I am trying to find the volume of a cone by integrating it in spherical coordinates, but elementary geometry suggests that my approach is incorrect.

The specifications of the cone are $0\le R \le 5$, $0 \le \theta \le \frac{\pi}{3}$, and $0 \le \phi \le 2\pi$. My reasoning was as follows:

The volume element in spherical coordinates is given by: $$dV = R^{2}\sin{\theta}dRd\theta d\phi$$

Simply integrate this over the specified region to obtain the total volume:

\begin{align} V & = \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{5}R^{2}\sin{\theta}dRd\theta d\phi = 2\pi\int_{0}^{\frac{\pi}{3}}\int_{0}^{5}R^{2}\sin{\theta}dRd\theta\\ & =2\pi\int_{0}^{\frac{\pi}{3}}[\frac{R^3}{3}]|_{0}^{5}\sin{\theta}d\theta\\ & =2\pi[\frac{125}{3}]\int_{0}^{\frac{\pi}{3}}\sin{\theta}d\theta = [\frac{125\pi}{3}] \end{align}

But when I frame the problem in terms of elementary geometry, with the slant height $R = 5$ and using the formula $V = \frac{1}{3}\pi r^2 h$ where $r$ is the radius of the base and $h$ is the height of the cone, I obtain a different answer.

The base radius $r$ and the cone height $h$ should be related to the slant height (which is $5$) by simple trigonometry:

$r = 5 \sin\theta = 5 \sin(\frac{\pi}{3}) = \frac{5\sqrt{3}}{2}$ $h = 5 \cos\theta = 5 \cos(\frac{\pi}{3}) = \frac{5}{2}$

Placing this into the elementary geometry formula $V = \frac{1}{3}\pi r^2 h$ one obtains:

$$V = \frac{1}{3}\pi(\frac{5\sqrt{3}}{2})^{2}(\frac{5}{2})= \frac{1}{3}\pi(\frac{25*3}{4})(\frac{5}{2}) = \frac{125\pi}{8}$$

Which one is correct, if either of them, and why?

  • $\begingroup$ If height = cos \theta then it isn't a cone with a straight slanted height side but a solid with circular height side. I think you have a sliced dome. About 1/3 of a sphere. $\endgroup$ – fleablood Jan 22 '16 at 22:49

Your problem is simple. The cone you described is an ice cream cone ie a cone with the a spherical cap since $R $ goes from 0 to 5, and it's volume cannot be given by $V=\frac{1}{3}\pi r^2 h $ (which is the volume of a cone without a cap), but by the integration you did.

Draw it and it will become more obvious.

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  • $\begingroup$ Thank you. I had realized this about 5 minutes after I posted this and forgot to remove the question. $\endgroup$ – Dudehahahacheckitout Jan 24 '16 at 18:47

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