I was having trouble showing that the following closed surface integral:
$$\iint_S F \cdot dS = 0$$
With the vector field $$F = \langle x,y,-2z\rangle$$ over the cone with a lid on top:
$$\ z^2 = x^2+y^2$$ $$\ 0≤z≤5 $$
I've tried parameterizing the cone as $$\ G(u,v) = (v\cos(u), v\sin(u),v) $$ and the normal vector $$\ n = (v\cos(u), v\sin(u), -v)$$
Dot multiplying $$\ F(G(u,v)) \cdot n $$
and doing the surface integral
$$\int_0^5\!\int_0^{2\pi}\ \ F(G(u,v)) \cdot\ n \ du\,dv$$
This does not end up being 0. can anyone explain why this isn't 0? Are my bounds correct? Does it have to do with the top lid? When I use divergence Theorem the answer is 0. Please help,Thanks in advance.