Surface Area to Volume Ratio

AS · 14 min

Why are cells so small, and why do large animals need lungs, guts and a heart while a single-celled organism manages with its cell surface membrane alone? The answer is a simple geometric fact: as an object gets bigger, its volume increases faster than its surface area, so its surface area to volume ratio falls. This note shows how to calculate surface areas, volumes and ratios for simple shapes, explains why the ratio matters for diffusion, and covers the agar block practical used to investigate it. Calculations of this kind appear in Paper 2 and the practical appears in Paper 3.

What the ratio means

Every cell exchanges substances with its surroundings across its surface: oxygen and glucose in, carbon dioxide and wastes out. The demand for exchange depends on the volume of the cell (more cytoplasm means more respiration and more reactions). The supply depends on the surface area available for diffusion. So the surface area to volume ratio (SA

) tells you how much surface there is to serve each unit of volume.

Definition

The surface area to volume ratio is the surface area of an object divided by its volume. It is written as a ratio, e.g. $6 : 1$ , or as a single number with units of $\text{length}^{-1}$ (e.g. $6\ \text{cm}^{-1}$ ). The larger an object is, the smaller its surface area to volume ratio.

Calculating surface area and volume

Key result

Shape	Surface area	Volume	SA
Cube of side $L$	$6L^2$	$L^3$	$6/L$
Cuboid $a \times b \times c$	$2(ab + bc + ac)$	$abc$	calculate each time
Sphere of radius $r$	$4\pi r^2$	$\tfrac{4}{3}\pi r^3$	$3/r$
Cylinder of radius $r$ , length $h$	$2\pi r^2 + 2\pi r h$ (two ends plus curved side)	$\pi r^2 h$	calculate each time

You will be given the formulae for spheres and cylinders if they are needed, but you must be able to use them.

Look at what happens to cubes as they get bigger:

Side of cube / cm	Surface area / cm²	Volume / cm³	SA
0.5	1.5	0.125	12 : 1
1	6	1	6 : 1
2	24	8	3 : 1
3	54	27	2 : 1
4	96	64	1.5 : 1

Doubling the side multiplies the surface area by four ( $2^2$ ) but the volume by eight ( $2^3$ ), so the ratio halves. This principle applies to every shape: SA

is inversely proportional to the linear size.

Calculating a surface area to volume ratio

Write down the dimensions, all in the same units.
Calculate the total surface area (remember every face: a cube has six; a cuboid has three pairs; a cylinder has two ends and a curved side).
Calculate the volume.
Divide surface area by volume.
Express as a ratio " $x : 1$ " (or as a number with units such as $\text{cm}^{-1}$ ), to a sensible number of significant figures.

Cubes and cuboids (routine)

(a) Calculate the surface area to volume ratio of a cube of side $3\ \text{mm}$ . (b) Calculate the surface area to volume ratio of a cuboid measuring $4\ \text{cm} \times 2\ \text{cm} \times 1\ \text{cm}$ , and compare it with a $2\ \text{cm}$ cube, which has the same volume.

Solution

(a) Surface area $= 6 \times 3^2 = 54\ \text{mm}^2$ ; volume $= 3^3 = 27\ \text{mm}^3$ ; ratio $= 54/27 = 2 : 1$ (or $2\ \text{mm}^{-1}$ ).

(b) Surface area $= 2(4 \times 2 + 2 \times 1 + 4 \times 1) = 2(8 + 2 + 4) = 28\ \text{cm}^2$ ; volume $= 4 \times 2 \times 1 = 8\ \text{cm}^3$ ; ratio $= 28/8 = 3.5 : 1$ .

The $2\ \text{cm}$ cube has surface area $24\ \text{cm}^2$ , volume $8\ \text{cm}^3$ , ratio $3 : 1$ . The cuboid has the same volume but a larger surface area and so a higher ratio: being flatter or more elongated increases SA

Cylinders and spheres (moderate)

(a) A root hair is modelled as a cylinder of radius $0.5\ \text{units}$ and length $4\ \text{units}$ , including both ends. Calculate its SA

. (b) Calculate the SA

of spherical cells of diameter

1\ \mu\text{m}

(a bacterium) and

100\ \mu\text{m}

(a large egg cell).

Solution

(a) Surface area $= 2\pi r^2 + 2\pi r h = 2\pi(0.5)^2 + 2\pi(0.5)(4) = 1.571 + 12.566 = 14.14$ ; volume $= \pi r^2 h = \pi (0.5)^2 (4) = 3.142$ . Ratio $= 14.14 / 3.142 = 4.5 : 1$ .

