Volume of a paraboloid

Consider a volume element of thickness $dz$ at a height $z$ above the XY plane. Treating $z$ as constant, the cross-section of this volume is an ellipse having equation:

$\frac{x^2}{A^2}+\frac{y^2}{B^2}=1 \ ; \ A = \sqrt{\frac{z}{a}} \ ; \ B = \sqrt{\frac{z}{b}}$

The cross-section area of this elementary volume is:

$S = \pi A B \implies S = \frac{\pi z}{\sqrt{ab}}$

The volume of this element is:

$dV = S \ dz \implies dV = \frac{\pi z}{\sqrt{ab}} \ dz$

The volume of the paraboloid below the plane $z=c$ is therefore:

$V = \int_{0}^{c} \frac{\pi z}{\sqrt{ab}} \ dz \implies \boxed{V = \frac{\pi c^2}{2\sqrt{ab}}}$