Surface Areas and Volumes

Square units, cubic units and capacity

A square centimetre covers a 1 cm by 1 cm square. A cubic centimetre fills a 1 cm by 1 cm by 1 cm cube. These are different units:

Since $1\text{ m}=100\text{ cm}$, a cubic metre contains

For capacity, $1\text{ cm}^3=1\text{ mL}$ and $1\text{ L}=1\,000\text{ cm}^3$. A tank with internal dimensions 40 cm by 25 cm by 30 cm has capacity $30\,000\text{ cm}^3$, or 30 L, when completely filled. A half-filled tank of that rectangular shape contains 15 L.

Convert all dimensions to compatible units before multiplying. Cubing a length conversion means cubing its numerical factor, not just changing the unit label. The SI system’s treatment of units is documented by the BIPM; the NIST volume reference also explains cubic units and litre conversions.

Rectangular prisms: layers and six faces

A rectangular prism with length $l$, width $w$ and height $h$ has volume

Each horizontal layer has area $lw$. Multiplying by height accounts for the stack of layers. The same product results if a different face is chosen as the base, provided its matching perpendicular height is used.

There are three pairs of congruent faces. Their areas give the closed prism’s total surface area:

For dimensions 8 cm, 5 cm and 3 cm,

A net unfolds the surface into flat faces. It provides an area inventory: two faces of 40 square centimetres, two of 24, and two of 15. A valid net must fold into the intended solid without overlapping faces. Merely drawing six rectangles does not guarantee that their edge lengths or arrangement will fit.

Open boxes and cubes

If the 8 cm by 5 cm top is absent, subtract its area from the closed-box total:

This counts one ideal surface for each of the five remaining faces. It does not count both the inside and outside of a material with thickness. Specify that different measurement separately.

For a cube of side $s$, the prism formulas simplify to $V=s^3$ and $SA=6s^2$. A cube of side 4 cm has volume $64\text{ cm}^3$ and surface area $96\text{ cm}^2$. These standard formulas are collected in OpenStax’s geometry reference.

Prisms with other bases

For any prism, volume is base area $B$ multiplied by perpendicular height $h$:

For a right prism, the lateral faces unfold into rectangles. Their total area is the base perimeter $p$ multiplied by height. Thus its closed surface area is

The volume formula also holds for an oblique prism when $h$ is the perpendicular separation of the bases. The convenient lateral-area expression $ph$ is for a right prism; the slanted lateral edges of an oblique prism need separate treatment. See the prism reference for this distinction.

Consider a right triangular prism whose base is a 3–4–5 right triangle and whose prism height is 10 cm. The base area is $B=\frac12(3)(4)=6\text{ cm}^2$, and its perimeter is $p=12$ cm. Therefore

The 10 cm measurement separates the triangular bases. It is not the altitude used to find the triangle’s area. Naming both roles prevents a common confusion when a problem uses the word “height” twice.

Cylinders: unroll the curved surface

For a right circular cylinder with radius $r$ and perpendicular height $h$, the base is a disk of area $\pi r^2$. Its volume is

The curved surface unrolls into a rectangle whose length is the base circumference, $2\pi r$, and whose width is $h$. The curved or lateral surface area is therefore $2\pi rh$. Adding two circular ends gives

A closed cylinder with radius 3 cm and height 8 cm has volume $72\pi\text{ cm}^3$ and total surface area $66\pi\text{ cm}^2$. Rounded to two decimal places, these are approximately 226.19 cubic centimetres and 207.35 square centimetres.

A label covering only the curved surface needs $48\pi\text{ cm}^2$, before any overlap allowance. A cylinder open at one end has surface area $57\pi\text{ cm}^2$. These are three different tasks, not alternative answers to the same question.

When a diameter is given, divide it by two before substituting for $r$. Squaring the diameter in a radius formula makes the circular area four times too large.

Pyramids and cones: why one third matters

A pyramid or cone with base area $B$ and perpendicular height $h$ has volume

It occupies one third of the volume of a prism or cylinder with the same base area and height. The factor is not one half: cross-sectional areas shrink as the solid narrows toward its apex. The pyramid reference derives the one-third factor from those changing cross-sections.

For a cone, $B=\pi r^2$, giving $V=\frac13\pi r^2h$. For a pyramid, calculate its polygonal base area first.