(b) For a sphere, SA

= \dfrac{4\pi r^2}{\tfrac{4}{3}\pi r^3} = \dfrac{3}{r}

Bacterium: $r = 0.5\ \mu\text{m}$ , ratio $= 3/0.5 = 6\ \mu\text{m}^{-1}$ ( $6 : 1$ ).
Egg cell: $r = 50\ \mu\text{m}$ , ratio $= 3/50 = 0.06\ \mu\text{m}^{-1}$ ( $0.06 : 1$ ).

The bacterium has a ratio 100 times greater, because its diameter is 100 times smaller.

Why the ratio matters

For cells

A cell depends on diffusion across its surface membrane for oxygen and for removing carbon dioxide. As a cell grows:

its volume (and so its demand for oxygen and its production of waste) increases faster than its surface area (supply);
the distance from the surface to the centre increases, so it takes longer for substances to diffuse to and from the middle of the cell.

Eventually diffusion across the surface cannot supply the cell's needs fast enough. This is one reason why cells are small (most are $10$ – $100\ \mu\text{m}$ ) and why cells divide rather than simply keep growing. A nucleus also limits cell size: one nucleus can control only a limited volume of cytoplasm.

Cells that need to exchange a lot of material have shapes or features that increase SA:V:

Microvilli on epithelial cells of the small intestine and kidney tubules: folds of the cell surface membrane that greatly increase surface area for absorption.
Root hair cells: long, thin extensions that increase surface area for water and mineral ion uptake.
Red blood cells: small and biconcave, giving a high SA
for rapid diffusion of oxygen, and no part of the cytoplasm is far from the surface.
Squamous epithelial cells lining the alveoli and capillaries are very thin and flat.

For organisms

A single-celled organism such as Amoeba has a high SA

and a short diffusion distance, so it can obtain all its oxygen by diffusion across its cell surface membrane.

A large multicellular organism has a low SA

. Its outer surface is far too small to supply all its cells by diffusion, and most of its cells are far from that surface: diffusion over more than about a millimetre is far too slow. So large organisms need:

specialised exchange surfaces with a large surface area, such as alveoli in the lungs, villi in the small intestine, gills in fish and spongy mesophyll in leaves;
transport systems (the blood circulatory system in mammals; xylem and phloem in plants) that carry substances by mass flow over long distances between exchange surfaces and the cells that need them.

Flatworms are the exception that proves the rule: they are multicellular but extremely flat, so no cell is far from the surface and they have no need of lungs or a circulation.

Beyond the syllabus: SA:V and heat loss

Heat is lost across the body surface and generated by the volume of respiring tissue. Small mammals such as shrews have a high SA

, lose heat very fast, and must have a very high metabolic rate. This is why there is a lower limit to the size of mammals and birds. Not examinable at AS, but a useful illustration of the same principle.

Investigating SA
with agar blocks

Practical skills

Effect of surface area to volume ratio on diffusion using agar blocks

Agar jelly is made up containing an indicator and an alkali, for example phenolphthalein and dilute sodium hydroxide, so it is pink. When a block is placed in dilute hydrochloric acid, the acid diffuses into the block, neutralises the alkali, and the pink colour disappears from the outside inwards. (Alternatives: agar with cresol red or universal indicator; agar containing a dye placed in water, measuring dye leaving.)

Method: cut cubes of agar of different sizes, e.g. $0.5$ , $1$ , $2\ \text{cm}$ sides, using a sharp scalpel and a ruler on a white tile. Place each cube in the same volume of acid of the same concentration (enough to cover the cube), at the same temperature, and start a stopwatch. Record the time taken for the pink colour to disappear completely. Alternatively, remove the blocks after a fixed time (e.g. 5 minutes), cut them in half and measure the distance the acid has penetrated.
Independent variable: size of block (and so SA
). Dependent variable: time for complete colour change, or distance penetrated, or percentage of the volume decolourised.
Standardised variables: concentration and volume of acid; concentration of alkali and indicator in the agar; temperature; agar from the same batch; the blocks fully covered by acid; stirring (or no stirring) the same for all.
Safety: hydrochloric acid and sodium hydroxide are irritants at these concentrations: wear eye protection. Scalpels: cut away from the body onto a tile.
Sources of error: blocks not cut precisely to size (a $0.1\ \text{cm}$ error in a $0.5\ \text{cm}$ cube is a 20% error in side and about 60% in volume); judging the exact end point of the colour change is subjective, especially for large blocks where the last pink core fades slowly; blocks touching the bottom of the beaker have less surface exposed; temperature not controlled.
Improvements: cut blocks with a template or cutting guide; stand blocks on a mesh so all faces are exposed; view against a white tile; use a water bath; repeat three times per size and calculate means.
Typical result: larger blocks (lower SA
) take longer to decolourise completely; the acid penetrates the same distance in a given time in all blocks (rate of diffusion is the same), but a larger block has a greater distance to its centre and a smaller proportion of its volume is reached.

Agar block results (moderate)

Cubes of pink agar were placed in hydrochloric acid and the time for the colour to disappear was recorded.

