Why the Volume Formulas Work

Formulas with reasons attached

Six formulas usually arrive as a list to memorize.

$$C = 2\pi r \qquad A = \pi r^2 \qquad V = \pi r^2 h$$ $$V = \tfrac{1}{3}Bh \qquad V = \tfrac{1}{3}\pi r^2 h \qquad V = \tfrac{4}{3}\pi r^3$$

Every one of them has a reason, and the reasons are short. The one third appearing twice is not a coincidence, and the $\tfrac{4}{3}$ is not arbitrary.

Circumference: π is a ratio

All circles are similar — each is a scaled copy of every other. Similar figures have matching ratios, so the ratio of a circle’s circumference to its diameter is the same number for all of them.

$$\frac{C}{d} = \pi$$

That is the definition of $\pi$, and it makes the formula a rearrangement rather than a discovery.

$$C = \pi d = 2\pi r$$

A circle’s area, from its wedges

Cut a circle into many thin wedges and lay them alternately point-up and point-down. The result is nearly a rectangle.

Its height is one radius. Its base is half the circumference, because half the wedge arcs went along the top and half along the bottom.

$$\text{base} = \frac{C}{2} = \pi r \qquad \text{height} = r$$ $$A = \pi r \cdot r = \pi r^2$$

Thinner wedges make the shape straighter, and the formula is the value it approaches. This is worked in more detail in area of a circle.

A cylinder is a stack of discs

A prism’s volume is base area times height, because it is a stack of congruent copies of its base. A cylinder is the same stack with a circular base.

$$V = (\text{base area})(\text{height}) = \pi r^2 h$$

The pyramid, and where the third comes from

Take a cube with edge $s$. Pick one vertex and draw the three pyramids that have that vertex as their apex, each sitting on one of the three faces the vertex does not touch.

Those three pyramids fill the cube exactly, with nothing left over and no overlap. They are congruent, so each holds a third of the cube.

$$V_{\text{pyramid}} = \frac{s^3}{3} = \frac{1}{3}(s^2)(s) = \frac{1}{3}Bh$$

The $s^2$ is the base and the second $s$ is the height, so this is $\tfrac{1}{3}Bh$ written out.

The result holds for every pyramid, not only this one. A pyramid’s volume depends on its base area and height alone, so tilting the apex sideways or changing the base’s shape leaves $\tfrac{1}{3}Bh$ intact.

The cone is a pyramid with a round base

Give a pyramid a base with more and more sides — square, hexagon, and onward. The base approaches a circle, and the solid approaches a cone.

$$V = \frac{1}{3}Bh \quad \longrightarrow \quad V = \frac{1}{3}\pi r^2 h$$

Nothing about the $\tfrac{1}{3}$ depends on the number of sides, so it survives. A cone is exactly a third of the cylinder it fits inside.

Why Cavalieri’s principle settles the sphere

The sphere resists every argument above. It has no flat base to stack and no pyramid decomposition. It needs a different tool.

Cavalieri’s principle. If two solids sit between the same two parallel planes, and every plane parallel to those two cuts both solids in regions of equal area, then the two solids have the same volume.

The picture behind it is a stack of coins. Push the stack sideways into a lean and no coin changes size, so the stack still holds what it held.

Now compare two solids of height $r$:

• a hemisphere of radius $r$, flat side down
• a cylinder of radius $r$ and height $r$, with a cone of radius $r$ and height $r$ removed, the cone’s point resting on the center of the base

Slice both at height $y$.

The hemisphere’s slice is a circle. Its radius comes from the right triangle with legs $y$ and the slice radius, and hypotenuse $r$.

$$\text{slice radius} = \sqrt{r^2 - y^2} \qquad \text{slice area} = \pi\left(r^2 - y^2\right)$$

The other solid’s slice is a ring. The cylinder contributes a full circle of radius $r$, and the cone removes a circle. Because the cone has height $r$ and top radius $r$, its radius at height $y$ is exactly $y$.

$$\text{ring area} = \pi r^2 - \pi y^2 = \pi\left(r^2 - y^2\right)$$

The two agree at every height. By Cavalieri, the solids have equal volume, and the second one is already computable.

$$V = \pi r^2 \cdot r - \frac{1}{3}\pi r^2 \cdot r = \pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3$$

That is the hemisphere. Doubling it gives the sphere.

$$V_{\text{sphere}} = \frac{4}{3}\pi r^3$$

The formulas together

Solid	Volume	Where it comes from
Cylinder	$\pi r^2 h$	a stack of congruent discs
Prism	$Bh$	a stack of congruent bases
Pyramid	$\tfrac{1}{3}Bh$	three fill a cube
Cone	$\tfrac{1}{3}\pi r^2 h$	a pyramid with a round base
Sphere	$\tfrac{4}{3}\pi r^3$	Cavalieri, against a cylinder minus a cone

Read the third column and the list stops being six facts. It is two facts — a stack and a third of a stack — plus one comparison.

Worked examples

Common mistakes

Practice problems