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Surface area to volume ratio

Laura Armstrong & Joe Wolfensohn

Teachers

Laura Armstrong Joe Wolfensohn

Contents

How to Calculate SA:V for Different Shapes

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Adaptations to Increase or Decrease SA

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Recall Questions

This topic requires prior knowledge of basic mathematical calculations involving area and volume, as well as diffusion principles from GCSE-level science. You can test your knowledge on these below.

Calculate the surface area and SA:V ratio of a cube with a side length of 2 cm.

  • SA = 6 × (2²) = 24 cm²

  • V = 2³ = 8 cm³

  • SA:V = 24 ÷ 8 = 3:1

True or False: As an organism’s size increases, its surface area to volume ratio also increases.

False. As size increases, SA:V decreases.

What structural adaptation can organisms have to increase their SA:V ratio for efficient diffusion?

Flattened shapes (e.g., leaves), folded surfaces (e.g., alveoli, villi), and projections (e.g., microvilli, root hair cells).

Topic Explainer Video

Check out this @JoeDoesBiology video that explains surface area to volume ratio or read the full notes below. Once you've gone through the whole note, try out the practice questions!

How to Calculate SA:V for Different Shapes

Cubes

  • Surface area (SA) = 6 × (side length)²

  • Volume (V) = (side length)³

  • SA:V ratio = SA ÷ V

Example

For a cube with a side length of 1 cm:

  • SA = 6 × (1²) = 6 cm²
  • V = 1³ = 1 cm³
  • SA:V = 6 ÷ 1 = 6:1

 

 

 

For a cube with a side length of 2 cm:

  • SA = 6 × (2²) = 24 cm²
  • V = 2³ = 8 cm³
  • SA:V = 24 ÷ 8 = 3:1

 

 


For a cube with a side length of 3 cm:

  • SA = 6 × (3²) = 54 cm²
  • V = 3³ = 27 cm³
  • SA:V = 54 ÷ 27 = 2:1

 

No answer provided.

Cuboids

  • SA = 2(lw + lh + wh)

  • V = l × w × h

  • SA:V ratio = SA ÷ V

Example

For a cuboid with dimensions 4 cm × 2 cm × 4 cm

  • SA = 2(4×4 + 4×2 + 4×2) = 64 cm²

  • V = 4 × 2 × 4 = 32 cm³

  • SA:V = 64 ÷ 32 = 2:1

No answer provided.

Cylinders

  • SA = 2πr² + 2πrh

  • V = πr²h

  • SA:V ratio = SA ÷ V

Example

For a cylinder with r = 2 cm and h = 5 cm:

  • SA = 2π(2²) + 2π(2)(5) = 8π + 20π = 28π ≈ 87.96 cm²

  • V = π(2²)(5) = 20π ≈ 62.83 cm³

  • SA:V ≈ 1.4:1

No answer provided.

Adaptations to Increase or Decrease SA

Increasing SA for Exchange

  • Flattened shape (e.g., leaves, flatworms) → More surface area for gas exchange by diffusion.
  • Folding of surfaces (e.g., villi in intestines, alveoli in lungs) → More efficient absorption.
  • Projections (e.g., root hair cells, microvilli) → Increased nutrient uptake.

Decreasing SA to Conserve Heat

  • Compact shape (e.g., Arctic fox, round body structure) → Reduces heat loss.
  • Fat insulation (e.g., blubber in whales) → Reduces heat loss.
  • Reduced extremities (e.g., smaller ears in polar bears) → Less surface area exposed to cold and therefore less heat loss.

Key Term

  • Surface Area to Volume Ratio (SA:V): A measure of how much surface area is available relative to the volume of an object.
No answer provided.

Exam Tip

Apply knowledge to real-life examples, such as explaining why desert animals have large ears (high SA:V for cooling).

No answer provided.

A cuboid has dimensions of 3 cm × 2 cm × 1 cm. Calculate its surface area and surface area to volume ratio.

  • SA = 2(3×2 + 3×1 + 2×1) = 2(6 + 3 + 2) = 22 cm²

  • V = 3 × 2 × 1 = 6 cm³

  • SA:V = 22 ÷ 6 = 3.67:1

Practice Question

Try to answer the practice question from the TikTok on your own, then watch the video to see how well you did!