= Solution
Use the <embedded surface parametrization>
$$
\mathbf X(r,\theta)=\left(r\cos\theta,r\sin\theta,a\left(1-\frac{r^2}{b^2}\right)\right).
$$
Its tangent vectors have <cross product>
$$
\mathbf X_r\times\mathbf X_\theta
=\left(\frac{2ar^2}{b^2}\cos\theta,
\frac{2ar^2}{b^2}\sin\theta,r\right),
$$
whose magnitude is $r\sqrt{1+4a^2r^2/b^4}$. Thus the <surface area> is
$$
A=\alpha\int_0^R r\sqrt{1+\frac{4a^2r^2}{b^4}}\,dr.
$$
With $u=1+4a^2r^2/b^4$, the radial integral becomes $b^4(8a^2)^{-1}\int u^{1/2}\,du$. Evaluating its endpoints gives
$$
\boxed{A=\frac{\alpha b^4}{12a^2}
\left[\left(1+\frac{4a^2}{b^2}\right)^{3/2}-1\right]}.
$$
This is the scalar-area part of the <paraboloid sector surface measures>.
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