<s>[INST] Volume of Cone By Integration 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?
 [/INST] 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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