Module 2 · L8 of 22~55 minMS-M1 · MEDIUM⚡ +95 XP available
Surface Area of Spheres and Composite Solids
A sphere needs only its radius. A hemisphere has two surfaces, not one. Composite solids hide faces at the join, so list what is exposed before calculating anything.
Today's hook, A beach ball is a sphere. A grain silo is a cylinder with a domed hemisphere on top. Before any formula, when the dome is joined to the silo, do you think you paint more surface, less, or the same as if the two sat apart?
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Recall, your gut answer first
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
A beach ball is a sphere. A grain silo is a cylinder with a domed hemisphere on top. Both need painting.
Before learning any formula when the dome is joined to the silo, do you think you paint more surface, less surface, or the same as if the two sat apart? Why?
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Surface area formulas, spheres, hemispheres and composites
+5 XP to read Two curved surfaces and one rule for joining them.
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Surface area formulas, spheres, hemispheres and composites
+5 XP to read
Two curved surfaces and one rule for joining them. The number one rule: a face at a join is hidden, so it is never part of the surface area.
Sphere: one continuous surface. $\text{SA} = 4\pi r^2$. Hemisphere: curved dome plus a flat circle. $\text{SA} = 3\pi r^2$. Composite: add the exposed faces only, and leave out both faces at the join.
$\text{SA} = 4\pi r^2$. Exactly four circles of the same radius laid flat. Only $r$ is needed.
Composite, list before you add
Write every face, mark it exposed or hidden, then add the exposed ones only. The listing step is what earns the method mark.
Sphere and hemisphere
Sphere: $4\pi r^2$. Hemisphere: $3\pi r^2$ (curved dome $2\pi r^2$ + flat circle $\pi r^2$). Never forget the flat base!
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What you'll master
Know
Key facts
SA formulas for spheres ($4\pi r^2$) and hemispheres ($3\pi r^2$)
That a hemisphere has two surfaces, the curved dome and the flat circle
How to handle composite solids with hidden faces
Understand
Concepts
Why a hemisphere SA $= 3\pi r^2$ (not $2\pi r^2$), the flat face must be included
Why hidden faces at composite joins must be excluded from SA
Why joining two solids always reduces the total surface, by twice the joined face
Can do
Skills
Calculate SA of any sphere or hemisphere from its radius
List exposed faces of composite solids and calculate total SA correctly
Identify which faces are hidden at a join and leave them out
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Key vocabulary
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Key vocabulary
SphereA solid where every point of the surface is the same distance $r$ from the centre. Like this: a beach ball of radius 6 cm has $\text{SA} = 4\pi(6)^2 = 144\pi \approx 452.39\text{ cm}^2$.
HemisphereExactly half a sphere, so it has a curved dome AND a flat circular face. Like this: radius 6 cm gives $2\pi(36) + \pi(36) = 108\pi \approx 339.29\text{ cm}^2$.
Curved surfaceThe rounded part of a solid, counted separately from any flat face. Like this: the dome of that hemisphere alone is $2\pi(36) = 72\pi \approx 226.19\text{ cm}^2$.
Composite solidTwo or more solids joined into one object. Like this: a hemisphere sitting on a cylinder of the same radius, such as a grain silo.
Hidden faceA face at the join between two solids, sealed inside and not part of the outer surface. Like this: joining a hemisphere of radius 4 cm to a cylinder hides two circles of $16\pi\text{ cm}^2$ each.
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The critical distinction: exposed faces vs hidden faces
This distinction is the source of more marks lost in this topic than anything else.
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The critical distinction: exposed faces vs hidden faces
core concept
This distinction is the source of more marks lost in this topic than anything else. Before adding anything up, decide which faces a tin of paint would actually reach.
Exposed face
On the outside of the finished object. Paint reaches it, so it counts towards the surface area.
Hidden face
Sealed inside the join between two solids. Paint never reaches it, so it is left out entirely.
Surface Area of a Sphere and Hemisphere
$$\text{SA}_\text{sphere} = 4\pi r^2$$
One of the most elegant results in geometry: the total surface area of a sphere equals exactly four circles of the same radius laid flat. A sphere needs nothing but $r$.
A hemisphere has two surfaces: curved outer dome $\frac{1}{2} \times 4\pi r^2 = 2\pi r^2$ plus flat circular base $\pi r^2$.
Most common hemisphere error: Writing $\text{SA} = \frac{1}{2}(4\pi r^2) = 2\pi r^2$, halving the sphere's SA and forgetting the flat circular base. For $r = 6$ that is $72\pi \approx 226.19\text{ cm}^2$ instead of $108\pi \approx 339.29\text{ cm}^2$, a shortfall of the whole $36\pi$ base. Think physically: if you dipped a hemisphere in paint, both the dome and the flat bottom would be coated.
Composite Solids, Hidden Faces at Joins
When two solids are joined, the faces at the join become internal. They are hidden from the outside and are not part of the surface area.
At any join between two solids, the joined faces are not included in the total SA. List which faces are exposed on the outer surface, then calculate only those.
Joining always hides TWO faces, one on each solid, never just one.
Process for composite solids: List all faces of both solids. Mark each as exposed or hidden. Calculate exposed faces only. For a hemisphere on a cylinder: hemisphere dome (yes), cylinder curved side (yes), cylinder base (yes), cylinder top (hidden at join), hemisphere flat face (hidden at join). This listing step takes 30 seconds and prevents multi-mark errors.
A hemisphere has TWO surfaces, the curved dome $2\pi r^2$ and the flat circle $\pi r^2$, giving $3\pi r^2$ in total. At a join between two solids, two faces are hidden, one from each solid, and neither counts towards the surface area.
