ydydx &=
\end{split}
$$\int_{r=0}^{r=\sqrt{\frac{3}{4}}}\int_{\theta=0}^{\theta=2\pi}4r^4\cos^2\theta \sin\theta + 2r^4\sin^3\theta - 2r^2\sin\theta d\theta dr = \boxed{0}\\
$$
 A: Let $$
\Omega: \frac{1}{2}\le z\le \sqrt{1-x^2-y^2}
$$ be the region enclosed by $S\bigcup S_1$ where $S_1 :z=\frac{1}{2}, x^2+y^2\le\frac{3}{4}$. We obtain
$$
\iiint_\Omega 2(x^2-y^2)\ \mathrm{d}x\mathrm{d}y\mathrm{d}z =\iint_{S\bigcup S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S
$$ by divergence theorem. Note that
$$
\iiint_\Omega x^2\ \mathrm{d}x\mathrm{d}y\mathrm{d}z=\iiint_\Omega y^2\ \mathrm{d}x\mathrm{d}y\mathrm{d}z
$$ by the rotational symmetry, hence the left-hand side equals $0$. This gives
$$\iint_{S\bigcup S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S=0.
$$ Since outward unit normal vector $\vec{n}=(0,0,-1)^T$ on $S_1$, it follows
$$\begin{eqnarray}
\iint_{S}\vec{F}\cdot \vec{n}\ \mathrm{d}S&=&-\iint_{S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S\\&=&-\frac{1}{2}\iint_{x^2+y^2\le \frac{3}{4}} y^2\ \mathrm{d}x\mathrm{d}y\\
&=&-\frac{1}{4}\iint_{x^2+y^2\le \frac{3}{4}} (x^2+y^2)\ \mathrm{d}x\mathrm{d}y\\
&=&-\frac{1}{4}\int_0^{2\pi}\int_{0}^{\frac{\sqrt{3}}{2}}r^3\ \mathrm{d}r\mathrm{d}\theta=-\frac{9\pi}{128}.
\end{eqnarray}$$
