= Paraboloid sector surface measures
For $h,R>0$ and $0<\alpha\leq2\pi$, the curved sector $z=h(1-r^2/R^2)$, $0\leq r\leq R$, $0\leq\theta\leq\alpha$ has <volume> $\alpha hR^2/4$ beneath it. Its <surface area> is
$$
A=\frac{\alpha R^4}{12h^2}\left[\left(1+\frac{4h^2}{R^2}\right)^{3/2}-1\right].
$$
To derive this, parametrize the <surface> by $(r\cos\theta,r\sin\theta,h(1-r^2/R^2))$. The upward <vector area element> is $(2hr^2\cos\theta/R^2,2hr^2\sin\theta/R^2,r)\,dr\,d\theta$; its magnitude integrates to the stated <surface area>. Its integral gives the <vector area>
$$
\mathbf S=\left(\frac{2hR}{3}\sin\alpha,\frac{2hR}{3}(1-\cos\alpha),\frac{\alpha R^2}{2}\right).
$$
In the shallow limit $h/R\to0$, 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.
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