How To Calculate Volume Of A Right Circular Cone's Hyperbola Segment?

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Learn more about Teams How to calculate volume of a right circular cone's hyperbola segment? Ask Question Asked 8 years, 9 months ago Modified 7 years, 11 months ago Viewed 5k times 6 $\begingroup$

PROBLEM I am working on calculating volumes of geometric solids. All shapes have been pretty basic until now. I am bewildered on how to attack the problem of calculating the volume of a slice of a right circle cone. Visualizing: The cone sits on a circular base with the apex directly above. The axis of symmetry passes through the apex and the center of the circle oriented normal to the circle. Now, when you pass a cutting plane parallel with the cone's axis of symmetry at a distance between the circle's center and the circle's perimeter, a segment is 'cleaved' off. I am looking to determine the volume of that segment. Note: All cone dimensions are known.

RESEARCH I have looked at numerous math and analytical geometry websites. The most information I can seem to find just about the equations of lines or basic formulas (volume, surface area, etc.) I have even dug up my ancient Calculus textbook (yes, an actual paper book) yet still cannot find an approach.

POSSIBLE SOLUTION I suspect that the solution will use DOUBLE INTEGRALS in some fashion but I am not sure. Frankly, even if it does use double integrals, I have been away from this for so long that I will need an example for how to work out the problem.

I want to apply calculus to more of my work but do not fully have the confidence to do so.

Thank you for any assistance or direction to a solution.

• calculus
• analytic-geometry

Share Cite Follow edited Feb 18, 2016 at 13:22 user142971 asked Feb 17, 2016 at 16:16 RikiTandoRikiTando 6111 silver badge22 bronze badges $\endgroup$ Add a comment |

3 Answers 3

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 4 $\begingroup$

I'll assume the following:

$$\text{Height of cone } = h \\ \text{Radius of the cone at its base } = R \\ \text{Distance from the cone axis to the slicing plane } = d,\quad 0\leq d\leq R.$$

Each horizontal slice of the solid is a circular segment. These segments vary in size due to the change of radius as we move vertically up or down the cone. Let's suppose at any such height $z$ the circle has radius $r$ and the segment subtends an angle $\theta$ at the circle centre (see the picture at the link below). Then, simple trigonometry/geometry shows that

$$\cos\frac{\theta}{2} = \frac{d}{r}, \\ \text{and the thickness of these slices will be: } dz = \dfrac{h}{R}dr.$$

(Actually, $r$ decreases as $z$ increases but this is handled by flipping the limits of integration below on $r$: $d$ to $R$ instead of $R$ to $d$.)

The area for the segment is:

\begin{align} A_s &= \dfrac{r^2}{2}\left(\theta - \sin\theta\right) \\ &= \dfrac{r^2}{2}\left(2\cos^{-1}\dfrac{d}{r} - \sin\left(2\cos^{-1}\dfrac{d}{r}\right)\right) \\ &= r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2} \\ & \qquad\text{using $\;\sin(2\alpha)=2\sin\alpha\cos\alpha\;$ and $\;\cos(\sin^{-1}(a/b))=\sqrt{b^2-a^2}/b$}. \end{align}

Therefore the volume of a thin horizontal slice of thickness $dz\;$ is

\begin{align} dV &= A_sdz \\ &= A_s\dfrac{h}{R}dr \\ &= \dfrac{h}{R}\left(r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2}\right)dr. \end{align}

Therefore the required volume is:

\begin{align} V &= \int_{r=d}^{R} \dfrac{h}{R}\left(r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2}\right)\;dr \\ &= \dfrac{h}{3R}\left[-2dr\sqrt{r^2-d^2} + d^3\ln\left(\sqrt{r^2-d^2}+r\right) + r^3\cos^{-1}\dfrac{d}{r} \right]_{r=d}^{R} \\ & \qquad\text{using Wolfram Alpha} \\ &= \dfrac{h}{3}\left[-2d\sqrt{R^2-d^2} + \dfrac{d^3}{R}\ln\left(\sqrt{R^2-d^2}+R\right) + R^2\cos^{-1}\dfrac{d}{R} - \dfrac{d^3}{R}\ln{d} \right]. \\ \end{align}

As a sanity check, setting $d=0$ gives $V=\pi R^2h/6,$ and setting $d=R$ gives $V=0,$ as expected.

Share Cite Follow edited Jun 17, 2016 at 8:31 answered Feb 18, 2016 at 12:59 Mick AMick A 10.3k22 gold badges1717 silver badges3434 bronze badges $\endgroup$ 5

• $\begingroup$ Thank you my friend. I am re-inspired by your excellent solution and explanation. Thank you again. @user36790 $\endgroup$ – RikiTando Commented Feb 18, 2016 at 22:59
• $\begingroup$ @RikiTando Glad to help. $\endgroup$ – Mick A Commented Feb 18, 2016 at 23:01
• $\begingroup$ @Blue Also, thank you people for correcting the title of my questions. I had not thought of it as yelling as in my field all text is capitalized. Thank you. $\endgroup$ – RikiTando Commented Feb 18, 2016 at 23:01
• 1 $\begingroup$ @ Mick A .. @jadsq seems to suggest following, but I have not checked it myself. $$ = \dfrac{h}{3} \left(-2d\sqrt{R^2-d^2} + \dfrac{d^3}{R} \ln \dfrac{\sqrt {R^2-d^2 }+R }{d} + R^2\cos^{-1}\dfrac{d}{R} \right) $$ $\endgroup$ – Narasimham Commented Jun 16, 2016 at 20:27
• $\begingroup$ @Narasimham Thanks! I've made the change. jadsq was right. $\endgroup$ – Mick A Commented Jun 17, 2016 at 8:34

Add a comment | 2 $\begingroup$

I'm sorry I don't have enougth reputation to comment on Mick A's answer but there is a small mistake in the last line of his answer (assuming the previous one is correct) :

$$ V=\dfrac{h}{3}\left[-2d\sqrt{R^2-d^2} + \mathbf{\dfrac{d^3}{R}}\ln\left(\sqrt{R^2-d^2}+R\right) + R^2\cos^{-1}\dfrac{d}{R} - \dfrac{d^3}{R}\ln{d} \right]. $$

Edit : to satisfy Irregular User , I'll add that the formula above is the correct answer to the question.

Share Cite Follow edited Jun 16, 2016 at 21:13 answered Jun 16, 2016 at 19:31 jadsqjadsq 12144 bronze badges $\endgroup$ 2

• $\begingroup$ technically the above commentators are right, it's not a full answer but still sort of gives an answer. As a compromise, could you just edit your answer in the way that you add the derivation of your result perhaps starting from the actual volume integral? $\endgroup$ – user190080 Commented Jun 16, 2016 at 21:34
• $\begingroup$ Thanks for the edit jadsq - will give you a +1 in 53 minutes when I can vote again :) $\endgroup$ – Irregular User Commented Jun 16, 2016 at 23:06

Add a comment | 0 $\begingroup$

i found a detailed derivation of generic formula to calculate volume of slice of cone of radius of base $r$ & normal height $h$ cut by a plane at distance $x$ from the symmetry axis.

$$volume \ of\ slice=\frac{h}{3r}\left(r^3\cos^{-1}\left(\frac xr\right)-rx\sqrt{R^2-x^2}+x^3\ln\left(\frac{r+\sqrt{r^2-x^2}}{x}\right)\right)$$

Share Cite Follow edited Dec 30, 2016 at 18:56 answered Dec 28, 2016 at 4:00 Bhaskara-IIIBhaskara-III 1,63133 gold badges1717 silver badges3131 bronze badges $\endgroup$ Add a comment |

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