Q 75. - Calculus

Question

Let $R$ be the radius of the base of a cone and $h$ be the height of the cone. Use spherical coordinates to set up and evaluate a triple integral proving that the volume of the cone is $\frac{1}{3}\pi R^{2}h$.

Step-by-Step Solution

Answer

This is done by finding the volume integral and substituting the limits of $\left(\rho ,\varphi ,\theta \right)$.

Step 1: Given Information

It is given that $R$ is radius of cone and $h$ is height of cone.

Step 2: Evaluation of limits

The limits of spherical coordinates are described as

$0\le \varphi \le {\tan }^{-1}\left(\frac{R}{b}\right),0\le \theta \le 2\pi ,0\le \rho \le h\sec \varphi$.

Step 3: Evaluating the Volume

Using spherical coordinates, the volume is evaluated as

$V={\int }_0^{{Tan}^{-1}\left(\frac{R}{h}\right)}{\int }_0^{2\pi }{\int }_0^{h\sec \varphi }{\rho }^2\sin \theta d\rho d\theta d\varphi$

$={\int }_0^{{Tan}^{-1}\left(\frac{R}{h}\right)}{\int }_0^{2\pi }\left({\int }_0^{h\sec \varphi }{\rho }^{2}d\rho \right)\sin \theta d\theta d\varphi$

$={\int }_0^{{Tan}^{-1}{\left(\frac{R}{h}\right)}^{2\pi }}{\int }_0{\left[\frac{{\rho }^3}{3}\right]}_0^{h\sec \varphi }\sin \phi d\theta d\varphi$

$={\int }_0^{{Tan}^{-1}{\left(\frac{R}{h}\right)}^{2\pi }}{\int }_0^{[}\left[\frac{(h\sec \varphi {)}^3}{3}-0\right]\sin \phi d\theta d\varphi$

$=\frac{h^3}{3}{\int }_0^{{Tan}^{-1}\left(\frac{R}{h}\right)}\left({\int }_0^{2\pi }d\theta \right)\frac{1}{{\cos }^3\varphi }\sin \varphi d\varphi$

$=\frac{h^3}{3}{\int }_0^{{Tan}^{-1}\left(\frac{R}{h}\right)}(\theta {)}_0^{2\pi }\frac{\sin \varphi }{{\cos }^3\varphi }d\varphi$

$=\frac{h^3}{3}{\int }_0^{{Tan}^{-1}\left(\frac{R}{n}\right)}\left(2\pi \right)\tan \varphi {\sec }^2\varphi d\varphi$

$=\frac{h^3}{3}\left(2\pi \right){\int }_0^{{Tan}^{-1}\left(\frac{R}{h}\right)}\tan \varphi {\sec }^2\varphi d\varphi$

$=\frac{h^3}{3}\left(2\pi \right){\left[\frac{(\tan \varphi {)}^2}{2}\right]}_0^{{Tan}^{-1}\left(\frac{R}{h}\right)}$

Using limits, we get

$V=\frac{h^3}{3}\left(2\pi \right)\left[\frac{\left.{\left({\mathrm{tantan}}^{-1}\left(\frac{R}{h}\right)\right)}^2\right]}{2}\right]$

$V=\frac{h^3}{3}\left(2\pi \right)\left[\frac{{\left(\frac{R}{h}\right)}^2}{2}\right]$

$V=\frac{1}{3}\pi R^{2}h$

Hence, proved.