Cylinder, Cone, and Sphere

Cylinders, cones, and spheres are the classical solids of revolution — bodies with circular symmetry generated by rotating a planar figure about an axis. They appear everywhere in engineering, architecture, and nature, and their surface-area and volume formulas are among the most important in all of mathematics.

Cylinder

📐Definition — Cylinder

A right circular cylinder is the solid bounded by two congruent, parallel circular bases of radius $R$ and a lateral surface connecting them.

Equivalently, it is the solid of revolution obtained by rotating a rectangle of width $R$ and height $h$ about one of its sides of length $h$ .

The height (or altitude) of the cylinder is the perpendicular distance $h$ between the two bases. The axis is the line segment connecting the centers of the two bases.

Surface Area of a Cylinder

The lateral surface of a cylinder, when unrolled, becomes a rectangle of width $2\pi R$ (the circumference of the base) and height $h$ :

$\boxed{S_{\text{lat}} = 2\pi R h}$

Adding both circular bases gives the total surface area:

$\boxed{S_{\text{total}} = 2\pi R h + 2\pi R^2 = 2\pi R(R + h)}$

Volume of a Cylinder

Since every cross-section parallel to the bases is a circle of radius $R$ and area $\pi R^2$ , by Cavalieri’s Principle:

$\boxed{V = \pi R^2 h}$

Cone

📐Definition — Cone

A right circular cone is the solid with a circular base of radius $R$ , an apex directly above the center of the base at height $h$ , and a lateral surface connecting the apex to the boundary of the base.

Equivalently, it is the solid of revolution obtained by rotating a right triangle with legs $R$ and $h$ about the leg of length $h$ .

The slant height $l$ is the distance from the apex to any point on the boundary of the base:

$l = \sqrt{R^2 + h^2}$

Surface Area of a Cone

The lateral surface of a cone, when unrolled, becomes a circular sector with radius $l$ and arc length $2\pi R$ (the base circumference). Its area is:

$\boxed{S_{\text{lat}} = \pi R l}$

The total surface area includes the circular base:

$\boxed{S_{\text{total}} = \pi R l + \pi R^2 = \pi R(R + l)}$

Volume of a Cone

⚡Theorem — Volume of a Cone (Theorem 27.1)

The volume of a cone equals one-third the volume of a cylinder with the same base and height:

$\boxed{V = \dfrac{1}{3}\pi R^2 h}$

Justification. A cross-section of the cone at height $t$ from the base is a circle of radius $R\!\left(1 - \dfrac{t}{h}\right)$ , so its area is $\pi R^2\!\left(1 - \dfrac{t}{h}\right)^{\!2}$ . Summing these areas (integrating from $0$ to $h$ ) gives $\dfrac{1}{3}\pi R^2 h$ . $\blacktriangleleft$

Sphere

📐Definition — Sphere and Ball

A sphere of radius $R$ centered at point $O$ is the set of all points in space at distance exactly $R$ from $O$ .

The ball (solid sphere) of radius $R$ centered at $O$ is the set of all points at distance at most $R$ from $O$ : it is the interior of the sphere together with the sphere itself.

A great circle of a sphere is any cross-section through the center; it has radius $R$ and area $\pi R^2$ .

Surface Area of a Sphere

$\boxed{S = 4\pi R^2}$

This remarkable formula — the surface area equals the area of four great circles — was proved by Archimedes: the lateral surface of the circumscribed cylinder of the same radius and height $2R$ equals $2\pi R \cdot 2R = 4\pi R^2$ , and Archimedes showed that the sphere’s surface equals this lateral surface.

Volume of a Ball

$\boxed{V = \dfrac{4}{3}\pi R^3}$

Cavalieri’s Principle derivation. Consider a hemisphere of radius $R$ and, beside it, a cylinder of radius $R$ and height $R$ with a cone of the same base and height removed (an “anti-cone”). At height $t$ from the base:

Hemisphere cross-section: circle of radius $\sqrt{R^2 - t^2}$ , area $= \pi(R^2 - t^2)$ .
Cylinder-minus-cone cross-section: annulus of outer radius $R$ , inner radius $t$ , area $= \pi R^2 - \pi t^2 = \pi(R^2 - t^2)$ .

