Lesson 46 – Surface Area and Volume

Quantum Leap Star Limited · GCSE Mathematics · Lesson 46 · Higher focus

Surface area covers the outside; volume fills the inside. This lesson explains how to calculate both, choose the correct dimensions and apply the methods to composite solids, capacity and similarity.

1. Measuring three dimensions

Surface area is the total area of the exposed faces and curved surfaces of a solid. Measure it in square units such as cm² or m². Volume is the space occupied by a solid, measured in cubic units such as cm³ or m³.

A face is a flat surface, an edge is where flat faces meet, and a vertex is a corner. A cube has 6 faces, 12 edges and 8 vertices. Cylinders and cones also have curved surfaces; distinguish these from flat faces.

Units and capacity

• 1 cm³ = 1000 mm³.
• 1 m³ = 1,000,000 cm³.
• 1 litre = 1000 cm³; 1 millilitre = 1 cm³.
• 1 m³ = 1000 litres.

Square length conversion factors for area and cube them for volume. Convert every dimension to the same unit before calculating. For example, 2.4 litres = 2400 cm³, whereas 2400 mm³ = 2.4 cm³.

A reliable method: identify the solid → decide which surfaces or volume are required → label the dimensions → choose a formula → calculate with matching units → state the units and round only at the end.

2. Cuboids, nets and prisms

For a cuboid with length l, width w and height h:

• Volume: V = lwh.
• Closed surface area: S = 2(lw + lh + wh).
• For a cube of side s: V = s³ and S = 6s².

Worked example: a box 8 cm × 5 cm × 3 cm

V = 8 × 5 × 3 = 120 cm³.
S = 2(40 + 24 + 15) = 158 cm².

If the 8 cm × 5 cm top is removed, the surface area becomes 158 − 40 = 118 cm², ignoring thickness.

Any prism

A prism has a constant cross section along its length. Volume = cross-sectional area × length. For a closed right prism, S = 2A + PL, where A is the cross-sectional area, P its perimeter and L the prism length.

For a triangular prism with a 3–4–5 cm right-triangle cross section and length 10 cm:
A = ½ × 3 × 4 = 6 cm².
V = 6 × 10 = 60 cm³.
S = 2 × 6 + (3 + 4 + 5) × 10 = 132 cm².

A net unfolds a solid into its faces. Count each exposed face once. When solids are joined, their touching faces become internal and are excluded from the outside surface area.

3. Cylinders

For radius r and perpendicular height h:

• Volume: V = πr²h.
• Curved surface area: 2πrh.
• Total closed surface area: 2πrh + 2πr².

The curved surface unwraps into a rectangle of width 2πr and height h. Each circular end has area πr². An open-top cylinder has only one end disc.

Worked example: r = 4 cm, h = 9 cm

V = π × 4² × 9 = 144π cm³ ≈ 452.39 cm³.
Curved area = 2π × 4 × 9 = 72π cm².
Closed total area = 72π + 2 × 16π = 104π cm² ≈ 326.73 cm².
Open-top area = 72π + 16π = 88π cm².

Reverse calculation: a cylinder has volume 180π cm³ and radius 3 cm. Then 180π = 9πh, so h = 20 cm.

4. Pyramids and cones

Volume uses perpendicular height. Cone surface area uses slant height.

• Pyramid volume: ⅓ × base area × perpendicular height.
• Cone volume: V = ⅓πr²h.
• Cone curved area: πrl.
• Cone total area: πrl + πr², where l is slant height.

For a right circular cone, l² = r² + h² by Pythagoras. The slant height is longer than the perpendicular height.

Worked example: cone with r = 5 cm and h = 12 cm

l = √(5² + 12²) = 13 cm.
V = ⅓π × 25 × 12 = 100π cm³.
Total area = π × 5 × 13 + π × 25 = 90π cm².

Square-based pyramid: base side 6 cm, perpendicular height 4 cm.
V = ⅓ × 36 × 4 = 48 cm³.
The triangular face height is √(4² + 3²) = 5 cm.
Total area = 36 + 4(½ × 6 × 5) = 96 cm².

5. Spheres, hemispheres and composite solids

• Sphere: V = (4/3)πr³ and S = 4πr².
• Hemisphere: V = (2/3)πr³.
• Hemisphere curved area: 2πr².
• Hemisphere total area, including its circular base: 3πr².