Height and slant height are different

The perpendicular height $h$ runs from the apex to the base plane. A slant height runs along a lateral surface. In a right circular cone,

The curved surface area is $\pi rs$, and the total area including the circular base is

For $r=3$ cm and $h=4$ cm, the slant height is 5 cm. The cone has volume $12\pi\text{ cm}^3$, lateral area $15\pi\text{ cm}^2$ and total surface area $24\pi\text{ cm}^2$. Using 5 instead of 4 in the volume formula would confuse the sloping surface with perpendicular height.

For a regular pyramid with base perimeter $p$ and a common triangular-face slant height $s$, the lateral surface area is $\frac12ps$. Add the base area for the total. A square pyramid with base side 6 cm and perpendicular height 4 cm has face slant height $\sqrt{3^2+4^2}=5$ cm. Its volume is 48 cubic centimetres and its total surface area is $36+\frac12(24)(5)=96\text{ cm}^2$.

The distance from apex to a base vertex is a sloping edge, not that triangular-face altitude. Do not substitute it for $s$ without checking the definition.

Spheres and hemispheres

For a sphere of radius $r$, the surface area and enclosed volume are

A sphere of radius 3 cm has surface area $36\pi\text{ cm}^2$ and enclosed volume $36\pi\text{ cm}^3$. Their numerical coefficients happen to match, but area uses square units and volume uses cubic units.

A hemisphere has half the sphere’s curved surface area and half its volume. If its flat circular base is included, the total surface area is

For radius 3 cm, the curved area is $18\pi\text{ cm}^2$, the area including the base is $27\pi\text{ cm}^2$, and the volume is $18\pi\text{ cm}^3$. A dome that is open underneath requires the curved area, not automatically the closed-hemisphere total.

Unlike a cylinder’s lateral surface, a sphere’s surface cannot be flattened into a plane without distortion. Its area formula should not be justified by pretending that a rectangular net covers it exactly.

Composite solids: add volumes, remove hidden interfaces

When solids meet along a face without overlapping in volume, their volumes add. Their external surface areas usually do not simply add, because the joining surfaces become internal.

Join a hemisphere of radius 3 cm to the top of a cylinder with the same radius and height 8 cm. Their interiors do not overlap. The combined volume is

Its exterior consists of the cylinder’s curved surface, the cylinder’s bottom disk, and the hemisphere’s curved surface:

Adding the two closed surface areas would give $66\pi+27\pi=93\pi$. Subtracting the two copies of the shared disk removes $18\pi$, leaving the correct $75\pi$. The interface is counted once on each original solid, which is why both copies must be removed.

For a hollow solid, subtract the removed volume but count any inner surfaces that the question asks you to measure. A bore drilled through a block adds an inner cylindrical wall even as it removes material.

Scale factors in three dimensions

For similar solids, multiplying every length by a positive factor $k$ multiplies surface area by $k^2$ and volume by $k^3$:

Doubling every dimension produces four times the surface area and eight times the volume. Tripling them produces nine times the surface area and twenty-seven times the volume. This difference explains why a larger similar container can hold substantially more without a proportional increase in its covering area.

These rules require uniform scaling. Doubling only a cylinder’s height doubles its volume, but its total surface area does not double because the circular ends remain the same size. Work from the actual dimensions when the shapes are not similar.

Find a missing dimension and check the result

Suppose a right circular cylinder has volume $200\pi\text{ cm}^3$ and radius 5 cm. Substituting into $V=\pi r^2h$ gives

Substituting back confirms $\pi(5^2)(8)=200\pi$. The units also work: volume divided by base area gives length.

Before accepting any result, check what surfaces are included, whether the height is perpendicular, whether a given circular measure is a radius or diameter, and whether all lengths use the same units. Keep exact expressions involving $\pi$ or square roots until the final rounding step. Those checks address the geometry, not just the arithmetic.

Formula reference

The surface-area formulas in the table describe closed solids unless otherwise stated. The prism lateral formula assumes a right prism; the cone surface formula assumes a right circular cone; the pyramid lateral formula assumes a regular pyramid with a common face slant height.

Related topics

Use Area and Perimeter of Shapes to review the base and face calculations. 3D Coordinate Axes and Planes introduces a precise way to locate points and measure distances within three-dimensional models.