What Fraction of The Volume was Removed?

Method 1:

Let the radius of the base be $R$ and the height be $h$ , then the semi-vertical angle $\theta_c$ is given by

$\theta_c = \tan^{-1} \dfrac{R}{h}$

If the origin is at the apex of the cone, then we can take the cutting plane to be $\vec{n} \cdot ( \vec{r} - \vec{r_0} ) = 0$ where the normal vector $\vec{n}$ is

orthogonal to the cutting plane which is parallel to ( \sin \theta c, 0, \cos \theta c ), hence we can take

$\vec{n} = (-\cos \theta_c, 0, \sin \theta_c )$

Now we'll slice the removed volume into slices parallel to the cutting plane, intersecting the base at x, and the volume will be

$V = \displaystyle \int_{0}^{t_{max}} A(t) dt$

the parameter $t$ represents the distance of the slice from the cutting plane

such that it is zero at $x = 0$ , and $t = t_{max}$ at $x = -R$ .

With the definition of the normal vector above, we can represent the slice in vector form as

$r = V u + r_0$ with

$u = (u_1, u_2)$ is the coordinate of $r$ with respect the two vectors in matrix $V$ which

are, respectively, (\sin \theta c, 0, \cos \theta c) and (0, 1, 0). And $r_0 = (x, 0, h )$

\( t is the distance of the slice plane from the cutting plane along the normal to the cutting plane,

hence, t = \vec{n} \cdot ( (x, 0, h) - (0, 0, h) ) = -( \cos \theta_c ) x

Next, we have to find the equation the parabola in the plane of the slice. For that we need the equation of

the cone, which is \( r^T Q r = 0 \), where $Q = \text{diag} \{ \cos^2 \theta_c , \cos^2 \theta_c, -sin^2 \theta_c \}$ .

Substituting $r = V u + r_0$ into the equation of the cone, results in

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

which when expanded becomes,

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

If use the notation $c = \cos \theta_c$ and $s = \sin \theta_c$ , then

$V^T Q V = \begin{bmatrix} s && 0 && c \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} c^2 && 0 && 0 \\ 0 && c^2 && 0 \\ 0 && 0 && -s&^2 \end{bmatrix} \begin{bmatrix} s && 0 \\ 0 && 1 \\ c && 0 \end{bmatrix}$

Evaluating the right-hand side, results in

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

In a similar fashion, we have

$V^T Q r_0 = \begin{bmatrix} sc(xc - sh) \\ 0 \end{bmatrix}$

and

${r_0}^T Q r_0 = c^2 x^2 - s^2 h^2$

Therefore, the equation of the parabola is

$c^2 {u_2}^2 + 2 u_1 s c (x c - s h) + c^2 x^2 - s^2 h^2 = 0$

Solving for the independent variable ( $u_1$ ) in terms of the dependent variable ( $u_2$ ), we have

$u_1 = - \dfrac{xc + sh}{2 sc} + \dfrac{c}{2 s(sh - xc)} {u_2}^2$

Note that $( sh - xc) = c ( h \tan \theta_c - x ) = c ( R - x )$ and similarly, (sh + xc) = c (x + R) )

Since $-R < x < 0$ , both terms are positive. The end points of the parabola corresponds to $u_1 = 0$ and are at

$u_2 = -a$ and $u_2 = a$ , where $a =\sqrt{ \dfrac{(s^2 h^2 - c^2 x^2) }{c^2}} = \sqrt{ R^2 - x^2 }$ .

Since the parabola is below the $u_2$ axis, the area is the negative of the integral between $-a$ and $a$ .

Performing the integral, results in,

$A = \dfrac{ x + R }{ s } \sqrt{R^2 - x^2} - \dfrac{1}{3} \dfrac{ (R^2 - x^2)^(3/2) }{ s(R - x) }$

Simplifying,

$A = \dfrac{2}{3} ( \dfrac{1}{s} ) (x + R) \sqrt{R^2 - x^2}$

Recall that the perpendicular distance of the slice from the cut is $t = (-\cos \theta_c) x$ , hence we can substitute for

$t$ in the volume integral, and integrate with respect to x instead.

$V = \displaystyle \int_{t = 0 }^{t = (\cos \theta_c) R} A(t) dt = \displaystyle \cos \theta_c \int_{x = -R}^{x = 0} A(x) dx$

The last integral is easy to perform, and is given by

$\dfrac{2c}{3s} ( \dfrac{-1}{3} R^3 + R^3 \dfrac{\pi}{4} ) = \dfrac{1}{3} \pi R^2 h ( \dfrac{1}{2} - \dfrac{2}{3 \pi})$

where in the last step we used the fact that $R = h \tan \theta_c$ .

The factor we're looking for is $( \dfrac{1}{2} - \dfrac{2}{3 \pi})$ which numerically evaluates to $0.28779341$ , making the answer $\boxed{2877}$ .

Method 2:

Here we will take the slices parallel to the base, one such slice is shown in the figure above with a red line segment. The slice take the shape of the red region with the circle on the bottom left.

The cut is at, $x = (\tan \theta_c) (h - z) = R (1 - z/h)$

The radius of circle of cut, $r = z \tan \theta_c = z R / h$

Area of cut, $A = 1/2 r^2 (2t - \sin 2 t)$

where $t = \cos^{-1} ( x / r )$

Therefore,

$A(z) = 1/2 (z R / h)^2 ( 2 \cos^{-1}( \dfrac{h - z}{z} ) - \sin ( 2 \cos^{-1}(\dfrac{h-z}{z} ) ) )$

The required volume, $V = \displaystyle \int_{z = h/2}^{h} A(z) dz$

To normalize the volume integral, we set $u = z / h$ , then

$V = R^2 h \displaystyle \int_{u = \frac{1}{2}}^{1} \dfrac{1}{2} u^2 ( 2 \cos^{-1}( \dfrac{1 - u}{u} ) - \sin ( 2 \cos^{-1}(\dfrac{1-u}{u} ) ) ) du$

Evaluating the integral numerically, it comes to, 0.3013765533760

Dividing by $\dfrac{1}{3} \pi$ gives the required ratio, $0.28779341$