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?
