[Math] How to evaluate volume of oblique frustum of right circular cone with elliptical section

geometry

oblique frustum of right circular cone

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!

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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.

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[Math] Minimum Volume of a circular, right cone, with a sphere inscribed in it.

As user WW1 says, the book must have a typo - your work is correct. Here's another way we could have calculated it giving the same answer:

The radius of the incircle of a triangle is $\frac{\triangle}{s}$ where $\triangle$ is the area of the triangle and $s$ is the semi-perimeter. Let's call $R$ the radius of the cone and $h$ the height of the cone. If $r$ is the radius of the sphere this gives us

$$\begin{align}&r=\frac{Rh}{\sqrt{h^2+R^2}+R} \\\implies &r\sqrt{h^2+R^2}+rR=Rh \\\implies &r\sqrt{h^2+R^2}=R(h-r) \\\implies &r^2\left(h^2+R^2\right)=R^2(h^2-2hr+r^2) \\\implies &r^2h^2=R^2\left(h^2-2hr\right) \\\implies &\frac{r^2h^2}{h^2-2hr}=R^2 \\\implies &\frac{r^2h}{h-2r}=R^2 \end{align}$$

Now the volume of the cone is $$V=\dfrac{\pi R^2h}{3}=\frac{\pi r^2h^2}{3(h-2r)}$$

Hence $$\frac{dV}{dh}=\frac{\pi\left(r^2h^2-2r^2h(h-2r)\right)}{3(h-2r)^2}=\frac{\pi\left(4r^3h-r^2h^2\right)}{3(h-2r)^2}$$

Setting the derivative to $0$, we get $$4r^3h-r^2h^2=0\hspace{5mm}\implies\hspace{5mm}4r-h=0\hspace{5mm}\implies\hspace{5mm}h=4r$$

Plugging this into our equation for $R^2$ we get $$R^2=\frac{r^2(4r)}{4r-2r}=\frac{4r^3}{2r}=2r^2$$ Hence $R=\sqrt{2}r$ and $h=4r$ minimizes the volume as you found.

[Math] Evaluate the eccentricity of the elliptical section of a right circular cone

Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $\theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $\beta$ so that $\beta = \alpha/2$ (this will help simplify the calculations later).

Clearly in $\Delta AOC$ we have $$\frac{OA}{\sin \angle OCA} = \frac{AC}{\sin \angle AOC}$$ so that $$b = AC = \dfrac{\sin \theta}{\sin(\theta + \beta)}\tag{1}$$ Similarly $$c = AB = \frac{\sin \theta}{\sin(\theta - \beta)}\tag{2}$$ and $$a = BC = \frac{b\sin 2\beta}{\sin(\theta - \beta)} = \frac{\sin\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}\tag{3}$$ Let the incircle of $\Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $\Delta ABC$ then we know that $$\tan \frac{C}{2} = \frac{r}{CD}$$ so that $$CD = \frac{r}{\tan(C/2)} = \dfrac{r}{\tan\left(\dfrac{\pi - \theta - \beta}{2}\right)} = r\tan\left(\frac{\theta + \beta}{2}\right)\tag{4}$$ Note further that the inradius $r$ is given by the formula \begin{align} r &= \frac{\Delta}{s}\notag\\ &= \frac{bc\sin 2\beta}{a + b + c}\notag\\ &= \dfrac{\dfrac{\sin^{2}\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}}{\dfrac{\sin\theta\{\sin(\theta + \beta) + \sin(\theta - \beta) + \sin 2\beta\}}{\sin(\theta + \beta)\sin(\theta - \beta)}}\notag\\ &= \frac{\sin\theta\sin 2\beta}{2\sin\theta\cos\beta + 2\sin\beta\cos\beta}\notag\\ &= \frac{2\sin\theta\sin\beta\cos\beta}{2\cos\beta\{\sin\theta + \sin\beta\}}\notag\\ &= \frac{\sin\theta\sin\beta}{\sin\theta + \sin\beta}\tag{5} \end{align} If $e$ is the eccentricity then we know that $$e = \frac{\text{distance of focus from center}}{\text{length of semi major axis}} = \frac{(a/2) - CD}{a/2} = \frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as \begin{align} e &= \dfrac{a - 2r\tan\left(\dfrac{\theta + \beta}{2}\right)}{a}\notag\\ &= 1 - \frac{2r}{a}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \frac{2\sin\theta\sin\beta}{\sin\theta + \sin\beta}\cdot\frac{\sin^{2}\theta - \sin^{2}\beta}{2\sin\theta\sin\beta\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \frac{\sin\theta - \sin\beta}{\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \dfrac{2\cos\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \dfrac{2\sin\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\notag\\ &= 1 - \frac{\cos \beta - \cos \theta}{\cos \beta}\notag\\ &= \frac{\cos \theta}{\cos \beta} = \frac{\cos\theta}{\cos(\alpha/2)}\tag{6} \end{align} Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < \beta = \alpha / 2 < \theta \leq \pi/2$. If $\beta = \alpha/2 = \theta$ then the section becomes a parabola and if $\beta = \alpha/2 > \theta$ then the section becomes a hyperbola. For the angles given in question we have $e = \cos 70^{\circ}/\cos 30^{\circ} = 0.39493084\ldots$.

Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = \cos\theta/\cos\beta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.

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Related Solutions

[Math] Minimum Volume of a circular, right cone, with a sphere inscribed in it.

[Math] Evaluate the eccentricity of the elliptical section of a right circular cone

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