Volume of Pyramids, Cones and Spheres

Points, cones and perfect balls

Ice-cream cones, party hats, the Great Pyramid of Giza, a football, a planet — the pointed and round solids look tricky, but they hide two of the most beautiful rules in all of geometry.

The first is almost unbelievable: a pyramid or a cone holds exactly one third of the box or can that surrounds it. Fill a cone with sand and tip it into the cylinder it fits snugly inside — you'll need to do it three times to fill the cylinder. The second is the elegant formula for a sphere, \tfrac{4}{3}\pi r^3, which lets you weigh a planet or blow up a balloon on paper. Two little formulas, and the whole rounded world opens up.

Pyramids and cones: one third

A pyramid rises from a flat base to a single point. Fill it and pour it into the prism with the same base and height, and it fills exactly one third — so its volume is one third of that prism's:

V = \tfrac{1}{3} \times \text{base area} \times \text{height}

A cone is just a pyramid whose base is a circle. The base area of a circle is \pi r^2, so the same "one third" rule gives:

V = \tfrac{1}{3}\pi r^2 h

Each is exactly a third of the prism or cylinder built on the same base and height. Whatever the base — square, triangle, circle, blob — if it narrows steadily to a point, its volume is a third of the "boxed" solid.

The sphere

A sphere is the perfectly round ball — every point on its surface the same distance r from the centre. It has its own volume formula, and a separate one for its surface area:

V = \tfrac{4}{3}\pi r^3 \qquad A = 4\pi r^2

Notice the volume uses r^3 (three r's multiplied — it's a 3D thing) while the surface area uses r^2 (a 2D skin). Doubling the radius makes the surface 4 times bigger but the volume 8 times bigger.

pyramid V = \tfrac{1}{3} \times \text{base area} \times h;
cone V = \tfrac{1}{3}\pi r^2 h;
sphere volume V = \tfrac{4}{3}\pi r^3;
sphere surface area A = 4\pi r^2;
the constant \pi \approx 3.142.

Worked example 1 — a cone is a third of its cylinder

A cone has radius 3\text{ cm} and height 10\text{ cm}. Find its volume, and compare it with the cylinder that boxes it.

The cylinder with the same base and height:

V_{\text{cyl}} = \pi r^2 h = \pi \times 3^2 \times 10 = 90\pi \approx 283\text{ cm}^3

The cone — one third of that:

V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times 90\pi = 30\pi \approx 94\text{ cm}^3

And 94 \times 3 = 282 \approx 283 — three cones fill the cylinder, exactly as promised. Leaving the answer as 30\pi\text{ cm}^3 is exact; 94\text{ cm}^3 is the rounded decimal.

Worked example 2 — a sphere

A ball has radius 6\text{ cm}. What is its volume?

V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi \times 6^3 = \tfrac{4}{3}\pi \times 216 = 288\pi\text{ cm}^3

The tidy step is \tfrac{4}{3} \times 216 = 288, so the exact volume is 288\pi\text{ cm}^3. As a decimal, using \pi \approx 3.142:

V \approx 288 \times 3.142 \approx 905\text{ cm}^3

Work out r^3 first, then the fraction, and save \pi for last — the numbers stay friendly.

Worked example 3 — a paper cup of water

A conical paper cup has radius 3\text{ cm} and height 8\text{ cm}. How many millilitres of water does it hold? (Recall 1\text{ ml} = 1\text{ cm}^3.)

V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi \times 3^2 \times 8 = \tfrac{1}{3}\pi \times 72 = 24\pi \approx 75\text{ cm}^3

So the cup holds about 75\text{ ml} — a decent gulp of water. The same method sizes an ice-cream cone, a funnel, or a witch's hat: it's always \tfrac{1}{3}\pi r^2 h.

See the three solids

Step through the figure: a cone, then a square-based pyramid, then a sphere. Each is labelled with its radius r (and the cone and pyramid with their height h).

See it in 3-D

A pyramid's volume is \tfrac{1}{3} \times \text{base area} \times \text{perpendicular height}. The dashed line is that perpendicular height — dropped straight down from the apex to the centre of the base, not along the slanted edge. Drag to rotate the solid and see the height standing up inside it.

Two mistakes cost more marks than any others here:

Never forget the \tfrac{1}{3}. A cone is not the same volume as its cylinder — it's a third. Writing \pi r^2 h for a cone (the cylinder's formula) triples the answer. The same trap waits for pyramids: it's \tfrac{1}{3} \times \text{base} \times h, not just base × height.
Use the perpendicular height, not the slant. For volume, h is the straight-down distance from the apex to the centre of the base — like a plumb line. The slant height (up the sloping side) is longer, and it's only used for surface area. Mixing them up quietly inflates every volume.

Over 2200 years ago — long before calculus existed — Archimedes proved that a sphere is exactly two thirds of the cylinder that just fits around it, and that a cone, a sphere and that cylinder (all sharing the same radius and height) have volumes in the neat ratio 1 : 2 : 3.

He was so proud of this that he asked for a sphere-inside-a-cylinder to be carved on his tombstone. He found these "curved" volumes with a dazzling slicing argument — imagining the solids sawn into countless thin discs and balancing them like a see-saw — that anticipated integration by almost two thousand years. A Roman writer later found his overgrown grave by looking for exactly that carving.