Pause, copy the hemisphere formula with both of its surfaces ($2\pi r^2$ curved plus $\pi r^2$ flat, so $3\pi r^2$) and the rule that a join hides two faces, not one, into your book.
Did you get this? True or false: the surface area of a hemisphere is exactly half the surface area of the whole sphere.
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Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
PROBLEM 1 · SPHERE SA
A beach ball is a sphere of radius 6 cm. Find its surface area, correct to 2 decimal places.
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Write the formula $\text{SA} = 4\pi r^2$
A sphere needs only $r$. There is no second height to find and nothing to hide at a join.
$3\pi r^2$ in total. Compare with $72\pi \approx 226.19\text{ cm}^2$, the answer you get by halving and stopping.
PROBLEM 3 · COMPOSITE SA (THE SILO)
A grain silo is a cylinder of radius 4 m and height 6 m, with a hemispherical dome of radius 4 m sitting on top. Find the total exposed surface area, correct to 2 decimal places.
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List every face, then mark it Exposed: dome, cylinder curved side, cylinder base Hidden: cylinder top, hemisphere flat face
Do the listing before any arithmetic. The join seals TWO circles, one from each solid, and neither is painted.
Counting the two hidden circles as well gives $128\pi \approx 402.12\text{ m}^2$, the most common wrong answer in this topic.
Quick check: A hemisphere has radius 5 cm. Its total SA is:
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Common errors · the 3 traps that cost marks
Common errors · the 3 traps that cost marks
Common errors · the 3 traps that cost marks
Trap 01
Halving the sphere and forgetting the hemisphere's flat face
Writing $\text{SA} = \tfrac{1}{2}(4\pi r^2) = 2\pi r^2$ and stopping. The cut creates a NEW flat circle that did not exist on the sphere. Every time you see a hemisphere, write both surfaces: $2\pi r^2$ curved plus $\pi r^2$ flat.
Trap 02
Hemisphere SA = half of sphere SA
The flat circular base ($\pi r^2$) appears when the sphere is cut, it must be included. SA of hemisphere $= 2\pi r^2 + \pi r^2 = 3\pi r^2$, not $2\pi r^2$. Option A in the MC above ($50\pi$) traps students who forget the flat base.
Trap 03
Including hidden faces in composite SA
Adding all faces of all component solids without removing the joined faces. Before calculating, list faces as exposed or hidden. Anything at the join is hidden. Writing "cylinder top and hemisphere flat face are hidden" at the top of your working signals clear method to the examiner.
Fill the gap: A hemisphere has radius 5 cm. Curved dome $= 2\pi(5)^2 =$ $\pi$ cm². Flat face $= \pi(5)^2 =$ $\pi$ cm². Total SA $=$ $\pi$ cm².
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Quick-fire practice · 11 calculations
Quick-fire practice · 11 calculations
Quick-fire practice · 11 calculations
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Sphere: radius 4 cm. Find total SA to 2 decimal places.
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Sphere: diameter 20 m. Find total SA to 2 decimal places.
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Hemisphere: radius 6 cm. Find the CURVED surface only, to 2 decimal places.
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Hemisphere: radius 3 cm. Find TOTAL SA to 2 decimal places.
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A sphere has total SA of $100\pi$ cm². Find its radius.
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Hemisphere: radius 10 m. Find TOTAL SA to 2 decimal places.
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Find SA of a sphere with $r = 9$ cm to 2 decimal places.
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Find SA of a sphere with diameter 14 m to 2 decimal places.
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Find total SA of a hemisphere with $r = 7$ cm to 2 decimal places.
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A capsule is a cylinder ($r = 2$ cm, height 9 cm) with a hemispherical dome on EACH end. Find total SA to 2 decimal places.
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A hemisphere ($r = 4$ cm) sits on top of a cylinder ($r = 4$ cm, height 10 cm). Find total SA to 2 decimal places.
Odd one out: Three of these statements about spheres, hemispheres and composite solids are correct. Which one is wrong?
Match each solid to its surface area formula:
Sphere
$3\pi r^2$
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Revisit your thinking
Earlier you predicted which solid would be hardest to calculate the area of. Look back at your initial response.
Most students expect the joined shape to need MORE paint, because it looks bigger. It needs less. Joining seals two circles, one on the cylinder and one on the hemisphere, and neither is ever painted. The sphere, meanwhile, is pure elegance: $4\pi r^2$, with nothing else to find first.
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Show what you have learned
Multiple choice, then short answer under exam conditions.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
ApplyBand 43 marks
SA 4. A hemisphere has radius 10 cm.
(a) Find the curved surface area, correct to 2 decimal places. (1 mark)
(b) Find the TOTAL surface area of the solid hemisphere, correct to 2 decimal places. (2 marks)
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ApplyBand 43 marks
SA 5. A sphere has a total surface area of $196\pi$ cm².
(a) Find its radius. (1 mark)
(b) The sphere is cut exactly in half. Find the total SA of one hemisphere, correct to 2 decimal places. (2 marks)
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AnalyseBand 54 marks
SA 6. A decorative ornament has a hemisphere (radius 3 cm) sitting on top of a cylinder (radius 3 cm, height 7 cm).
(a) Find the curved SA of the hemisphere. (1 mark)
(b) Find the exposed SA of the cylinder (top hidden, base included). (2 marks)
(c) Find the total SA correct to 2 decimal places. (1 mark)
SA 6(c): $18\pi + 51\pi = 69\pi = \mathbf{216.77\text{ cm}^2}$
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Consolidate and move on
Sit the module quiz, then close the lesson off.
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Boss battle · Asteroid Blaster
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