How to find the flux through the surface of a solid bounded by the cone $y^2= z^2 + x^2$ and the planes $y=1$ and $y=4$.

Question

The field $F(x,y,z)= \langle \sqrt{z^2+4y},yz^2, zx^2+5y\rangle$. I found the divergence of $F$ to equal $\langle 0,z^2,x^2 \rangle$. Using this in the divergence integral $\int Fdr$ and the range of $y$ from $1$ to $4$, I converted the integral to cylindrical coordinates where $z$ ranges from $0$ to $r$, $r$ ranges from $0$ to $4$ and $\theta$ from $0$ to $2\pi$. This was to give me the flux through the surface of the cone which I calculated as $255\pi/5$. I then added the flux through each circular area of radius $1$ and radius $4$ and found that to be $63\pi$ which when added to the surface flux gave me $570\pi/5$. However the answer is supposed to be $1023\pi/10$. I don't see how to get this answer.