Volume of a Cone

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

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

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

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 $$\begin{aligned} V&=\pi\int_0^hR^2(y)dy\\ &=\pi\int_0^h\left(-\dfrac{r}{h}y+r\right)^2dy\\ &=\pi\int_0^h\left(\dfrac{r^2}{h^2}y^2-\dfrac{2r^2}{h}y+r^2\right)dy\\ &=\left. \pi\left(\dfrac{r^2}{3h^2}y^3-\dfrac{r^2}{h}y^2+r^2y\right)\right|^h_0\\ &=\dfrac{1}{3}\pi r^2h.\ _\square \end{aligned}$$

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 oblique instead of right.

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. $_\square$

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

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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 $24$.

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Let $h,l$ denote the height and slant height of the cone respectively, then

$$\begin{aligned} l&=\sqrt {h^2 + r^2}\\ 25^2&= \sqrt {h^2 + 24^2}\\ h^2&=49\\ h& = 7. \end{aligned}$$

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?

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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:

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