Finding the flux of the vector $ \vec F=z\hat k$ across the boundary of a sphere centered around the origin having positive orientation

Question

I'm trying to find the flux of the vector $ \overrightarrow F=z\hat k$ across the boundary of a sphere $S$ centered around the origin with radius $\\a\\$ and having positive orientation. The answer is obviously the volume of the sphere $\frac{4 \pi a^3}{3}\\$ via the divergence theorem, but I'm trying to compute it directly using an appropriate surface integral. I observed that $S=\{f(x,y,z)=a^2\}$ where $\\f(x,y,z)=x^2+y^2+z^2\\$ with gradient $\nabla{f}=2x\hat {i}+2y\hat {j}+2z\hat {k}$. The flux SHOULD be the double integral of $\overrightarrow F \cdot \nabla{f}=2z^2=2(a^2-x^2-y^2)\\$ over the projection of $S$ onto the $(x,y)$ plane $R=\{0\leq r\leq a;0\leq \theta \leq 2\pi\}$. But I keep getting $\pi a^4$! I do, however, get the correct answer if I don't include the $r$ in $dA=rdrd\theta$, which makes me suspect it's not necessary since the vector field $\overrightarrow F\\$ is only depends on $z$ and not $r$ and $\theta$. I feel very inadequate for not being able to figure out this triviality on my own, so let me down easy. Thank you.