Evaluation of a surface integral

Question

How do we evaluate $\int_S A.n dS $ over the entire surface of the region above the $x$-$y$ plane bounded by the cone $x^2+y^2=z^2$ and the plane $z=4$ if $A=4xz\vec i + xyz^2 \vec j + 3z \vec k$ ?

Answer 1

The surface in question bounds a solid cone $C$ of radius $4$ and height $4$, standing on his tip. Assume that the surface normal ${\bf n}$ points outward. Since $${\rm div}({\bf A})={\partial\over\partial x}(4xz)+{\partial\over\partial y}(xyz^2)+{\partial\over\partial z}(3z)=4z+xz^2+3$$ we obtain by Gauss' theorem $$\Phi:=\int_{\partial C}{\bf A}\cdot{\bf n}\>d\omega=\int_C(4z+xz^2+3){\rm d}(x,y,z)\ .\tag{1}$$ The cone $C$ is symmetrical with respect to the plane $x=0$. Therefore the middle term on the right hand side in $(1)$, being odd in $x$, gives no contribution. The remaining terms are rotationally invariant. Introducing cylindrical coordinates we therefore obtain $$\Phi=2\pi\int_0^4 \left( \int_r^4(4z+3)\ dz\right)\ r\> dr=320\pi\ .$$