The cross-sections match at every level, so by Cavalieri’s Principle the volumes are equal. The cylinder-minus-cone has volume $\pi R^2 \cdot R - \tfrac{1}{3}\pi R^2 \cdot R = \tfrac{2}{3}\pi R^3$ . Doubling for the full sphere: $V = \dfrac{4}{3}\pi R^3$ . $\blacktriangleleft$

Summary Table

Solid	Lateral Surface Area	Total Surface Area	Volume
Cylinder (radius $R$ , height $h$ )	$2\pi Rh$	$2\pi R(R+h)$	$\pi R^2 h$
Cone (radius $R$ , height $h$ , slant $l$ )	$\pi Rl$	$\pi R(R+l)$	$\dfrac{1}{3}\pi R^2 h$
Sphere (radius $R$ )	—	$4\pi R^2$	$\dfrac{4}{3}\pi R^3$

Worked Examples

✎Example — Example 1 — Cylinder

A cylinder has base radius $R = 5$ cm and height $h = 12$ cm. Find the lateral surface area, total surface area, and volume.

Solution.

$S_{\text{lat}} = 2\pi Rh = 2\pi \times 5 \times 12 = 120\pi \approx 376{.}99 \text{ cm}^2$

$S_{\text{total}} = 2\pi R(R + h) = 2\pi \times 5 \times (5 + 12) = 170\pi \approx 534{.}07 \text{ cm}^2$

$V = \pi R^2 h = \pi \times 25 \times 12 = 300\pi \approx 942{.}48 \text{ cm}^3$

Answer: $S_{\text{lat}} = 120\pi$ cm², $S_{\text{total}} = 170\pi$ cm², $V = 300\pi$ cm³. $\blacktriangleleft$

✎Example — Example 2 — Cone

A cone has base radius $R = 9$ cm and height $h = 12$ cm. Find the slant height and the lateral surface area.

Solution.

Slant height:

$l = \sqrt{R^2 + h^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \text{ cm}$

Lateral surface area:

$S_{\text{lat}} = \pi R l = \pi \times 9 \times 15 = 135\pi \approx 424{.}1 \text{ cm}^2$

Answer: $l = 15$ cm, $S_{\text{lat}} = 135\pi$ cm². $\blacktriangleleft$

✎Example — Example 3 — Sphere Inside a Cube

A sphere of radius $R$ fits exactly inside a cube (the sphere is inscribed in the cube, so the cube edge equals $2R$ ). Find the ratio of the volume of the sphere to the volume of the cube.

Solution.

Volume of the sphere: $V_{\text{sphere}} = \dfrac{4}{3}\pi R^3$ .

Volume of the cube: The cube edge is $a = 2R$ , so $V_{\text{cube}} = (2R)^3 = 8R^3$ .

Ratio:

$\dfrac{V_{\text{sphere}}}{V_{\text{cube}}} = \dfrac{\dfrac{4}{3}\pi R^3}{8R^3} = \dfrac{4\pi}{24} = \dfrac{\pi}{6} \approx 0{.}5236$

The sphere occupies approximately $52{.}4\%$ of the cube.

Answer: $\dfrac{V_{\text{sphere}}}{V_{\text{cube}}} = \dfrac{\pi}{6}$ . $\blacktriangleleft$

Exercises

✏Exercise

A cone is inscribed in a cylinder of radius $R = 6$ cm and height $h = 8$ cm (they share the same base and the apex of the cone touches the top base of the cylinder). Find the ratio of the volume of the cone to the volume of the cylinder.

✏Exercise

A sphere has surface area $S = 100\pi$ cm². Find: (a) the radius $R$ of the sphere; (b) the volume of the ball.