Compute Flux (Divergence Theorem) Given that $\textbf{F} =  \langle x\cos^2z,  y\sin^2z,  \sqrt{x^2+y^2}\:z \rangle $ and let E be the solid cone above the $xy$-plane and inside z = $1 -\sqrt{x^2+y^2}.$
I'm trying to use the divergence theorem to compute the flux.
$$\iint_{D} \textbf{F} \cdot \textbf{N} \: dS = \iiint_E \nabla \cdot \textbf{F}\:d\textbf{V}$$

Attempt:
$$\text{div}\textbf{F} \:=\: \nabla \cdot \textbf{F} 
=\cos^2z + \sin^2z + \sqrt{x^2+y^2} \:
= z + r \:
$$ 
Since solid cone is above the $xy$-plane, $\text{z} \ge 0$ and $z = 1-\sqrt{x^2+y^2} = 1-r$. Hence $ 0 \le \text{z} \le 1-r$. Is this bound right for z?
\begin{align*}
\iiint_E \nabla \cdot \textbf{F}\:d\textbf{V} &= \int_0^{2\pi} \int_0^{1} \int_0^{1-r} (z+r) r\: dz\: dr\: d\theta \\
&= 2\pi \int_0^{1} \left(\frac{z^2r}{2} + r^2z\right)\bigg\rvert_0^{1-r} dr \\
&= 2\pi \cdot \frac{1}{2} \int_0^1 r-r^3\, dr \\
&= \pi \left(\frac{r^2}{2}-\frac{r^4}{4}\right)\bigg\rvert_0^1 \\
&= \frac{\pi}{4}.
\end{align*}
The answer given was $\frac{\pi}{2}.\:$ 
Perhaps I'm doing something wrong. One thing that I'm not sure of is picturing the surface that they're looking for. Also, not sure if my bounds are right. Where am I going wrong here? Would appreciate some help! Thank you.

Since $\cos^2z + \sin^2z = 1$, you must have $\nabla \cdot \mathbf{F} = 1+r$. In this case, your integral becomes
$$
\iiint_E\nabla \cdot \mathbf{F} \, \mathrm{d}V = \int_0^{2\pi} \int_0^1 \int_0^{1-r} (1+r)r\, \mathrm{d}z\, \mathrm{d}r\, \mathrm{d}\theta = \frac{\pi}{2}.
$$