Side of cube / cm	0.5	1.0	2.0
Time to decolourise / s	95	380	1500

(a) Calculate the SA

of each cube. (b) Describe the relationship between SA

and the time taken. (c) Explain why the time increases much more than the size of the cube.

Solution

(a) Using $6/L$ : $0.5\ \text{cm}$ : $12 : 1$ ; $1.0\ \text{cm}$ : $6 : 1$ ; $2.0\ \text{cm}$ : $3 : 1$ .

(b) As SA

decreases, the time to decolourise increases: each halving of SA

increases the time about four-fold.

(c) In a larger cube there is less surface area for each unit of volume, so acid enters at a slower rate relative to the volume to be neutralised; and the distance from the surface to the centre is greater (0.25, 0.5 and 1.0 cm). Diffusion is fast over short distances but becomes increasingly slow over longer ones, so doubling the distance to the centre roughly quadruples the time.

Percentage of a block reached by diffusion (moderate)

A $2\ \text{cm}$ cube of pink agar was left in acid for 5 minutes, then cut in half. The acid had penetrated $0.3\ \text{cm}$ from every face. Calculate the percentage of the cube's volume that had been reached by the acid.

Solution

The pink core that remains is a cube of side $2.0 - (2 \times 0.3) = 1.4\ \text{cm}$ .

Volume of core $= 1.4^3 = 2.744\ \text{cm}^3$ . Total volume $= 2^3 = 8\ \text{cm}^3$ .

Volume reached by acid $= 8 - 2.744 = 5.256\ \text{cm}^3$ .

Percentage $= \dfrac{5.256}{8} \times 100 = 65.7\%$ .

(A $1\ \text{cm}$ cube in the same time would have a core of side $1.0 - 0.6 = 0.4\ \text{cm}$ and volume $0.064\ \text{cm}^3$ , so $93.6\%$ of its volume would be reached. The smaller cube, with the higher SA

, is almost entirely supplied in the same time.)

Exam-hard: why cells are flat (6 marks)

A cell is modelled as a cuboid $20\ \mu\text{m} \times 20\ \mu\text{m} \times 2\ \mu\text{m}$ .

(a) Calculate its SA

. (b) A cube has the same volume. Calculate the length of its side and its SA

. (c) Use your answers to explain why squamous epithelial cells lining the alveoli are flattened.

Solution

(a) Surface area $= 2(20 \times 20 + 20 \times 2 + 20 \times 2) = 2(400 + 40 + 40) = 960\ \mu\text{m}^2$ . Volume $= 20 \times 20 \times 2 = 800\ \mu\text{m}^3$ . SA

= 960/800 = 1.2 : 1

(b) Side $= \sqrt[3]{800} = 9.28\ \mu\text{m}$ . Surface area $= 6 \times 9.28^2 = 517\ \mu\text{m}^2$ . SA

= 517/800 = 0.65 : 1

(or

6/9.28 = 0.65

(c)

The flat cell has almost twice the SA
of a cube-shaped cell of the same volume ( $1.2$ vs $0.65$ ).
A larger surface area per unit volume allows faster diffusion of oxygen and carbon dioxide across the cell.
The flat cell is only $2\ \mu\text{m}$ thick, so the diffusion distance is very short (compared with up to $4.6\ \mu\text{m}$ from the centre of the cube).
Squamous cells form the wall of the alveolus (and of capillaries), so gas exchange between air and blood is rapid;
this maintains a steep concentration gradient because gases do not accumulate.

Watch out

As size increases, SA
decreases. Students who say "larger organisms have a larger surface area, so they need lungs" miss the point: the surface area is larger, but the ratio is smaller.
Count all six faces of a cube or cuboid, and both ends of a cylinder (unless told otherwise).
Keep units the same throughout: do not mix mm and cm.
In the agar practical, the rate at which the acid penetrates (distance per unit time) is about the same in every block. What differs is the distance to the centre and the proportion of the volume reached.

Exam tip

Show your working for SA
calculations: surface area, then volume, then the ratio. Each is often a separate mark.
Give the ratio as " $x : 1$ " unless the question asks for something else.
When explaining why large organisms need transport systems or exchange surfaces, use three ideas: low SA
; long diffusion distances (diffusion too slow over more than a few cells' width); high metabolic demand (many cells needing oxygen).
Questions often link SA
to the structure of red blood cells (biconcave), microvilli, root hairs or alveoli: say how the feature increases the surface area for diffusion, absorption or exchange.

Summary

Surface area to volume ratio = surface area ÷ volume; it decreases as size increases.
Cube: $6/L$ ; sphere: $3/r$ ; doubling linear size halves the ratio.
A small SA
means less surface for exchange per unit volume and longer diffusion distances.
Cells stay small and specialised cells have features that increase SA:V: microvilli, root hairs, biconcave shape, flattening.
Large multicellular organisms have low SA
, so they need specialised exchange surfaces and transport systems.
Agar blocks with indicator: larger blocks (lower SA
) take longer to change colour completely; control acid concentration, volume and temperature; cut blocks accurately.

