Joining two solids: which faces count

Press two solids together

Take a ball of clay and press it on the top of a wooden block. Now paint the new object. Is the paint area the area of the block plus the area of the ball? No. The place where they are pressed together is not painted. It is hidden. Both solids lose that piece.

The rule. The surface area of a combined solid is the area of the faces that stay on the outside. Never add the total surface areas of the parts. That adds each hidden joining face twice.

A good habit is to walk round the solid with a paint brush. Add every face that the brush touches, and nothing else.

A toy: a cone on a hemisphere

A cone with radius $r$ stands on a hemisphere with the same radius. The flat base of the cone and the flat face of the hemisphere are the same circle. This circle is hidden.

What stays outside? The curved surface of the cone, $\pi r l$, and the curved surface of the hemisphere, $2\pi r^2$. So

Here $l = \sqrt{r^2 + h^2}$, where $h$ is the height of the cone only, not of the whole toy.

Example 1. A toy is a cone on a hemisphere. Both have radius $5\ \text{cm}$. The whole toy is $17\ \text{cm}$ high. Find its surface area in terms of $\pi$.

1. The cone's height. The hemisphere is $5\ \text{cm}$ high, so the cone is $17\ \text{cm} - 5\ \text{cm} = 12\ \text{cm}$ high.
2. The slant height. $l = \sqrt{5^2 + 12^2}\ \text{cm} = 13\ \text{cm}$.
3. The cone. $\pi \times 5\ \text{cm} \times 13\ \text{cm} = 65\pi\ \text{cm}^2$.
4. The hemisphere. $2\pi \times (5\ \text{cm})^2 = 50\pi\ \text{cm}^2$.
5. The total. $65\pi + 50\pi = 115\pi\ \text{cm}^2$, which is about $361.3\ \text{cm}^2$.

A capsule: a cylinder with two hemispheres

A medicine capsule is a cylinder with a hemisphere stuck on each end. The two flat faces are hidden. What stays outside is the curved surface of the cylinder, $2\pi r h$, and two curved hemispheres.

Two hemispheres make one whole sphere. So

Take care with the length. The total length of the capsule is $h + 2r$, because each hemisphere adds a length $r$. A question often gives the total length. Then the cylinder part is $h = \text{length} - 2r$.

Example 2. A capsule is $20\ \text{mm}$ long in all and has a diameter of $6\ \text{mm}$. Find its surface area in terms of $\pi$.

1. The radius. $r = 3\ \text{mm}$.
2. The cylinder part. $h = 20\ \text{mm} - 2 \times 3\ \text{mm} = 14\ \text{mm}$.
3. The cylinder. $2\pi \times 3\ \text{mm} \times 14\ \text{mm} = 84\pi\ \text{mm}^2$.
4. The ends. $4\pi \times (3\ \text{mm})^2 = 36\pi\ \text{mm}^2$.
5. The total. $120\pi\ \text{mm}^2$, about $377.0\ \text{mm}^2$.

Other joins

A cylinder with a hemisphere on top. A cup of ice cream or a silo has a flat bottom. The bottom circle is outside, and so is the curved cylinder. The hemisphere adds its curved surface, and its flat face is hidden.

Joins of different sizes. What if the cone is narrower than the cylinder it stands on? Then a ring of the cylinder's top face is still outside. Its area is $\pi R^2 - \pi r^2$, where $R$ is the radius of the cylinder and $r$ is the radius of the cone. The hidden part is only a circle of radius $r$.

Example 3. A cone with radius $3\ \text{cm}$ and slant height $5\ \text{cm}$ stands on the top of a cylinder with radius $4\ \text{cm}$ and height $6\ \text{cm}$. Find the total surface area in terms of $\pi$.

1. The cylinder's bottom. $\pi \times (4\ \text{cm})^2 = 16\pi\ \text{cm}^2$.
2. The cylinder's side. $2\pi \times 4\ \text{cm} \times 6\ \text{cm} = 48\pi\ \text{cm}^2$.
3. The ring on top. $\pi \times (4\ \text{cm})^2 - \pi \times (3\ \text{cm})^2 = 7\pi\ \text{cm}^2$.
4. The cone. $\pi \times 3\ \text{cm} \times 5\ \text{cm} = 15\pi\ \text{cm}^2$.
5. The total. $16\pi + 48\pi + 7\pi + 15\pi = 86\pi\ \text{cm}^2$.

Practise with new numbers

Try it in Python

This program works out the surface area of a toy for five different cone heights. It also shows the wrong answer you would get by adding the two total surface areas.

Common mistakes

• Adding the total surface areas of the two parts. The joining circle is counted twice, and it should not be counted at all.
• Using the height of the whole toy as the height of the cone. The hemisphere takes the radius $r$ out of the height.
• Using the whole length of a capsule as the cylinder part. The two ends take away $2r$.
• Forgetting the ring when the upper solid is narrower than the lower one.

Remember these

• The surface area of a combined solid is the area of the faces that stay outside.
• A cone on a hemisphere: $\pi r l + 2\pi r^2$. A capsule: $2\pi r h + 4\pi r^2$.
• The total length of a capsule is $h + 2r$.
• A hemisphere on top of a cylinder hides the flat face of the hemisphere but keeps the cylinder's bottom.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two cubes of edge $3\ \text{cm}$ are glued face to face to make a cuboid. What is its total surface area?

2. A cone with slant height $10\ \text{cm}$ and radius $6\ \text{cm}$ stands on a hemisphere of radius $6\ \text{cm}$. Find the surface area in terms of $\pi$.

3. A capsule has radius $2\ \text{mm}$ and a cylinder part $5\ \text{mm}$ long. What is its total length?

4. A cylinder of radius $7\ \text{cm}$ and height $10\ \text{cm}$ has a hemisphere of radius $7\ \text{cm}$ on its top. Which faces are outside?

5. Why is the surface area of a toy smaller than the sum of the total surface areas of its two parts?