Mathematics Standard • Year 11 • Module 2 • Lesson 8
Surface Area of Spheres and Composite Solids
Apply sphere, hemisphere and composite SA to realistic Australian contexts: weather balloons, pavilion domes, beach balls and joined tanks.
Problem 1, Weather balloon (sphere)
A spherical weather balloon has a diameter of 3.6 m. The fabric is sold by the m².
Set up: What are we solving for?
(i) State the radius. 1 mark
(ii) Find the surface area. Give the exact value and the value to 2 decimal places. 2 marks
(iii) The manufacturer orders 10% extra fabric for wastage, at $18 per m². Find the total cost to the nearest cent. 2 marks
Problem 2, Pavilion dome (curved surface only)
A garden pavilion has a hemispherical dome roof of radius 5 m. Only the curved outside is painted, because the flat underside is the ceiling inside the pavilion.
Set up: What are we solving for?
(i) Explain in one sentence why the flat circular face is NOT included, even though a solid hemisphere has one. 1 mark
(ii) Find the area to be painted, exact and to 2 decimal places. 2 marks
(iii) One litre of paint covers 12 m². How many whole litres must be bought? 2 marks
Problem 3, Beach ball material
A beach ball is spherical with diameter 40 cm. A manufacturer cuts the surface from a PVC sheet, allowing 8% extra material for seams and waste.
Set up: What are we solving for?
(i) Find the radius. Then find the SA of one ball in cm², to 2 d.p. 2 marks
(ii) Add 8% for seams and waste. Find the PVC required per ball, to 2 d.p. 1 mark
(iii) A production run makes 500 beach balls. Find the total PVC required in m², to 2 d.p. 2 marks
Problem 4, Storage tank: hemisphere on cylinder
A grain storage tank consists of a cylinder of radius 1.5 m and height 4 m, with a hemisphere of the same radius sealing the top. The tank sits on the ground (the cylinder base is also painted on the underside as it is visible during cleaning).
Set up: What are we solving for?
(i) List which faces are exposed and which are hidden at the join between the cylinder and hemisphere. 1 mark
(ii) Calculate the total external SA (cylinder curved + cylinder base + hemisphere dome), in exact form (in terms of π) and to 2 d.p. 3 marks
(iii) Anti-rust paint covers 8 m² per litre. Find the number of full 1-litre tins required. 1 mark
Problem 5, Capsule-shaped tank (composite)
A capsule-shaped water tank is a cylinder of radius 2 m and length 7 m, with a hemispherical cap on EACH end. The whole outside is sealed.
Set up: What are we solving for?
(i) List the exposed surfaces, and state which faces are hidden at the two joins. 2 marks
(ii) Find the total surface area to be sealed, exact and to 2 decimal places. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1, Weather balloon (sphere)
Set up. A sphere has one surface, so only the radius is needed.
(i) r = 3.6 ÷ 2 = 1.8 m.
(ii) SA = 4πr² = 4π(1.8)² = 4π(3.24) = 12.96π m² ≈ 40.72 m².
(iii) With 10% extra: 40.72 × 1.1 ≈ 44.79 m². Cost = 44.79 × $18 = $806.22. (Common error: halving after squaring, or using the diameter as the radius, which multiplies the area by four.)
Problem 2, Pavilion dome (curved surface only)
Set up. Curved dome only, so 2πr² and not the full 3πr².
(i) The flat circular face is the ceiling inside the pavilion. It is not on the painted outside, so it is excluded. A solid hemisphere sitting on a bench WOULD include it.
(ii) Curved SA = 2π(5)² = 50π m² ≈ 157.08 m².
(iii) 157.08 ÷ 12 = 13.09, and part-tins cannot be bought, so 14 litres. (Rounding down to 13 leaves about 1.08 m² unpainted.)
Problem 3, Beach ball
Set up. Sphere SA = 4πr², then add 8%, then scale to 500 balls, then convert cm² → m².
(i) r = 40 ÷ 2 = 20 cm. SA = 4π(400) = 1600π ≈ 5026.55 cm².
(ii) +8%: 5026.55 × 1.08 ≈ 5428.67 cm² per ball.
(iii) 500 balls: 500 × 5428.67 = 2 714 335 cm². Convert: 2 714 335 / 10 000 ≈ 271.43 m².
Problem 4, Storage tank: hemisphere on cylinder
Set up. Hidden = cylinder top + hemisphere flat face (at the join). Exposed = cylinder curved + cylinder base + hemisphere dome.
(i) Hidden: cylinder top, hemisphere flat face. Exposed: cylinder curved (2πrh), cylinder base (πr²), hemisphere curved dome (2πr²).
(ii) r = 1.5, h = 4. Cylinder curved = 2π(1.5)(4) = 12π. Cylinder base = π(2.25) = 2.25π. Hemisphere dome = 2π(2.25) = 4.5π.
Total = 12π + 2.25π + 4.5π = 18.75π m² ≈ 58.90 m².
(iii) Tins = 58.90 ÷ 8 = 7.36... ⇒ 8 tins (always round up).
Problem 5, Capsule-shaped tank (composite)
Set up. Two joins, and each join seals one face on each solid.
(i) Exposed: the cylinder's curved side and BOTH hemispherical domes. Hidden: four circles in total, the cylinder's two ends and the flat face of each cap. No flat circle is exposed anywhere.
(ii) Cylinder curved = 2π(2)(7) = 28π. Two domes = 2 × 2π(2)² = 16π. Total = 44π m² ≈ 138.23 m². (Adding the two end circles as well gives 52π ≈ 163.36 m², counting surfaces that are sealed inside.)