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!    
 A: Possibly useful hint: the volume depends only on the height $h$ and on the area of the base circle. In this example you need $\alpha$ to find the radius (and hence the area) of the base.
A: 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 $
A: The cross section at distance $h$ is defined by the cone axis hitting the plane at angle $\theta$.
[The cross section is a plane section through a circular cone, hence an ellipse.]
The long axis of the ellipse is defined by the smallest and longest distances of the cone apex
to the plane. The shortest line from the apex to the plane intersects the plane at angle $\theta+\alpha$, the
longest line at the angle $\theta-\alpha$. The horizontal distances from M are therefore
$$m_{\pm}=h\cot (\theta\pm \alpha),$$
and the long axis of the ellipse is the difference
$$
2a=m_--m_+ = h[\cot(\theta-\alpha)-\cot(\theta+\alpha)],
$$
where we have denoted as usual the half-axis by $a$.
The center of the ellipse is the arithmetic mean of the two long axis end points at distance
$$
\bar m = \frac12(m_++m_-)
$$
from M.
The place where the cone axis hits the plane is $(c,0,0)$ where
$$\cot\theta = c/h.$$
The short half-axis $b$ of the ellipse is computed by maintaining the correct cone angle at the point $(\bar m,b,0)$ in
the plane. The short axis runs though O'.
A vector from there to the cone apex $(0,0,h)$ must have an angle $\alpha$ relative to the
axis. The vectors from the cone axis are $(\bar m,b,0)-(0,0,h)=(\bar m,b,-h)$ and $(c,0,0)-(0,0,h)=(c,0,-h)$,
and the angle between must be by the usual dot product formula
$$
\cos\alpha = \frac{(\bar m,b,-h)\cdot (c,0,-h)}{\sqrt{\bar m^2+b^2+h^2}\sqrt{c^2+h^2}}
 = \frac{\bar m c+h^2}{\sqrt{\bar m^2+b^2+h^2}\sqrt{c^2+h^2}}.
$$
Solving this equation for $b$,
$$
 \sqrt{\bar m ^2+b^2+h^2}
 = \frac{\bar m c+h^2}{\cos\alpha\sqrt{c^2+h^2}}.
$$
$$
\bar m^2+b^2+h^2
 = \frac{(\bar mc+h^2)^2}{\cos^2\alpha(c^2+h^2)}.
$$
We are only interested in the absolute value $|b|$ to compute the area, so we may
ignore the solution with negative $b$:
$$
b = \sqrt{\frac{(\bar m c+h^2)^2}{\cos^2\alpha(c^2+h^2)}-h^2-\bar m^2}
$$
$$
 = \sqrt{\frac{(\frac{h}{2}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]h\cot\theta+h^2)^2}{\cos^2\alpha(h^2\cot^2\theta+h^2)}-h^2-\frac{h^2}{4}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]^2}
$$
$$
 = h\sqrt{\frac{(\frac{1}{2}[\cot(\theta+\alpha)-\cot(\theta+\alpha)]\cot\theta+1)^2}{\cos^2\alpha(\cot^2\theta+1)}-1-\frac{1}{4}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]^2}.
$$
The area of the ellipse is $ab\pi$, and to obtain the volume this must be integrated along $h$:
$$
V
=\pi\int_0^h dz  ab
$$
$$
=\pi\int_0^h dz  \frac{z}{2}[\cot(\theta-\alpha)-\cot(\theta+\alpha)] z\sqrt{\frac{(\frac{1}{2}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]\cot\theta+1)^2}{\cos^2\alpha(\cot^2\theta+1)}-1-\frac{1}{4}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]^2}
$$
$$
=\pi\frac{h^3}{6}[\cot(\theta+\alpha)-\cot(\theta+\alpha)] \sqrt{\frac{(\frac{1}{2}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]\cot\theta+1)^2}{\cos^2\alpha(\cot^2\theta+1)}-1-\frac{1}{4}[\cot(\theta-\alpha)+\cot(\theta+\alpha)]^2}.
$$
Plugging in $h=20$ cm, $\theta=\pi/3$, $\alpha=\pi/6$ gives $V=2000\sqrt{2/3}\pi$ cm$^3$.
A: 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.
