Grade 12

Spheres, cylinders, cones

The three round surfaces of grade 12: how rotating a line or a curve sweeps out a cylinder, a cone, or a sphere, and where their area and volume formulas come from.

IntuitionFrom flat solids to round ones

A cube or a pyramid is bounded by flat faces. Spin a straight line or an arc around an axis instead, and the swept surface is curved everywhere: this is a surface of revolution. The three simplest ones — cylinder, cone, sphere — already appeared in middle school as shapes; here we build them from a precise rotation rule and derive their formulas instead of memorizing them.

Sweeping a generating line through 270° around the axis

SchoolThree surfaces of revolution

Definition: Cylinder

Let $\Delta$ be a line (the axis). Let $\ell$ be a line parallel to $\Delta$ , at distance $r$ . Rotating $\ell$ around $\Delta$ sweeps out a cylindrical surface of radius $r$ . Two planes perpendicular to $\Delta$ , a distance $h$ apart, cut out a cylinder of radius $r$ and height $h$ .

Definition: Cone

Let $O$ be a point. Let $\Delta$ be a line through $O$ (the axis). Let $\ell$ be a line through $O$ , making a fixed angle $\alpha$ with $\Delta$ . Rotating $\ell$ around $\Delta$ sweeps out a conical surface with apex $O$ and half-angle $\alpha$ . A plane perpendicular to $\Delta$ , at distance $h$ from $O$ , cuts out a cone: apex $O$ , height $h$ , base radius $r = h\tan\alpha$ , slant height $l = \sqrt{r^2+h^2}$ .

Definition: Sphere

Let $O$ be a point and $r$ a positive number. The sphere of center $O$ and radius $r$ is the set of points at distance $r$ from $O$ — equivalently, the surface swept by rotating a semicircle of radius $r$ a full turn around its diameter. The ball is the solid region bounded by the sphere, $\{M : OM \le r\}$ ; the sphere itself is only that region's boundary.

Surface area and volume of the three solids
Solid	Lateral area	Total area	Volume
Cylinder (r, h)	$S_{xq}=2\pi rh$	$S_{tp}=2\pi r(r+h)$	$V=\pi r^2h$
Cone (r, h, slant l)	$S_{xq}=\pi rl$	$S_{tp}=\pi r(r+l)$	$V=\frac{1}{3}\pi r^2h$
Sphere (r)	—	$S=4\pi r^2$	$V=\frac{4}{3}\pi r^3$

Theorem: Volume and surface area of a sphere

For a sphere of radius $r$ : volume $V_{\text{sphere}}=\frac{4}{3}\pi r^3$ and surface area $S=4\pi r^2$ .

Why is it true?

Archimedes compared a ball of radius $r$ to the cylinder of radius $r$ and height $2r$ that exactly contains it: the ball fills exactly $\frac{2}{3}$ of the cylinder, so $V_{\text{sphere}}=\frac{2}{3}\cdot(2\pi r^3)=\frac{4}{3}\pi r^3$ . He considered this his best result and asked for the cylinder-and-sphere figure to be carved on his tombstone.

Proof

Applied to the sphere: compare a hemisphere of radius $r$ to a cylinder of radius $r$ and height $r$ with a cone (apex at the bottom center, base radius $r$ ) removed. At height $x$ , the hemisphere's cross-section is a disk of area $\pi(r^2-x^2)$ ; the cylinder-minus-cone's cross-section is an annulus of area $\pi r^2-\pi x^2$ — the same. So the hemisphere has the same volume as the cylinder minus the cone: $\pi r^2\cdot r-\frac{1}{3}\pi r^2\cdot r=\frac{2}{3}\pi r^3$ . Doubling gives $V_{\text{sphere}}=\frac{4}{3}\pi r^3$ , and differentiating $V_{\text{sphere}}(r)$ with respect to $r$ gives $S=4\pi r^2$ .

The sphere as a rotated semicircle

Example: Cone, sphere, cylinder: the ratio 1 : 2 : 3

Take radius $r$ and height $2r$ for all three: a cone (apex up, base radius $r$ ), a sphere of radius $r$ , and a cylinder (radius $r$ , height $2r$ ). Compare their volumes.

Solution

$V_{\text{cone}}=\frac{1}{3}\pi r^2(2r)=\frac{2}{3}\pi r^3$ , $V_{\text{sphere}}=\frac{4}{3}\pi r^3$ , $V_{\text{cylinder}}=\pi r^2(2r)=2\pi r^3$ . Dividing by $\frac{2}{3}\pi r^3$ gives the ratio $1 : 2 : 3$ — independent of $r$ . The same cylinder that circumscribes the sphere also has lateral area $2\pi r\cdot 2r=4\pi r^2$ , exactly the sphere's surface area $S=4\pi r^2$ .

Example: Volume of a pressurized gas tank with hemispherical caps

A cylindrical propane tank has a cylindrical barrel of radius $r = 0.30\text{ m}$ and length $L = 1.20\text{ m}$ , capped at both ends by hemispherical domes of the same radius $r$ . Find the total volume of the tank.

Solution

Step 1 — Identify the components. The tank consists of one cylinder (radius $r$ , height $L$ ) plus two hemispherical caps, which together form one complete sphere of radius $r$ .

Step 2 — Compute each volume. $V_{\text{cylinder}} = \pi r^2 L = \pi(0.30)^2(1.20) = 0.108\pi\text{ m}^3$ . $V_{\text{sphere}} = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi(0.30)^3 = 0.036\pi\text{ m}^3$ .

Step 3 — Sum the parts. $V_{\text{total}} = V_{\text{cylinder}} + V_{\text{sphere}} = (0.108 + 0.036)\pi = 0.144\pi \approx 0.452\text{ m}^3$ . This is the standard formula for a "capsule" shape: $V = \pi r^2(L + \tfrac{4}{3}r)$ .

UndergraduateProving the volume formula with an integral

Theorem: Cavalieri's principle

If two solids of the same height have cross-sections of equal area at every level, then the two solids have equal volume.

Why is it true?

Volume is the integral of cross-sectional area over height, $V=\int_0^h S(x)\,dx$ ; if $S_1(x)=S_2(x)$ for every $x$ , the two integrals are equal.

Proof

V_{\text{cylinder}} = \pi r^2 h, \qquad V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h, \qquad V_{\text{sphere}} = \tfrac{4}{3}\pi r^3

V_{\text{sphere}} = 2\int_0^r \pi\left(r^2-x^2\right)dx = 2\cdot\frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3

AdvancedBeyond the sphere: other quadrics of revolution

The sphere is the surface of revolution of a circle. Rotating other conics around an axis gives more quadric surfaces: an ellipse gives a spheroid, a parabola gives a paraboloid of revolution, and a hyperbola rotated around its transverse axis gives a hyperboloid of revolution of one sheet — doubly ruled (it contains two families of straight lines through every point), used for cooling towers and gear shapes because it is both curved and buildable from straight beams. Unlike the sphere, its curvature is not constant: this is where the topic connects to differential geometry (see Differential geometry).

Hyperboloid of one sheet

A cylinder has radius $r$ and height $h$ . Which is its volume?

The sphere of center $O$ and radius $r$ is best described as:

A ball of radius $r$ sits exactly inside a cylinder of radius $r$ and height $2r$ (touching top, bottom, and side). What fraction of the cylinder's volume does the ball fill?

Cavalieri's principle justifies the sphere volume formula because:

References

Archimedes, translated by T. L. Heath (2002). The Works of Archimedes