For radius 6 cm, a sphere has volume 288π cm³ and area 144π cm². A hemisphere has volume 144π cm³ and total area 108π cm².

Worked example: a cone joined to a hemisphere

A cone of radius 3 cm and perpendicular height 4 cm is joined to a hemisphere of radius 3 cm along their circular bases. The cone slant height is √(3² + 4²) = 5 cm.

Volume = cone + hemisphere
= ⅓π × 3² × 4 + ⅔π × 3³
= 12π + 18π = 30π cm³.

Exposed area = cone curved area + hemisphere curved area
= π × 3 × 5 + 2π × 3²
= 15π + 18π = 33π cm².

The joined circular faces are internal. Keep π for exact answers; for decimals, use the calculator π key and round once at the end. Doubling a sphere’s radius multiplies its area by 4 and volume by 8.

6. Frustums

A frustum is formed when a cone is cut parallel to its base. The removed cone and original cone are similar. Find missing dimensions using matching length ratios, then subtract the smaller cone’s volume.

Worked example

An original cone has radius 6 cm and height 18 cm. The removed cone has height 6 cm, so its radius is (6/18) × 6 = 2 cm. The frustum height is 18 − 6 = 12 cm.

Original volume = ⅓π × 6² × 18 = 216π cm³.
Removed volume = ⅓π × 2² × 6 = 8π cm³.
Frustum volume = 208π cm³ ≈ 653.45 cm³.

For large radius R, small radius r and frustum height h, the direct formula is V = ⅓πh(R² + Rr + r²). Here, ⅓π × 12(36 + 12 + 4) = 208π cm³.

For a right circular frustum:

• Slant height s = √[h² + (R − r)²].
• Curved area = π(R + r)s.
• Closed total area = π(R + r)s + πR² + πr².

In this example, s = √(12² + 4²) = 4√10 cm.
Closed total area = (32√10 + 40)π cm² ≈ 443.57 cm².

For an open container, omit the missing top disc. Use the frustum height, not the original cone height.

7. Capacity, flow rates and density

At constant flow, volume = flow rate × time, so time = required volume ÷ flow rate. Convert units first. For a partly full tank, calculate only the remaining volume.

Example: a cylinder of radius 20 cm and height 50 cm is filled at 5 litres per minute.
Capacity = π × 20² × 50 = 20,000π cm³ = 20π litres.
Time from empty = 20π ÷ 5 = 4π minutes ≈ 12 min 34 s.
To fill it three-quarters full takes 3π minutes ≈ 9 min 25 s.

Volume–time graphs

The gradient is the rate of change of volume. With litres vertically and seconds horizontally, its units are litres per second. Use a tangent to estimate an instantaneous rate on a curved graph. A falling line represents a net loss of water.

Density and recasting

Density = mass ÷ volume; mass = density × volume.
A metal cuboid 8 cm × 5 cm × 3 cm with density 2.7 g/cm³ has volume 120 cm³ and mass 2.7 × 120 = 324 g.

When the same material is melted and recast without loss and its density is unchanged, its volume stays the same. A sphere of volume 288π cm³ is recast into cones of radius 3 cm and height 8 cm. Each cone has volume 24π cm³, so the number of cones is 288π ÷ 24π = 12.

8. Similarity, plans and elevations

For similar solids with length scale factor k, the area scale factor is k² and the volume scale factor is k³. Mass follows the volume scale factor only when density is unchanged.

• A solid of volume 40 cm³ enlarged with length scale factor 3/2 has volume 40 × (3/2)³ = 135 cm³.
• Area ratio 9 : 25 → length ratio 3 : 5 → volume ratio 27 : 125.
• Volume ratio 8 : 27 → length ratio 2 : 3 → area ratio 4 : 9.

A plan is the view from above. A front elevation is the view from directly in front, and a side elevation is the view from the specified side. Follow any direction arrows and draw flat views without perspective.

For a cuboid 6 cm wide, 4 cm deep and 3 cm high: the plan is 6 cm × 4 cm; the front elevation is 6 cm × 3 cm; the side elevation is 4 cm × 3 cm.

For cube stacks, the plan records the footprint. An elevation records the greatest height in each viewing column; it does not reveal hidden internal cubes. Different solids can have the same three views.

Final check: use radius rather than diameter, distinguish slant and perpendicular heights, exclude hidden joining faces, and write square units for area and cubic units for volume.