volume of a truncated cone that is not a frustum

Question

I know the formula for a conical frustum is

$$\frac{\pi h}{3}\left( r^2+rR+R^2 \right) $$

What would the formula be for the area of a truncated right circular cone be where the top is not parallel to the base.

With the plane truncating the cone at an angle $\theta$ from parallel and $\theta <.5 \pi$.

Answer 1

A plane that meets one nappe of a right circular cone in an ellipse defines an oblique cone with an elliptical base. If the plane lies at distance $d$ from the cone's vertex, and if the base has semi-axes $a$ and $b$, the volume is $\frac{1}{3} \pi abd$.

If the cone has vertex half-angle $\phi$, the cutting plane crosses the cone axis at distance $h$ from the vertex, and the cutting plane makes angle $\theta$ with the flat base, with $0 \leq \theta < \frac{\pi}{2} - \phi$, then the volume of the truncated cone is $$ \frac{\pi}{3}\, \frac{h^{3} \cot\phi}{(\cot^{2} \phi - \tan^{2} \theta)^{3/2}}. $$ (If $\theta \geq \frac{\pi}{2} - \phi$, the volume is infinite.)

For brevity, let $k = \cot\phi$ denote the slope of the cone generator and $m = \tan\theta$ denote the slope of the cutting plane.

In Cartesian coordinates with the vertex at the origin and the cone opening around the positive $z$-axis, the cone and cutting plane have respective equations \begin{align*} z^{2} &= k^{2}(x^{2} + y^{2}), \tag{1a} \\ z &= mx + h. \tag{1b} \end{align*} The plane and cone meet where $k^{2}(x^{2} + y^{2}) = z^{2} = (mx + h)^{2}$, or $$ (k^{2} - m^{2}) x^{2} - 2mhx - h^{2} + k^{2} y^{2} = 0. \tag{2} $$ (This is the equation of the elliptical "shadow" of the base, which does not directly give the shape of the base.)

In the longitudinal plane $y = 0$ (shown), the cone and cutting plane meet when $$ (k^{2} - m^{2}) x^{2} - 2mhx - h^{2} = 0. $$ The quadratic formula gives the roots $$ x_{\pm} = \frac{h(m \pm k)}{k^{2} - m^{2}}. \tag{3} $$ The semi-major axis is the distance between the corresponding points on the cone, $$ a = \tfrac{1}{2} \sqrt{1 + m^{2}}(x_{+} - x_{-}) = \sqrt{1 + m^{2}}\, \frac{hk}{k^{2} - m^{2}} = \sec\theta\, \frac{hk}{k^{2} - m^{2}}. \tag{4a} $$ The semi-minor axis of the slant base is the semi-minor axis of the ellipse (2), namely the positive value of $y$ in equation (2) when $$ x = \frac{x_{-} + x_{+}}{2} = \frac{hm}{k^{2} - m^{2}}. $$ For this $x$, we have $mx + h = \dfrac{hk^{2}}{k^{2} - m^{2}}$, so \begin{align*} y^{2} &= \frac{1}{k^{2}} \bigl[(mx + h)^{2} - k^{2} x^{2}\bigr] \\ &= \frac{1}{k^{2}} \left[\frac{h^{2} k^{4}}{(k^{2} - m^{2})^{2}} - \frac{k^{2} h^{2} m^{2}}{(k^{2} - m^{2})^{2}}\right] \\ &= \frac{h^{2}}{k^{2} - m^{2}}. \end{align*} The semi-minor axis is therefore $$ b = \frac{h}{\sqrt{k^{2} - m^{2}}}. \tag{4b} $$ Trigonometry shows the distance from the vertex to the cutting plane is $d = h\cos\theta$. Combining with equations (4a) and (4b), the volume of the slant cone is $$ \frac{\pi}{3} abd = \frac{\pi}{3} \left(\sec\theta\, \frac{hk}{k^{2} - m^{2}}\right) \frac{h}{\sqrt{k^{2} - m^{2}}}\, (h\cos\theta) = \frac{\pi}{3}\, \frac{h^{3} k}{(k^{2} - m^{2})^{3/2}}, $$ as claimed.

Answer 2

Here is an answer using a double integral. I use the same set up and notation as in Andrew D. Hwang's answer, but in cylindrical coordinates.

The equation of the cone is $z = kr$; of the plane, $z = m r \cos\theta + h$; and therefore of the elliptical shadow, $r = h/(k - m\cos\theta)$. Then the volume is\begin{align} & \int_0^{2\pi} \int_0^{h/(k - m\cos\theta)} (z_\text{plane} - z_\text{cone})r\,dr\,d\theta\\ & = \int_0^{2\pi} \int_0^{h/(k - m\cos\theta)} (m r \cos\theta + h - kr)r\,dr\,d\theta\\ & = \frac{h^3}6 \int_0^{2\pi} \frac{d\theta}{(k - m \cos\theta)^2}\\ & = \frac{h^3}6 \frac{2 \pi k}{(k^2 - m^2)^{3/2}}\\ & = \frac{\pi h^3 k}{3(k^2 - m^2)^{3/2}}.\end{align} For the second integration, see How to evaluate $\int_0^{2\pi} \frac{d\theta}{(A+B\cos\theta)^2}$?