Assume that a plane and a cone intersect, e.g., $(z-z_0)^2=x^2 + y^2$, and $z=-y+f(t)$. I can find that the intersection of these two objects would be equal to $y=\frac{x^2-[f(t)]^2+2z_0f(t)-z_0^2}{2z_0-2f(t)}$.
I am trying to find an expression for the area within the parabola which would be projected on the $z=-y+f(t)$ plane, and as that plane moves relative to the cone (hence the f(t) term).
This problem is similar to this Area of the intersection of two cones problem, where my plane would be the plane through the points $p_3$, $p_4$, and the origin.
In diagram below, line $OC$ is the $z$-axis and line $OE$ is the $y$-axis.
From your data it follows that $CO=OD=OE=z_0$ and $BO=OH=f(t)$. It is then quite easy to compute $$ VH={z_0+f(t)\over\sqrt2} \quad\hbox{and}\quad AH=\sqrt{HD\cdot HE}=\sqrt{z_0^2-f(t)^2}. $$ Finally, as Archimedes of Syracuse proved many years ago, the area of the parabola segment is given by $$ area={4\over3}AH\cdot VH= {4\over3}\sqrt{z_0^2-f(t)^2}{z_0+f(t)\over\sqrt2}. $$