Volume of Right Circular Cone

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The volume $V$ of a right circular cone is given by:

$V = \dfrac 1 3 \pi r^2 h$


$r$ is the radius of the base
$h$ is the height of the cone, that is, the distance between the apex and the center of the base.


Construct the following triangle:


Let $A$ be located at the origin of the $xy$-plane.

By definition of tangent, the line segment $\overline {AB}$ can be described by the equation $y = \dfrac {BC} {AC} x$.

By Euclid's definition of a cone, the solid of revolution generated by rotating $\triangle ABC$ about the $x$-axis is a right circular cone:

whose axis is $\overline {AC}$
whose base consists of the circle whose center is $C$, whose radius is $BC$ and whose plane is perpendicular to $\overline {AC}$.

As $\overline {AC}$ is perpendicular to the base of the cone, the height of the cone is $AC$.

Let $h = AC$ denote the height and $r = BC$ denote the radius of the base.


This proof utilizes the Method of Disks and thus is dependent on Volume of Cylinder.

From the Method of Disks, the volume of the cone can be found by the definite integral:

$\ds (1): \quad V = \pi \int_0^{AC} \paren {\map R x}^2 \rd x$

where $\map R x$ is the function describing the line which is to be rotated about the $x$-axis in order to create the required solid of revolution.

In this example, $\map R x$ describes the line segment $\overline {AB}$, and so:

$\map R x = \dfrac r h x$

We have also defined:

$\overline {AC}$ as the axis of the cone, whose length is $h$
$A$ as the origin.

So the equation $(1)$ is interpreted as:

\(\ds V\) \(=\) \(\ds \pi \int_0^h \paren {\frac r h x}^2 \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\pi \paren {\frac r h}^2 \frac {x^3} 3} {x \mathop = 0} {x \mathop = h}\) Linear Combination of Definite Integrals, Integral of Power
\(\ds \) \(=\) \(\ds \frac 1 3 \pi r^2 h\)