Verifying Gauss's divergence theorem on a upside down truncated cone

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.

$$V = \frac{\pi}{3}\left(R_1^2 h_1 - R_2^2 h_2\right)$$
where $$R_1 = 2 = h_1$$ (the large cone) and $$R_2 = 1 = h_2$$ (removed tip), so the result is
$$V = \frac{7\pi}{3}$$