One model for your solid is to take a plane region $D$ of area $A$, to take a real-valued (and, say, continuous) function $f$ with domain $D$, and to let the cylinder be the set of points $(x, y, z)$ such that
$$
(x, y) \in D,\qquad
f(x, y) \leq z \leq f(x, y) + H.
$$
The volume of the cylinder is $AH$, independently of $f$, as you've observed.
One natural geometric interpretation of this formula is the height over an arbitrary point of $D$ multiplied by the area of the projection of the cylinder on the $(x, y)$-plane.
This type of formula holds in greater generality. If $f$ is a continuous function of $(n - 1)$ variables, then the "pointwise shearing" homeomorphism
$$
(x_{1}, \dots, x_{i}, \dots, x_{n}) \mapsto
(x_{1}, \dots, x_{i} + f(x_{1}, \dots, x_{i-1}, x_{i+1}, \dots, x_{n}), \dots, x_{n})
\tag{1}
$$
preserves $n$-dimensional volume. If you start with the closure $C$ of an arbitrary bounded open subset of $\mathbf{R}^{n}$ and repeatedly apply transformations of the type (1), the resulting image has the same volume as $C$.
Your cylinder with non-flat base is the image of a "standard" cylinder $C = D \times [0, H]$ under
$$
(x, y) \mapsto \bigl(x, y, z + f(x, y)\bigr).
$$
Best Answer
The solution in the polar coordinates system where the lemniscate is given by the formula $r^{2}=2a^{2}\cos2\phi$.
Surface area can be obtained by using the formula $A=2\pi\int_a^br(\phi)\sin\phi\sqrt{r^{2}(\phi)+[dr(\phi)/d\phi]^{2}}d\phi$: $$A=2\pi a^{2}\sqrt{a}\int_a^b\sin\phi\sqrt{\cos2\phi +sin^{2}2\phi } d\phi$$ The integral above seems to have no analytical solutions.
The formula for the volume is pretty simple and can be obtained directly from the formula $V=\pi\int_a^b r^2(\phi)d\phi$ : $$V=2a^{2}\pi\int_a^b cos2\phi d\phi$$