Q 73. - Calculus

Question

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

Step-by-Step Solution

Answer

This can be proved by taking an equation of cone that gives it a point at the origin that opens upward and downward.

Step 1: Given Information

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

Step 2: Evaluate the limits

Let us assume equation $a^2{z}^2=h^2{x}^2+h^2{y}^2$ and it gives it a point at the origin that opens upward and downward and $h=z$. It gives $a$ as radius of circle.

$\le \theta \le 2\pi ,0\le r\le R,\frac{h}{R}\le z\le h$ is the region bounded by curves.

Step 3: Calculating Volume of cone

The required volume is given by

$V={\int }_0^{2\pi }{\int }_0^R{\int }_{\frac{hr}{R}}^{h}rdzdrd\theta$

$V={\int }_0^{2\pi }{\int }_0^R(z{)}_{\frac{hr}{R}}^{h}rdrd\theta$

$V={\int }_0^{2\pi }{\int }_0^R\left(h-\frac{hr}{R}\right)rdrd\theta$

$V={\int }_0^{2\pi }\left(h{\int }_0^{R}rdr-\frac{h}{R}{\int }_0^R{r}^{2}dr\right)d\theta$

Simplifying further

$V={\int }_0^{2\pi }\left(\frac{h{R}^2}{2}-\frac{h}{R}\left(\frac{R^3}{3}\right)\right)d\theta$

$V={\int }_0^{2\pi }\left(\frac{h{R}^2}{2}-\frac{h{R}^2}{3}\right)d\theta$

$V={\int }_0^{2\pi }\left(\frac{h{R}^2}{6}\right)d\theta$

$=\frac{h{R}^2}{6}(\theta {)}_0^{2\pi }$

$V=\frac{h{R}^2}{6}\left(2\pi \right)$

Hence, required volume is $V=\frac{\pi R^{2}h}{3}$