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!    

First, assign an $xyz$ coordinate system.  For simplicity, let $M$ be the origin, and let the positive $x$-axis point to the right, and the positive $y$-axis point into the screen, while the $z$-axis points upward.
Next, find the unit vector in this coordinate system of the axis of the cone.  This can be found quite easily to be
$ a = ( - \cos 60^\circ, 0 , \sin 60^\circ ) = \left( - \dfrac{1}{2}, 0, \dfrac{\sqrt{3}}{2} \right) $
Now the equation of the surface of the cone is
$ (r - O)^T Q (r - O) = 0 $
where $r = [x, y, z]^T$, and $O$ is the tip of the cone, and
$ Q = \cos^2 \alpha I - a a^T = \dfrac{3}{4} I - \begin{bmatrix} -\dfrac{1}{2} \\ 0 \\ \dfrac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} -\dfrac{1}{2} && 0 && \dfrac{\sqrt{3}}{2} \end{bmatrix} $
Upon evaluting the above expression, one finds that
$ Q = \begin{bmatrix} \dfrac{1}{2} && 0 && \dfrac{\sqrt{3}}{4} \\ 0 && \dfrac{3}{4} && 0 \\ \dfrac{\sqrt{3}}{4} && 0 && 0 \end{bmatrix} $
Since we want to find the area of the elliptical base, we set $z=0$ in the equation of the cone, and this leads to the following equation in $x$ and $y$ (Note that the coordinates of point $O$ are $(0, 0, 20) $
$\dfrac{1}{2} x^2 + \dfrac{3}{4} y^2 + \dfrac{\sqrt{3}}{2} (-20 x)  = 0 $
Multiplying through by $4$,
$ 2 x^2 + 3 y^2 - 40\sqrt{3} x = 0 $
Completing the square in $x$,
$ 2 (x - 10\sqrt{3})^2 + 3 y^2 = 600 $
Dividing through by $600$
$ \dfrac{(x-10\sqrt{3})^2}{300} + \dfrac{ y^2 }{200} = 1 $
Hence the elliptical base has a semi-major axis along the $x$-axis of length $ = \sqrt{300} = 10\sqrt{3} \text{ cm} $ and semi-minor axis along the $y$-axis of length $=\sqrt{200} = 10\sqrt{2} \text{ cm} $.
The area of the base is therefore $A = \pi a b = 100 \sqrt{6} \pi \text{ cm}^2 $
And the volume is $ V = \dfrac{1}{3} A h  = 2000 \sqrt{\dfrac{2}{3}} \pi \text{ cm}^3 $
