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Here is the question:

To find the volume of a right circular cone with base radius $r$ and height $h$, the cone is divided into $n$ frustums of equal heights. The volume of each frustum is approximated as if it were a circular cylinder, having the larger of the tow plane surfaces as the base of the cylinder. Show that the approximate volume of the cone is $V_n =(1/6)(2πrh)(1+1/n)(2+1/n)$. Hence, find the actual volume of the cone.

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I am stuck for a few hours and i really need ur help, thanks!! :)

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  • $\begingroup$ Can you tell what you've tried so far? $\endgroup$ – Naive Aug 2 '17 at 9:28
  • $\begingroup$ Write the volumes of the individual slices and sum them. If needed, use the Faulhaber fomulas. en.wikipedia.org/wiki/Faulhaber%27s_formula $\endgroup$ – Yves Daoust Aug 2 '17 at 9:59
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The start of solution in handwriting below the task is correct. Here is the complete solution:

The sum of volumes of the circular cylinders is: $$V_n=\pi \cdot \left(\frac{r}{n}\right)^2\cdot \frac{h}{n}+\pi \cdot \left(\frac{2r}{n}\right)^2\cdot \frac{h}{n}+\cdots+\pi \cdot \left(\frac{nr}{n}\right)^2\cdot \frac{h}{n}=$$ $$\frac{\pi r^2h}{n^3}\left(1+2^2+\cdots+n^2\right)=\frac{\pi r^2h}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{1}{6}\cdot \pi r^2h\cdot\left(1+\frac1n\right)\left(2+\frac1n\right).$$ When $n\to+\infty$, it is: $$V=\lim_\limits{n\to+\infty}V_n=\frac{1}{6}\cdot \pi r^2h\cdot1\cdot 2=\frac{1}{3}\pi r^2h=\frac{1}{3}S_{base}h.$$

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The diameters of the frustrums (frustra) are decreasing linearly, hence the volumes quadratically.

$$v_n=\frac Vn\left(\frac{n-k}n\right)^2,$$ where $\dfrac{V}{n}$ denotes the volume of the corresponding cylindrical slices.

Then the total volume

$$V'=\frac Vn\sum_{k=0}^{n-1}\left(\frac {n-k}n\right)^2=\frac Vn\sum_{k=1}^{n}\frac{k^2}{n^2}=\frac V{n^3}\frac{n(n+1)(2n+1)}6=V\frac{(n+1)(2n+1)}{6n^2}.$$

The ratio tends to $\dfrac13.$

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Hints:

  • Let's find the volume $V_{n,1}$ of the smallest cylinder:

Its height is $h/n$, it's radius is $r/n$ (by Thales), so its volume is $V_{n,1}=\pi*(r/n)^2*(h/n)=\frac{hr^2}{n^3}\pi$

  • Can you similarly calculate teh Volume of the $k$-th cylinder ?

  • Then you have to sum the volumes of the $n$ cylinders to find an approximation by excess of the volume of the cone,

  • and make $n\rightarrow\infty$ to find its exact value.

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