I have a surface with the geometric equation $$(z+1)^2=x^2+y^2$$ where $1\leq z\leq 0$ which give a sort of upside down truncated cone like this
I want to verify Gauss's divergence theorem for this volume with $\mathbf{F}=(x,y,0)$.
So I first went about parameterizing the surfaces for the curved part we have $$(u,v) \rightarrow (ucos(v),usin(v),u-1) ,\quad 1\leq u \leq 2, \quad 0\leq v \leq 2\pi $$ the other surfaces are just this same parameterization with z component $z=0$ and $0\leq u \leq 1$ for the bottom circle and z component $z=1$ with $0\leq u \leq 2$ for the top circle.
When I calculate the surface integrals for the each surface flux integral for the top and bottom circles I get that they are just $0$, since for the top circle we get $n=(0,0,1)$ as the outward normal and for the bottom circle the outward normal is $n=(0,0,-1)$ and we only really need the normal vector not the unit normal since the area element $dS=\lvert n \rvert dudv$. So $F \cdot n = 0$ and so the surface integrals for thee top and bottom circle are also $0$.
When I calculate the surface flux integral for the curved surface I find that the normal $n=(-ucos(v),-usin(v),u)$ by finding the tangent vector with respect to u and the tangent vector with respect to v and taking their cross product. So $F.n=-u^2$, so I evaluate my surface integral $$\int_{v=0}^{v=2\pi}\int_{u=1}^{u=2}-u^2dS$$ get that it is $\frac{-14\pi}{3}$ so the divergence theorem says that $$\int\int\int_V \nabla \cdot F dV = \int\int_S F \cdot \hat{n} dS$$
$\nabla \cdot F = 2$ so our volume integral is just 2 time the volume of V using the formula for area of a truncated cone I get that this is $\frac{10\pi}{3}$ which isn't right so i know i made a mistake somewhere I've went through my working like 100 times I really cant tell where i went wrong though.
Problem is in the last step, the volume of the cone is
$$ V = \frac{\pi}{3}\left(R_1^2 h_1 - R_2^2 h_2\right) $$
where $R_1 = 2 = h_1$ (the large cone) and $R_2 = 1 = h_2$ (removed tip), so the result is
$$ V = \frac{7\pi}{3} $$