The volume of a cone

Integration reveals why the volume of a cone is one third of the base area times the height¹. We revolve the part of the line $y = \dfrac{x}{2}$ with $0 \leq x \leq 6$ about the $x$ -axis.

Adding up the sections

The resulting solid is a cone of base radius $3$ and height $6$ . The section cut at $x$ is a circle of radius $\dfrac{x}{2}$ , of area $\pi \dfrac{x^2}{4}$ .

V = \pi \int_0^6 \frac{x^2}{4}\,dx = \frac{\pi}{4} \left[ \frac{x^3}{3} \right]_0^6 = \frac{\pi}{4} \times 72 = 18\pi

Checking against the formula, $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \times 9 \times 6 = 18\pi$ , in agreement.

Where the one third comes from

The cross-sectional area grows in proportion to $x^2$ , so integrating gives $\dfrac{x^3}{3}$ , and the division by $3$ is what remains.

Solid	Cross-sectional area	Volume
Cylinder	constant	$\pi r^2 h$
Cone	proportional to $x^2$	$\dfrac{1}{3}\pi r^2 h$

The general case

A cone of radius $r$ and height $h$ is the solid obtained by revolving $y = \dfrac{r}{h}x$ from $0$ to $h$ .

V = \pi \int_0^h \frac{r^2}{h^2}x^2\,dx = \frac{\pi r^2}{h^2} \cdot \frac{h^3}{3} = \frac{1}{3}\pi r^2 h

The formula itself falls out.

The same for a pyramid

Even when the section is a square rather than a circle, the area of similar figures is proportional to the square of a length, so the cross-sectional area is again of the form $x^2$ . Integrating produces the same one third. That every cone and pyramid carries a factor of $\dfrac{1}{3}$ is due to that square.

The same method for a sphere

Revolve $y = \sqrt{r^2 - x^2}$ from $-r$ to $r$ .

V = \pi \int_{-r}^{r} \left( r^2 - x^2 \right) dx = \frac{4}{3}\pi r^3

The two lines spreading from the origin on the graph are $y = \pm\dfrac{x}{2}$ , the outline before revolving. The large dots are the rim of the base.

Cone, Wikipedia