Divergence theorem integrating over a cylinder Problem: Calculate $\int \int_S \langle F,n \rangle dS$ where $S$ is the half cylinder $y^2+z^2=9$ above the $xy$-plane, and $F(x,y,z) = (x,y,z).$
My working: I did this using a surface integral and the divergence theorem and got different results. First, using a surface integral:
Write $z=h(x,y)=(9-y^2)^\frac{1}{2}$. So the normal is given by $N=(h_x, h_y, -1)$. Calculating the partial derivatives gives $h_x =0, h_y = \frac{-y}{(9-y^2)^\frac{1}{2}}$.
The unit normal is $n=\frac{(9-y^2)^\frac{1}{2}}{3}(0,\frac{-y}{(9-y^2)^\frac{1}{2}},-1) = (0, \frac{-y}{3}, \frac{-(9-y^2)^\frac{1}{2}}{3})$ .
So $\langle F,n \rangle = \frac{-y^2}{3} - \frac{9-y^2}{3}$ and $\int \int_S \langle F,n \rangle dS = \int_0^3 \int_{-3}^3 \frac{-y^2}{3} - \frac{9-y^2}{3} dy dx =  0$.
However, using the divergence theorem, we have $$\int \int_S \langle F,n \rangle dS = \int_{x=0}^{x=3} \int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=3} 3 r dr d\theta dx = \frac{81\pi}{2}$$
Where have I made a mistake?

For the surface $z=h(x,y)=(9-y^2)^{\frac1{2}}$ the outward unit normal vector is
$$n = \Big(0,\frac{y}{3},\frac{(9-y^2)^{\frac1{2}}}{3}\Big)$$
and the differential surface area element is
$$dS = \sqrt{1+h_x^2+h_y^2}dxdy= 3(9-y^2)^{\frac{-1}{2}}dxdy$$
so the surface integral over the top is 
$$\int F \cdot n dS= \int_{0}^{3}\int_{-3}^{3}\Big[y\frac{y}{3}+(9-y^2)^{\frac1{2}}\frac{(9-y^2)^{\frac1{2}}}{3}\Big]3(9-y^2)^{\frac{-1}{2}}dxdy=3\int_{-3}^{3}9(9-y^2)^{\frac{-1}{2}}dy=27\pi$$
There is also a contribution of $\frac{27\pi}{2}$ from the surface integral over the end where $x=3$. The sum is $\frac{81\pi}{2}$ which matches the volume integral of the divergence.
