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A paraboloid is inside a cylinder as follows:
The goal is to prove that the volume of the paraboloid is exactly one-half that of the cylinder. So, I did the proper integration required, and got $\pi ph^2$. But what is $p$?
In the meantime, let's mention that for the parabola $y^2=px$, $p$ is the distance between the focus $F$ and your directrix $d$ called focal parameter. The paraboloid of the figure has equation $x=p(y^2+z^2)$ (where $z$ is the axis coming out of the page) and is a surface of revolution about the $y$ axis of the section of the paraboloid $y^2=2px$.
In the meantime, let's mention that for the parabola $y^2=px$, $p$ is the distance between the focus $F$ and your directrix $d$ called focal parameter. The paraboloid of the figure has equation $x=p(y^2+z^2)$ (where $z$ is the axis coming out of the page) and is a surface of revolution about the $y$ axis of the section of the paraboloid $y^2=2px$.