Find the flux of the vector field $F(x,y,z)=z\vec{k}$ across the outward oriented sphere $x^2+y^2+z^2=a^2$ 
Find the flux of the vector field $F(x,y,z)=z\vec{k}$ across the outward oriented sphere $x^2+y^2+z^2=a^2$

$G(x,y,z)=x^2+y^2+z^2-a^2=0$ hence $\nabla G=⟨2x,2y,2z⟩$ and $F.\nabla G=2z^2=2(a^2-x^2-y^2)$
\begin{align}
\int\int_\sigma F.N\:dS&=\int\int_R F.\nabla G\:dA\\
&=2\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} (a^2-x^2-y^2)dydx\\
&=2\int _0^{2\pi }\int _0^a\left(a^2-r^2\right)rdrd\theta\\ 
&=\pi a^4
\end{align}
But if I use parametrization $$r(\phi,\theta)=a\sin\phi\cos\theta\vec{i}+a\sin\phi\sin\theta\vec{j}+a\cos\phi\vec{k}\quad (0\leq\phi\leq\pi,0\leq \theta\leq 2\pi)$$ $$F.\left(\frac{\partial r}{\partial \phi}\times\frac{\partial r}{\partial \theta} \right)=a^3\sin\phi\cos^2\phi$$
\begin{align}
\int\int_\sigma F.N\:dS&=\int\int_R F.\left(\frac{\partial r}{\partial \phi}\times\frac{\partial r}{\partial \theta} \right)\:dA\\
&=\frac{4\pi a^3}{3}
\end{align}
I knew that my second approach give the correct answer but I was wondering why the first approach give me wrong answer$?$ Am I doing it wrongly or missing something$?$
Thanks in advances.
 A: The first calculation contains three errors. First, $\mathrm dA$ is not $\mathrm dx\mathrm dy$. You're missing the factor $\frac az$ in $\mathrm dA$ that accounts for the slope of the surface. Second, you need to normalize the outward normal vector, which divides it by a factor of $2a$. Third, you're only calculating the result for the upper hemisphere; adding the identical result for the lower hemisphere yields a factor of $2$.
The second calculation also contains an error, I'm not sure how you got the correct result out of it. The dimensions in $F\cdot\left(\frac{\partial r}{\partial \phi}\times\frac{\partial r}{\partial \theta} \right)\mathrm dA$ are wrong – you're integrating $F$ over an area, so the result should have the dimensions of $F$ times length squared, but here you have the dimensions of $F$ times length to the fourth. Instead of $\mathrm dA$ you need $\mathrm d\phi\,\mathrm d\theta$. If you substitute that, the result comes out right.
In any case, you don't really need all this integration stuff. In three dimensions, the sphere has the special property that slices of equal height have equal surface areas. So you just need to average $F\cdot\frac{\nabla G}{|\nabla G|}=\frac{z^2}a$ over $z$ from $-a$ to $a$, yielding $\frac a3$, and multiply by the total surface are $4\pi a^2$ of the sphere.
