Only exposed faces count
Surface area is the total area you could paint. When two solids are joined, the parts that press together are sealed inside and can never be painted, so they do not count. This is the one idea that separates compound surface area from compound volume: for volume you simply add, but for surface area you must leave out the join.
- Exposed face
- A surface on the outside of the finished solid. Any surface where two pieces touch is hidden and is left out of the surface area.
A method that works
A reliable routine keeps these questions under control:
1. List every face of every piece as if the pieces were separate. 2. Find where the pieces meet and identify the two faces that touch. 3. Remove both touching faces from your list. 4. Add the areas of everything that remains.
Every join hides two faces
When one piece sits on another, the join removes area from both pieces, not one. Miss this and your total comes out too big. If a cone caps a cylinder, the cylinder loses its top circle and the cone loses its base circle, so neither appears in the answer.
Spotting the hidden faces
The touching faces are always equal in shape where the fit is exact: a cone base on a cylinder top both being circles of the same radius, or a smaller cube resting flat on a larger one. For a solid capped by a hemisphere, remember the curved part of the hemisphere is \(2\pi r^2\), which is half a sphere's \(4\pi r^2\), and its flat circular face is the one that gets hidden against the solid below.
A worked example
A solid is a cylinder of radius 3 cm and height 10 cm with a cone of radius 3 cm and slant height 5 cm fixed on top. Find the total surface area, to 1 decimal place.
Where this is assessed
Knowing which face formulas to use is Criterion A (Knowing and understanding). Listing the exposed faces clearly and explaining which are hidden is Criterion C (Communicating). Costing the paint or wrapping for a real object is Criterion D (Applying maths in context). Keep every area in square units and double-check you removed both faces at each join.
Check yourself
1. Two cubes of edge 4 cm are stacked to form a 4 cm by 4 cm by 8 cm tower. Find its total surface area. +
Each cube alone has \(6 \times 4^2 = 96\) cm\(^2\), so two give \(192\) cm\(^2\). The join hides one face on each cube, that is \(2 \times 16 = 32\) cm\(^2\). Total \(= 192 - 32 = \mathbf{160}\) cm\(^2\). (Check: the tower is a cuboid, \(2(4\times4 + 4\times8 + 4\times8) = 2(16+32+32) = 160\) cm\(^2\).)
2. A cube of edge 6 cm has a small cube of edge 2 cm sitting flat and fully on the centre of its top face. Find the total surface area. +
Large cube: \(6 \times 6^2 = 216\) cm\(^2\), but \(2 \times 2 = 4\) cm\(^2\) of its top is covered, leaving \(216 - 4 = 212\) cm\(^2\). Small cube: its bottom \(4\) cm\(^2\) is hidden, so its exposed area is \(6 \times 2^2 - 4 = 24 - 4 = 20\) cm\(^2\). Total \(= 212 + 20 = \mathbf{232}\) cm\(^2\).
3. A cylinder of radius 2 cm and height 10 cm has a hemisphere of radius 2 cm fixed on top. Find the total surface area (to 1 d.p.). +
Bottom circle: \(\pi r^2 = 4\pi\). Curved wall: \(2\pi r h = 2\pi \times 2 \times 10 = 40\pi\). The cylinder's top circle is hidden by the hemisphere's flat face. Curved hemisphere: \(2\pi r^2 = 2\pi \times 4 = 8\pi\). Total \(= 4\pi + 40\pi + 8\pi = 52\pi = 163.36\ldots \approx \mathbf{163.4}\) cm\(^2\).
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