 Paraboloid sector surface measures

For $h,R>0$ and $0<\alpha \le 2\pi$, the curved sector $z=h(1-r^2/R^2)$, $0\le r\le R$, $0\le \theta \le \alpha$ has volume $\alpha h{R}^2/4$ beneath it. Its surface area isTo derive this, parametrize the surface by $(r\cos \theta ,r\sin \theta ,h(1-r^2/R^2))$. The upward vector area element is $(2h{r}^2\cos \theta /R^2,2h{r}^2\sin \theta /R^2,r)drd\theta$; its magnitude integrates to the stated surface area. Its integral gives the vector areaIn the shallow limit $h/R\to 0$, the surface area is $\alpha R^2/2+\alpha h^2/2+O(h^4/R^2)$, approaching the area of its planar sector projection.