Practice questions

Question

Calculate the SA
of a cube of side $5\ \text{mm}$ .
A cube of side $1\ \text{cm}$ is cut into eight cubes of side $0.5\ \text{cm}$ . Calculate the total surface area before and after, and explain the effect on SA
.
Explain why a single-celled organism does not need a specialised gas exchange system. (2 marks)
Calculate the SA
of a spherical cell of radius $15\ \mu\text{m}$ .
A cylindrical block of agar has a radius of $0.5\ \text{cm}$ and a length of $2\ \text{cm}$ . Calculate its SA
.
State three variables that should be controlled in an investigation into the effect of block size on diffusion in agar.
Suggest why it is difficult to measure accurately the time for very small agar cubes to decolourise. (2 marks)
Explain why large multicellular animals need a transport system. (3 marks)
A $3\ \text{cm}$ cube of agar is left in acid until the acid has penetrated $0.5\ \text{cm}$ from each face. Calculate the percentage of the cube's volume that remains pink.
A student claimed: "A cell with twice the diameter of another needs twice as much surface area to supply it." Evaluate this claim using calculations for spherical cells of diameter $10\ \mu\text{m}$ and $20\ \mu\text{m}$ , and explain the implications for cell size. (6 marks)

Answers

SA $= 6 \times 25 = 150\ \text{mm}^2$ ; V $= 125\ \text{mm}^3$ ; ratio $= 1.2 : 1$ .
Before: $6 \times 1^2 = 6\ \text{cm}^2$ . After: $8 \times 6 \times 0.5^2 = 8 \times 1.5 = 12\ \text{cm}^2$ . The volume is unchanged ( $1\ \text{cm}^3$ ), so SA
doubles from $6 : 1$ to $12 : 1$ : dividing a volume into smaller units increases the surface area available.
It has a large SA
, so its surface is large enough relative to its volume to supply its oxygen needs by diffusion; the diffusion distance to the centre of the cell is short.
$3/r = 3/15 = 0.2 : 1$ (or $0.2\ \mu\text{m}^{-1}$ ). Check: SA $= 4\pi(15)^2 = 2827\ \mu\text{m}^2$ ; V $= \tfrac{4}{3}\pi(15)^3 = 14\,137\ \mu\text{m}^3$ ; ratio $= 0.20$ .
SA $= 2\pi(0.5)^2 + 2\pi(0.5)(2) = 1.571 + 6.283 = 7.854\ \text{cm}^2$ ; V $= \pi(0.5)^2(2) = 1.571\ \text{cm}^3$ ; ratio $= 5.0 : 1$ .
Any three: concentration of acid; volume of acid; temperature; concentration of alkali/indicator in agar; agar from same batch; whether blocks are stirred / all surfaces exposed.
The colour change is very fast, so reaction time in starting and stopping the stopwatch is a large proportion of the time; the end point (last trace of pink) is hard to judge; small errors in cutting the cube make a large percentage difference to its volume.
Large animals have a small SA
, so the body surface is too small to supply all cells by diffusion; the distances from the surface to cells deep inside the body are too great for diffusion to be fast enough; they have a high metabolic rate / many cells requiring oxygen and glucose; so a mass-flow transport system (blood) is needed to carry substances between exchange surfaces and cells.
Pink core side $= 3 - 1 = 2\ \text{cm}$ ; core volume $= 8\ \text{cm}^3$ ; total $= 27\ \text{cm}^3$ ; percentage pink $= 8/27 \times 100 = 29.6\%$ .
Diameter 10 µm ( $r = 5$ ): SA $= 4\pi(25) = 314\ \mu\text{m}^2$ , V $= \tfrac{4}{3}\pi(125) = 524\ \mu\text{m}^3$ , ratio $0.6$ . Diameter 20 µm ( $r = 10$ ): SA $= 1257\ \mu\text{m}^2$ , V $= 4189\ \mu\text{m}^3$ , ratio $0.3$ . Doubling the diameter increases the volume (and so the demand) eight-fold but the surface area only four-fold. So the claim is wrong: the larger cell would need eight times the supply, but has only four times the surface, so it has half as much surface per unit volume. Also the diffusion distance to its centre doubles. This means cells cannot simply keep growing: beyond a certain size diffusion across the surface cannot meet the cell's needs, so cells divide, stay small, or have shapes/features (flattening, microvilli) that increase SA
.

What the ratio means

Calculating surface area and volume

Why the ratio matters

For cells

For organisms

Investigating SA with agar blocks

Practice questions

Console

Investigating SA
with agar blocks