Cutting a cone in half

Let $R = 5$ be the base radius, and $H = 10$ be the height.

Set up a coordinate system such that the origin is at the apex (bottom).

Equation of cone: $r^T Q r = 0$

with $Q = \text{diag}\{ \cos^2 \theta , \cos^2 \theta , -\sin^2 \theta \}$

, where $\tan \theta = \dfrac{R} { H} = \frac{1}{2}$

Now the cutting plane passes through $r_0 = (r_1, 0, H)$ and its unit normal is $(-\cos \theta, 0, \sin \theta)$ .

A matrix of unit vectors spanning the cutting plane is V , where,

$V = \begin{bmatrix} 0 && s \\ 1 && 0 \\ 0 && c \end{bmatrix}$

where $s = \sin \theta , c = cos \theta$

points on the cutting plane are given by $r = r_0 + V u$

Let's substitute this into the equation of the cone; this will result in the equation of the points common to both the cone and the cutting plane.

$(r_0 + V u)^T Q (r_0 + V u) = 0$

expanding, this becomes,

$r_0^T Q r_0 + 2 u^T V^T Q r_0 + u^T V^T Q V u = 0$

Evaluating the above expression, term by term, we obtain,

$V^T Q = \begin{bmatrix} 0 && c^2 && 0 \\ s c^2 && 0 && -c s^2 \end{bmatrix}$

so that,

$V^T Q r_0 = \begin{bmatrix} 0 \\ s c ( r_1 c - H s ) \end{bmatrix}$

and,

$V^T Q V = \begin{bmatrix} c^2 && 0 \\ 0 && 0 \end{bmatrix}$

and finally, $r_0^T Q r_0 = r_1^2 c^2 - s^2 H^2$

so now the equation of the cut reads,

$r_1^2 c^2 - s^2 H^2 + 2 u_2 s c (r_1 c - H s) + c^2 u_1^2 = 0$

divide by $c^2$ , then

$r_1^2 - R^2 + 2 u_2 s (r_1 - R) + u_1^2 = 0$

and this is a parabola in the $u_1$ - $u_2$ plane with axis of symmetry extending along the $u_2$ -axis

Next, we want to find the area of the cut, so we have to find the limits of u1

We note that at $z = H$ , $u_2 = 0$ , and subbing this in the equation of the parabola

yields $u_1 = \pm \sqrt{ R^2 - r_1^2 }$

Re-arranging the equation of the parabola, we get

$u_2 = \dfrac{ 1}{ 2 s (R - r_1) } ( u_1^2 - (R^2 - r_1^2) )$

The area of the cut is given by $A = \displaystyle -\int u_2 \hspace{6pt} d u_1$

Integrating is straightforward, and we get,

$\text{Area} = A(r_1) = \dfrac{2}{ 3 s ( R - r1) } (R^2 - r_1^2)^{(3/2)}$

Now to find the volume we have to integrate the area $A(r_1) dt$ , where $dt$ is the infinitesimal thickness normal to the plane of the cut

$dt = \cos \theta dr_1$

The volume cut out is ,

$V = \displaystyle \int_{r = -R}^{ r = r_1} A(r) \cos \theta \hspace{6pt} dr$

$= \displaystyle \frac{2}{3} \cot \theta \int_{-R}^{r_1} (R^2 - r^2)(3/2) / (R - r) \hspace{6pt} dr$

let $r = R \sin t$ , then the integral becomes,

$\displaystyle \int_{-\frac{\pi}{2}}^{\alpha} \dfrac {R^3 \cos^3 t}{ R (1 - \sin t)} \cdot R \cos t \hspace{6pt} dt$

where $\alpha = \sin^{-1} \frac{r_1}{R}$

Thus the volume is given by,

$V = \frac{2}{3} \cot \theta R^3 \displaystyle \int_{-\frac{\pi}{2}}^{\alpha} \dfrac{\cos^4 t}{ (1 - \sin t) } \hspace{6pt} dt$

but $\cos^4 t = (1 - \sin^2 t )^2 = (1 - \sin t ) (1 + \sin t ) (1 - \sin^2 t)$

hence, $\dfrac{ \cos^4 t}{ (1 - \sin t) } =(1 + \sin t ) (1 - \sin^2 t) = -\sin^3 t - \sin^2 t + \sin t + 1$

$= \frac{2}{3} \cot \theta R^3 \displaystyle \int_{-\frac{\pi}{2}}^{\alpha} - \sin^3 t - \sin^2 t + \sin t + 1 \hspace{6pt} d t$

But,

$\displaystyle \int \sin^3 t \hspace{6pt}dt = - \cos t + \frac{1}{3} \cos^3 t$

$\displaystyle \int \sin^2 t \hspace{6pt}dt = \frac{1}{2} t - \frac{1}{4} \sin 2 t$

Hence, the volume is given by,

$V = \frac{2}{3} \cot \theta R^3 ( \cos t - \frac{1}{3} \cos^3 t - \frac{1}{2} t + \frac{1}{4} \sin 2 t - \cos t + t )\displaystyle |_{- \pi/2}^{\alpha} = \frac{1}{2} \cdot \frac{1}{3} \pi R^2 H$

Noting that $\cot \theta R^3 = H R^2$ , we get,

$\frac{2}{3} ( -\frac{1}{3} \cos^3 t + \frac{1}{2} t + \frac{1}{4} \sin 2 t )\displaystyle |_{- \pi/2}^{\alpha} = \frac{ \pi}{6}$

Evaluating the expression between the two limits, we end up with,

$\frac{2}{3}(- \frac{1}{3} \cos^3 \alpha + \frac{1}{2} \alpha + \frac{1}{4} \sin 2 \alpha + \frac{\pi}{4} ) = \frac{\pi}{6}$

so that,

$- \frac{1}{3} \cos^3 \alpha + \frac{1}{2} \alpha + \frac{1}{4} \sin 2 \alpha = 0$

Solving this equation by numerical methods, yields, $\alpha = 0.299555$

Hence, $r_1 = R \sin \alpha = 0.29509544 R$

Therefore, the required length is $r_1 - (-R) = 1.29509544 R = 1.29509544(5) = 6.475$