Elementary question concerning flux integral I was asked to evaluate the flux integral $\int\int_{S}F\cdot dS$. Here $F=\langle x,y,z^{4}\rangle$ and $S$ is part of the cone $z=\sqrt{x^{2}+y^{2}}$ beneath the plane $z=1$, oriented downward. 
My thought is to evaluate this by capping off a disk at $z=1$. Then by divergence theorem we have 
$$
\begin{align}
\int^{2\pi}_{0}\int^{1}_{0}\int^{r}_{0}[\nabla\cdot F] r\,dr\,dz\,d\theta
&= \int^{2\pi}_{0}\int^{1}_{0}\int^{r}_{0} \left(2+4z^{3} \right)r\,dr\,dz\,d\theta \\
&= \int^{2\pi}_{0}\int^{1}_{0} \left(2r^{2}+r^{5} \right)dr\,d\theta \\
&= 2\pi\left(\frac{2}{3}+\frac{1}{6} \right)=\frac{5}{3}\pi
\end{align}
$$
While the top part can be "easily" calculated by noticing $F=\langle x,y,1\rangle$, and $F\cdot n=1$. So we have $$\int\int_{S}F\cdot dS=\pi$$since the unit disk has area $1$ in the $z$-plane. So sum up we should have the result to be $$\frac{5}{3}\pi-\pi=\frac{2}{3}\pi$$
But this answer is incorrect. Because a standard flux integral by $-Pf_{x}-Qf_{y}+R$ formula gives the answer to be $-\frac{1}{3}\pi$ instead. I want to ask where did I made a mistake(conceptual or arithmetic). I checked my computation several times but did not detect a problem. It is kind of embarrassing to ask this as a grad student. 
 A: Your error lies in your evaluation of the volume integral.
Assuming that you have made a typo in the order of the integration variables, so that it was meant to read
$$
\int^{2\pi}_{0}\int^{1}_{0}\int^{r}_{0} \left(2+4z^{3} \right)dz\,r\,dr\,d\theta
$$
then you are describing the area between the cone and the $x$-$y$ plane, rather than between the cone and the plane $z=1$. Instead, you want the first integral to run from $r$ to $1$. If you had the order of integration variables right, then your limits were even more incorrect.
With this correction, you get
\begin{align}
\int^{2\pi}_{0}\int^{1}_{0}\int^{1}_{r} \left(2+4z^{3} \right)dz\,r\,dr\,d\theta
&= \int_0^{2\pi}\int_0^1\left(3r-2r^2-r^5\right)dr\,d\theta\\
&= 2\pi \left[\frac{3r^2}2-\frac{2r^3}3-\frac{r^6}6\right]_0^1 = \frac{4\pi}3
\end{align}
Note that your "correct" answer is also incorrect. Simple geometric considerations (vector, for $z<1$, will point outwards compared with surface of cone) indicate that the flux must be positive. I haven't checked, but I assume that orientation wasn't accounted for.
A: 
The volume integral of divergence can be evaluated as 
$$\int_0^1\int_0^{2\pi}\int_r^1 (2+4z)r dzd\theta dr = \frac{4 \pi }{3 }$$
Also surface can parametrized as $S(r, \theta) = (r \cos (\theta ),r \sin (\theta ),r)$ and integral can be evaluated as 
$$\int_0^1 \int_0^{2\pi }\langle r \cos \theta, r\sin\theta,r^{4}\rangle \cdot (-1)\left( \frac{\partial S}{\partial r} \times\frac{\partial S}{\partial \theta} \right) d\theta dr = \frac \pi 3 $$
$(-)$ because $\displaystyle \frac{\partial S}{\partial r} $ is radially tangent and $\displaystyle \frac{\partial S}{\partial \theta}$ is circularly tangent anticlockwise giving the flux through inner surface.
