Volume of Cube/Cone - Part 3

The volume of the cone inside the cube $(V)$ is equal to the volume of a truncated cone with a top circle of $\frac {\sqrt 2}2$ in diameter and the bottom circle base of $\sqrt 2$ in diameter $(V_t)$ subtracts four of the solid shaded green in the figure $(V_s)$ . That is $V=V_t - 4V_s$ .

To find $V_s$ we consider $A(z)$ is the cross-sectional area of the solid at $z$ . We note that the radius of the circular arc is given by $\dfrac r{2-z} = \dfrac {\frac {\sqrt 2}2}2 \implies r = \dfrac {2-z}{\sqrt 8}$ . And $A(z) = \dfrac {\theta r^2}2 - \dfrac {r^2 \sin \theta}2$ , where $\theta = 2\cos^{-1} \dfrac 1{2r}$ . Then we have:

$\begin{aligned} V_s & = \int_0^{2-\sqrt 2} \frac {r^2(\theta - \sin \theta)}2 dz & \small \blue{\text{Since }\sqrt 8 r = 2 - z \implies \sqrt 8 \ dr = - dz} \\ & = \sqrt 2 \int_\frac 12^\frac 1{\sqrt 2} r^2(\theta - \sin \theta)\ dr \\ & = \frac 1{\sqrt 2} \int_\frac 12^\frac 1{\sqrt 2} \left(4 r^2 \cos^{-1} \frac 1{2r} - \sqrt{4r^2 -1} \right) dr \\ & \approx 0.032337 \end{aligned}$

Therefore $V = V_t - 4V_s = \dfrac \pi 3 \left(2\times \left(\frac 1{\sqrt 2} \right)^2 - \left(\frac 1{2\sqrt 2} \right)^2 \right) - 4 \times 0.032337 \approx \boxed{0.787}$ .

In the first problem of this series , it was determined that the volume of the cone lying outside the cube is given by

$V_1 = \dfrac{9 \pi}{24} - \dfrac{4}{3} + \dfrac{\sqrt{2}}{3} \ln ( \sqrt{2} + 1 )$

It follows that the volume of the cone lying inside the cube is

$V = \dfrac{1}{3} \pi \left( \dfrac{1}{\sqrt{2}} \right)^2 (2) - V_1 = -\dfrac{\pi}{24} + \dfrac{4}{3} - \dfrac{\sqrt{2}}{3} \ln ( \sqrt{2} + 1 ) = \boxed{ 0.787 }$

$\text{figure 1}$ The volume that is inside the cube and inside the cone ( figure 1 ) can be calculated by the relation $V={{V}_{truncated\text{ }cone}}-4\cdot {{V}_{cone\text{ }segment}}$ By “cone segment” we mean the part of the cone that is outside the cube, bounded by the cord $AD$ , the hyperbolic arc and the arc of the cone’s circular base that corresponds to the cord $AD$ . We proceed in calculating the volume of this segment.

In the animated figure 2 , $M$ is the midpoint of $\overset\frown{AD}$ (assume $M\in {x}'x$ ) and $ZSW$ is the hyperbolic intersection of a plane through a point $V$ of $OM$ perpendicular to $OM$ . Denote $SV=h$ and $MV=t$ . The triangles $\triangle SMV$ , $\triangle KMO$ are similar right triangles, hence, $\dfrac{SV}{MV}=\dfrac{OK}{OM}\Rightarrow \dfrac{h}{t}=\dfrac{2}{\frac{\sqrt{2}}{2}}\Rightarrow h=2\sqrt{2}t$ $\text{figure 2}$

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If we project the hyperbolic segment on the $yz$ plane the vertex of the hyperbola is projected on point $R$ and its asymptotes are the lines $KP$ , $KQ$ ( figures 2 and 3 ).

$\text{figure 3}$

The area enclosed by a hyperbola with equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the line $l:$ $x=k$ is given by the formula $A=b\left[ k\sqrt{{{\left( \dfrac{k}{a} \right)}^{2}}-1}-a\cdot \text{arcosh}\left( \dfrac{k}{a} \right) \right]$ (for a proof see here )

On the $yz$ plane, consider a 2D coordinate system with the origine at $K$ and the line $KO$ as the x-axis (positive direction the direction of $\overrightarrow{KO}$ ). In this setting, line $l$ has equation $x=2$ .

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Moreover,
$a=KR=KO-h=2-2\sqrt{2}t$ ,
$b=RT=RS=OV=OM-MV=\dfrac{\sqrt{2}}{2}-t$

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Substituting we get $A=\left( \dfrac{\sqrt{2}}{2}-t \right)\left[ 2\sqrt{{{\left( \dfrac{2}{2-2\sqrt{2}t} \right)}^{2}}-1}-\left( 2-2\sqrt{2}t \right)\cdot \text{arcosh}\left( \dfrac{2}{2-2\sqrt{2}t} \right) \right]$ Now, the volume of the cone segment is ${{V}_{cone\text{ }segment}}=\int\limits_{0}^{\frac{\sqrt{2}-1}{2}}{A\cdot dt}=\int\limits_{0}^{\frac{\sqrt{2}-1}{2}}{\left( \dfrac{\sqrt{2}}{2}-t \right)\left[ 2\sqrt{{{\left( \dfrac{2}{2-2\sqrt{2}t} \right)}^{2}}-1}-\left( 2-2\sqrt{2}t \right)\cdot \text{arcosh}\left( \dfrac{2}{2-2\sqrt{2}t} \right) \right]\cdot dt}\approx 0.0323369$

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Next, we go for the volume of the truncated cone.
The big cone with base on plane $xy$ and height $2$ is similar to the small one with base $\left( J,JT \right)$ and height $1$ .
Since the ratio of the heights is $\dfrac{KJ}{KO}=\dfrac{1}{2}$ , the ratio of their volumes is $\dfrac{1}{8}$ .
Hence,
${{V}_{truncated\text{ }cone}}={{V}_{big\text{ }cone}}-{{V}_{small\text{ }cone}}=\frac{7}{8}\times {{V}_{big\text{ }cone}}=\frac{7}{8}\times \frac{1}{3}\times \pi \times {{\left( \frac{\sqrt{2}}{2} \right)}^{2}}\times 2=\frac{7\pi }{24}$ Finally, $\begin{aligned} V& ={{V}_{truncated\text{ }cone}}-4\cdot {{V}_{cone\text{ }segment}} \\ & =\frac{7\pi }{24}+4\times 0.0323369 \\ & \approx \boxed{0.787} \\ \end{aligned}$