# Volume of a Cone

The volume of a cone is \( \dfrac { 1 } { 3 } \pi r ^{ 2 } h \), where \(r\) denotes the radius of the base of the cone, and \(h\) denotes the height of the cone.

## Proof

The proof of this formula can be proven by volume of revolution.

Let us consider a right circular cone of radius \(r\) and height \(h\).

The equation of the slant height is \(y=\dfrac{r}{h}x\).Then the volume of the cone is

\[\begin{align} S&=\int_0^h \pi y^2dx\\ &=\pi\int_0^h \left(\dfrac{r}{h}x\right)^2dx\\ &=\dfrac{\pi r^2}{h^2}\int_0^h x^2dx\\ &=\left. \dfrac{\pi r^2}{h^2}\cdot\dfrac{x^3}{3}\right|^h_0\\ &=\dfrac{\pi r^2}{h^2}\cdot\dfrac{h^3}{3}\\ &=\dfrac{1}{3}\pi r^2h.\ _\square \end{align}\]

We can also use the disk method to proof this formula:

We draw a representative disk that has radius \(R\) and width \(\Delta y\). The height of our representative disk is \(y\). The volume of the representative disk is \(V_\text{disk}=\pi R^2\Delta y\).

We need to have \(R\) in terms of \(y\), so we must find the relationship between \(R\) and \(y\), that is, find \(R(y)\).

As the drawing suggests, \(R\) is a linear function of \(y\), so \(R(y)=my+b\).

We know that \(R(0)=r\) and \(R(h)=0\). Thus, \(m=\dfrac{\Delta R}{\Delta y}=\dfrac{r-0}{0-h}=-\dfrac{r}{h}\).

Then the function is \(R(y)=-\dfrac{r}{h}y+r\).Therefore, the total volume is \[\begin{align} V&=\pi\int_0^hR^2(y)dy\\ &=\pi\int_0^h\left(-\dfrac{r}{h}y+r\right)^2dy\\ &=\pi\int_0^h\left(\dfrac{r^2}{h^2}y^2-\dfrac{2r^2}{h}y+r^2\right)dy\\ &=\left. \pi\left(\dfrac{r^2}{3h^2}y^3-\dfrac{r^2}{h}y^2+r^2y\right)\right|^h_0\\ &=\dfrac{1}{3}\pi r^2h.\ _\square \end{align}\]

We can generalize the notion of a cone so that any simple closed curve, circular or not, can be the base of a cone. Thus pyramids are cones as well. We can further liberalize the definition to allow that the tip of a cone needn't be directly above the center of its base; that is, a cone can be

obliqueinstead ofright.Cones of the same height whose bases have the same area also have the same volume, because their cross-sectional slices have the same area at every height (where height means distance from the plane of the base; this is an application of Cavalieri's principle). Thus we can derive a formula for the volume of a cone of any shaped base if we can do so for some one shaped base. And it's easy to do that in the case of a square.

Six pyramids of height \(h\) whose bases are squares of length \(2h\) can be assembled into a cube of side \(2h\). Thus the volume of each is \(\frac{8h^3}{6}\) or \(\frac{1}{3}Ah\), where \(A\) is the area of its base. Scaling any object in a single dimension affects its volume linearly, so a square-based pyramid of any height \(H\) with base area \(A\) has volume \(\frac{1}{3}Ah\) times the scaling factor \(\frac Hh\), yielding \(\frac{1}{3}AH\).

It follows that a cone of any shape has as its volume one third times the area of its base times its height. \(_\square\)

## Examples

A right cone has a volume of \(200\pi\) and a height of \(6\). What is the radius of its base?

Let \(r\) denote the radius of the base. Then

\[\dfrac{1}{3}\pi r^2\times6=200\pi\implies r^2=100\implies r=10.\ _\square\]

Find the volume of a cone having slant height \(25\) and radius of the base \(24\).

Let \(h,l\) denote the height and slant height of the cone respectively, then

\[\begin{align} l&=\sqrt {h^2 + r^2}\\ 25^2&= \sqrt {h^2 + 24^2}\\ h^2&=49\\ h& = 7. \end{align}\]

So the volume of the cone is

\[ \frac {1}{3}\pi\times 24^2\times7= 1344\pi.\ _\square\]

You must answer a trivia question before the sand in the timer falls to the bottom. The sand falls at a rate of 50 cubic millimeters per second. How much time do you have to answer the question?

Use the formula for the volume of the cone to find the volume of the sand in the timer:

\[V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi\cdot10^2\cdot24=800\pi.\]

The volume of the sand is \(800\pi\) cubic millimeters. To find the amount of time you have to answer the question, multiply the volume by the rate at which the sand falls:

\[800\pi\times\dfrac{1}{50}=16\pi\approx 50.265\text{ (seconds)}.\ _\square\]

**Cone**is made from a circular sheet by cutting out a sector and gluing the cut edges of the remaining sheet together, as shown in the above figure. \(\theta \) is the angle of the sector cut out for which the volume of the cone is maximized.

If the value of \(\theta\) can be written as \( a\left(1-\frac { \sqrt { b } }{ \sqrt { c } } \right)\pi\) , where \(a,b,\) and \(c\) are positive integers and \(b\) and \(c\) are not divisible by the square of any prime, determine the value of \(a+b+c\).