# Proving the volume of a cone

Abstract definition of a cone: locus of points of line segments join a single point $v$, called vertex with set of coplanar points called base.

Height of cone is distance from vertex to plane.

If I take strange set of co-planer point such as complicated design for example, as base then volume of the set join with some vertex is still given : $\tfrac13$ (base)(height).

How can I prove this in calculus?

I have tried a few it is too late to wake my sister, thank you.

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Deno te $A$ the base area and $H$ the height of the cone. now we establish the coordinate system by setting the origin as the vertex of the cone, and the X-axis the line starting from the origin and pependicular to the base, i.e., the X-axis is along the direction of the height. Then take variable x as the distance from any point on X-axis on the interval [0,H]. Denote $A_t$ the area of the transverse section which is defined as the intersection of the plane $x=t$ and the cone. Then the volume of the cone shall be $$V = \int_0^H A_t dt$$ Note that we have $$\frac{A_t}{A} = \frac{t^2}{H^2}$$ Thus we have $A_t = \frac{A}{H^2} t^2$. Combining these two formula together we get $$V = \frac{A}{H^2} \int_0^H t^2 dt = \frac{1}{3} AH.$$

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I must find out more about At/A = t^2/H^2... it seems like it could be true, but what is it called? how is that relation shown to be true? – Amy Pry May 10 '11 at 5:17
Also thank you so much. – Amy Pry May 10 '11 at 5:17
@Amy Pry: If two plane figures $A$ and $B$ are similar, and $A$ is obtained from $B$ by multiplying all (linear) dimensions of $B$ by the scaling factor $s$, then the area of $A$ is $s^2$ times the area of $B$. (In your situation, any slice parallel to the base is the base scaled by $t/h$, where $h$ is the height of the cone and $t$ is the perpendicular distance of the slice from the apex.) – André Nicolas May 10 '11 at 5:46

You can assume the cone to be constructed of disks of infinitesimal thickness stacked one on top of the other, with the largest disk having radius $R$ at height $h=H$ and the smallest having radius $0$ at height $h=0$.

The radius of any disk in between, $r$ can be written as

$$r=R\frac{h}{H}$$

The volume of a single disk, $dV$ is

$$dV=\pi r^2 dh = \pi r^2 \frac{H}{R}dr$$

and the volume of the cone is then obtained by integrating (summing) the volume of each of the disks

$$V=\int dV=\pi\frac{H}{R}\int_0^R r^2 dr=\frac{1}{3}\pi R^2 H$$

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The cone need not be right circular. The idea however is correct. – user17762 May 10 '11 at 5:06
@Sivaram: Yes, you're right. My intention at first was to leave the general case to the OP as the question seemed like a HW. However, I forgot to mention that in my answer. – user4423 May 10 '11 at 5:13
Why can I assume it is discs... It may not be ... base come in any shape. – Amy Pry May 10 '11 at 5:13
@Amy: see my comment above. This was only to illustrate the approach. No matter what the base shape, the idea is the same: infinitesimal volume of a slice = area of slice $\times$ infinitesimal thickness. – user4423 May 10 '11 at 5:14
Thank you. I have done a lot of examples with the base as different shapes, square, rectangle, circle, etc. What I do not see is how to generalize this since in each A(x) is different. – Amy Pry May 10 '11 at 5:23

The derivation below is true for any cone (need not be right circular). Let the base area of the cone be $A$ and the height of the cone be $h$. Note that the area at any height $y$ is given by $A(y) = A \times \left( \frac{y}{h} \right)^2$ (Why?). Now the volume of the cone is nothing but $$V = \int_0^h A(y) dy \text{ (Why?)}$$ $$V = \int_0^h A \times \left( \frac{y}{h} \right)^2 dy = \frac1{3} A h$$

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The "why?" you are asking is just the part I do not understand! LOL. I have the rest. – Amy Pry May 10 '11 at 5:13
@Amy: Here is one explanation for the first "Why?". The area element is given by $r dr d \theta$. What can you say about your $r$ as a function of the height $y$ since you are joining the base to the vertex by straight lines? For the second "Why?", the reason is that the volume can be thought of as stacking up surfaces with area $A(y)$ of height $dy$ – user17762 May 10 '11 at 5:20
Is r a radius? The shapes I have in mind may have no radius. I am confused. – Amy Pry May 10 '11 at 5:21
@Amy: What is the area element in polar coordinate system? Just to help you sort out. How will you find the area of a square in polar coordinate system? – user17762 May 10 '11 at 5:22
Forgive me, but where is the square in the the polar coordinates? I would say to find the area of a square we must find the length of a side... but I do not think this is what you want. – Amy Pry May 10 '11 at 5:28

$$V = \int_{V}{\rm d}V = \int_{V}{\nabla\cdot\vec{r} \over 3}\,{\rm d}V = {1 \over 3}\int_{S}\vec{r}\cdot{\rm d}\vec{S}\,\qquad\mbox{( according Gauss )}$$

Whith the cone vertex as the origin, $\vec{r}\cdot{\rm d}\vec{S} \not= 0$ in the base. Then, we get

$$V = {1 \over 3}\int_{\mbox{base}}h\left\vert{\rm d}\vec{S}\right\vert = {1 \over 3}hA$$

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