Solution
= Solution
The <volume element> in <cylindrical coordinates> is $r\,dz\,dr\,d\theta$. Integrating from the base to the curved <surface> gives
$$
V=\int_0^\alpha\!d\theta\int_0^R\!r\,dr\int_0^{a(1-r^2/b^2)}\!dz
=\alpha a\left[\frac{r^2}{2}-\frac{r^4}{4b^2}\right]_0^R.
$$
Hence
$$
\boxed{V=\frac{\alpha ab^2}{4}}.
$$
For a complete rotation this is half the <volume> of the enclosing cylinder.