Here it's a question actually on integration but I simply don't know how to calculate the volume of the cone enclosed between the dotted planes. It might be very trivial but I could get up to the only realisation that it's going to be probably an ellipsoid not a circular area. The figure is actually a right circular cone with semi vertical angle alpha.
Any help is greatly appreciated.
With the clarifications from the comments the problem becomes tractable. Especially the fact that the element $AV$ of the cone is perpendicular to the cutting plane will be used.
To find the volume we need first to calculate the area of the cross section of the cone with the plane, which is an ellipse. Without loss of generality we can assume that the cutting planes are perpendicular to the plane of your figure (PoF) and the cone axis lies in the PoF. Let $d$ be the distance between the planes. As it is just a scaling factor we may for simplicity assume $d=1$.
With this assumption one easily computes $$\begin{align} a&=EB=\frac12\tan2\alpha,\\ x&=DE=\frac12\tan2\alpha-\tan\alpha\\ y&=DF=\frac{\tan\alpha}{\cos\alpha}\\ b&=EE'=\frac y{\sqrt{1-\frac{x^2}{a^2}}} =\frac{\sin\alpha}{\sqrt{\cos2\alpha}} \end{align} $$
Recalling that the area of an ellipse is $S=\pi a b$ one finally obtains: $$ V=\frac13Sd=\frac{\pi d^3}6\frac{\tan2\alpha\sin\alpha}{\sqrt{\cos2\alpha}}. $$