How to evaluate volume of oblique frustum of right circular cone with elliptical section? 
Let there be an oblique frustum, with an elliptical section, of a right circular cone with apex point O & cone angle $2\alpha=60^{o}$. It is obtained by cutting the cone by a plane at a normal distance $OM=h=20 cm$ & making an angle $\theta=60^{o}$ with the axis OO' of the frustum (cone) (as shown in the above diagram). How to evaluate the volume of this oblique frustum with elliptical section? 
Any help is greatly appreciated!    

We will not only find the volume but prove that the cross-section is indeed an ellipse.

We look for the intersection (see figure) of the cone:
$$
x^2+y^2=z^2\,\tan^2\alpha\tag1
$$
with the plane:
$$
z=\frac d{\sin\theta}+\frac x{\tan\theta}\tag2
$$
Substituting $(2)$ into $(1)$ results in:
$$\begin{align}
&x^2+y^2=\left(\frac d{\sin\theta}+\frac x{\tan\theta}\right)^2\tan^2\alpha\\
&\iff
\lambda^2\left(x-\frac {d\,\tan^2\alpha}{\lambda^2\sin\theta\tan\theta}\right)^2+y^2=
\frac{d^2\tan^2\alpha}{\sin^2\theta}\left(1+\frac{\tan^2\alpha}{\lambda^2\tan^2\theta}\right)\tag3
\end{align}
$$
with $\displaystyle\lambda=\sqrt{1-\frac{\tan^2\alpha}{\tan^2\theta}}$.
The expression $(3)$ is the equation of an ellipse with
$$
b^2=\frac{d^2\tan^2\alpha}{\sin^2\theta}\left(1+\frac{\tan^2\alpha}{\lambda^2\tan^2\theta}\right)=\frac{d^2\tan^2\alpha}{\lambda^2\sin^2\theta},\quad
a=\frac b\lambda
$$
and the corresponding area
$$
A=\pi ab=\pi\frac{b^2}\lambda
=\pi\frac{d^2\tan^2\alpha}{\lambda^3\sin^2\theta}\tag4.
%=\frac{\pi d^2\tan^2\alpha\sin\theta}{\left(\sin^2\theta-\cos^2\theta\tan^2\alpha\right)^\frac32},\tag4
$$
Observe that this is the area of the ellipse in the plane $z=0$ (marked red in the figure). The area of the corresponding ellipse (marked blue in the figure) in the plane $(2)$ is
$$
A'=\frac A{\sin\theta}\tag5,
$$
so that finally
$$
V=\frac13A'd=\frac{\pi d^3\tan^2\alpha}{3\lambda^3\sin^3\theta}
=\frac{\pi d^3\sin^2\alpha\cos\alpha}{3\left[\sin(\theta+\alpha)\sin(\theta-\alpha)\right]^\frac32}.\tag6
$$
Substituting in $(6)$ the given values of $d,\theta,\alpha$ one obtains:
$$
V=2000\sqrt\frac23\pi~\text{cm}^3.
$$
PS. $d=h$ from the question.
